Get month number from month name - java

I have this.
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String monthName = br.readLine();
How to get month number which contain in monthName variable?
Thanks!

Use Java's Calendar class. It can parse any given string into a valid calendar instance.
Here is an example (assuming that the month is in english).
Date date = new SimpleDateFormat("MMMM", Locale.ENGLISH).parse(monthName);
Calendar cal = Calendar.getInstance();
cal.setTime(date);
println(cal.get(Calendar.MONTH));
You can specify the language in SimpleDateFormat:
String monthName = "März"; // German for march
Date date = new SimpleDateFormat("MMMM", Locale.GERMAN).parse(monthName);
Calendar cal = Calendar.getInstance();
cal.setTime(date);
println(cal.get(Calendar.MONTH));
By default, Java uses the user's local to parse the string.
Keep in mind that a computer starts counting at 0. So, January will be 0. If you want a human readable date, you should format the calendar instance:
SimpleDateFormat inputFormat = new SimpleDateFormat("MMMM");
Calendar cal = Calendar.getInstance();
cal.setTime(inputFormat.parse(monthName));
SimpleDateFormat outputFormat = new SimpleDateFormat("MM"); // 01-12
println(outputFormat.format(cal.getTime()));

Solution in Java 8 for English month names.
private int getMonthNumber(String monthName) {
return Month.valueOf(monthName.toUpperCase()).getValue();
}

Another approach using java.text.DateFormatSymbols is this:
public static int monthAsNumber(
String month,
Locale locale,
boolean abbreviated,
boolean caseInsensitive
) {
DateFormatSymbols dfs = new DateFormatSymbols(locale);
String[] months = (abbreviated ? dfs.getShortMonths() : dfs.getMonths());
if (caseInsensitive) {
for (int i = 0; i < 12; i++) {
if (months[i].equalsIgnoreCase(month)) {
return i; // month index is zero-based as usual in old JDK pre 8!
}
}
} else {
for (int i = 0; i < 12; i++) {
if (months[i].equals(month)) {
return i; // month index is zero-based as usual in old JDK pre 8!
}
}
}
return -1; // no match
}
The proposed signature of seach method illustrates the many possible variations. Example:
System.out.println(monthAsNumber("MÄRZ", Locale.GERMANY, false, true));
// output: 2 (zero-based!)
If you want a month number starting with 1 then just add 1 to the result (more intuitive and also my recommendation).
Starting with Java 8 you have a new variation, too, namely stand-alone months. While in English these month names are identical in other languages they are not always identical (for example in Czech language "leden" (January) instead of "ledna"). To achieve these stand-alone forms you can use Month.getDisplayName(...) (not tested):
public static int monthAsNumber(
String month,
Locale locale,
boolean abbreviated,
boolean caseInsensitive,
boolean standAlone
) {
TextStyle style;
Month[] months = Month.values[];
if (abbreviated) {
style = standAlone ? TextStyle.SHORT_STANDALONE : TextStyle.SHORT;
} else {
style = standAlone ? TextStyle.FULL_STANDALONE : TextStyle.FULL;
}
if (caseInsensitive) {
for (int i = 0; i < 12; i++) {
if (months[i].getDisplayName(style, locale).equalsIgnoreCase(month)) {
return i; // month index is zero-based as usual in old JDK pre 8!
}
}
} else {
for (int i = 0; i < 12; i++) {
if (months[i].getDisplayName(style, locale).equals(month)) {
return i; // month index is zero-based as usual in old JDK pre 8!
}
}
}
return -1; // no match
}

If you have month name and you want integer number corresponding to that month
try this
try{
Date date = new SimpleDateFormat("MMM", Locale.ENGLISH).parse(monthName);//put your month name in english here
Calendar cal = Calendar.getInstance();
cal.setTime(date);
monthNumber=cal.get(Calendar.MONTH);
System.out.println(monthNumber);
}
catch(Exception e)
{
e.printStackTrace();
}

Using Java 8 API
final DateTimeFormatter dtFormatter = DateTimeFormatter.ofPattern("MMM").withLocale(Locale.ENGLISH);
final TemporalAccessor temporalAccessor = dtFormatter.parse(month);
final int monthNumber = temporalAccessor.get(ChronoField.MONTH_OF_YEAR);

java.time
In March 2014, modern date-time API API supplanted the error-prone java.util date-time API and their formatting API, SimpleDateFormat. Since then it has been highly recommended to stop using the legacy API.
Solution using java.time, the modern Date-Time API:
Use a DateTimeFormatter with the pattern MMMM to parse the given date. If the name of the month can be in different cases, you should build and use a case-insensitive DateTimeFormatter.
Demo:
import java.time.Month;
import java.time.format.DateTimeFormatter;
import java.time.format.DateTimeFormatterBuilder;
import java.util.Locale;
class Main {
public static void main(String[] args) {
// A sample month
String monthName = "March";
DateTimeFormatter parser = DateTimeFormatter.ofPattern("MMMM", Locale.ENGLISH);
Month month = Month.from(parser.parse(monthName));
int monthNumber = month.getValue();
System.out.println(monthNumber);
// A case-insensitive formatter
DateTimeFormatter parserIgnoreCase = new DateTimeFormatterBuilder()
.parseCaseInsensitive()
.appendPattern("MMMM")
.toFormatter(Locale.ENGLISH);
System.out.println(Month.from(parserIgnoreCase.parse("march")).getValue());
}
}
Output:
3
3
Note: Never use SimpleDateFormat or DateTimeFormatter without a Locale.
Learn more about the modern Date-Time API from Trail: Date Time.

Related

Convert two digit year to four digits and also support one or two digit month

I want to convert two digits year to four digits and also it is possible to come 4 digits
final Integer year = 2020;
final Integer month = 12;
final DateFormat originalFormat = new SimpleDateFormat("MMyy", Locale.US);
final Date monthAndYear = originalFormat.parse(month + String.valueOf(year));
final DateFormat formattedDate = new SimpleDateFormat("yyyy-MM", Locale.US);
System.out.println(formattedDate.format(monthAndYear));
This code fails if the input is 2-2020, which not parsing one digit month.
I want to pass the code by below conditions
year | month || expeected
2020 | 12 || "2020-12"
30 | 2 || "2030-02"
41 | 05 || "2041-05"
You can use YearMonth for this like so:
final DateTimeFormatter YEAR_FORMAT = DateTimeFormatter.ofPattern("[yyyy][yy]");
YearMonth yearMonth = YearMonth.of(
Year.parse(year, YEAR_FORMAT).getValue(),
month);
Note: The year should be a String
Outputs:
2020-12
2030-02
2041-05
I don’t think I’d want to use parsing for this, though it is an option. Your year and month come as numbers, so I find it more natural to handle them as such. Like the others I recommend using YearMonth from java.time, the modern Java date and time API.
final Integer year = 2020;
final Integer month = 12;
YearMonth monthAndYear;
if (0 <= year && year < 100) { // two digits
YearMonth currentMonthAndYear = YearMonth.now(ZoneId.systemDefault());
// Truncate current year to whole centuries
int fullYear = currentMonthAndYear.getYear() / 100 * 100 + year;
monthAndYear = YearMonth.of(fullYear, month);
if (monthAndYear.isBefore(currentMonthAndYear)) { // in the past, wrong
monthAndYear = monthAndYear.plusYears(100);
}
} else {
monthAndYear = YearMonth.of(year, month);
}
System.out.println(monthAndYear);
Output from this example snippet:
2020-12
It’s wordier than the code in the other answers. The upside is that it gives us precise control over which century to use for two digit years. Year 20 and moth 4 give us 2120-04, while year 20 and month 9 give us 2020-09. So both are in the future.
You want the result as a string in the format from your question? We’re lucky, YearMonth.toString() gives us that.
System.out.println(monthAndYear.toString());
The format is ISO 8601.
Link: Wikipedia article: ISO 8601
Use single M for a month in the format. A single M holds good for any allowed number of digits for a month in the modern date-time API. You should stop using the broken and outdated java.util date-time API and switch to the modern date-time API.
import java.time.YearMonth;
import java.time.format.DateTimeFormatter;
public class Main {
public static void main(String[] args) {
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("M-[uuuu][uu]");
// Test strings (month-year)
String[] ymStrArr = { "2-2020", "2-20", "02-2020", "02-20" };
for (String ymStr : ymStrArr) {
YearMonth ym = YearMonth.parse(ymStr, formatter);
System.out.println(ym);
}
}
}
Output:
2020-02
2020-02
2020-02
2020-02
String pattern = "";
if(year1.length() == 4){
pattern = "yyyy";
}elif(year1.length() == 2){
pattern = "yy[yy]";
}
YearMonth yearMonth = YearMonth.of(Year.parse(year1, DateTimeFormatter.ofPattern(pattern)).getValue(),month);
System.out.println(yearMonth);

How can i know if my date is valid in java using simple Date Format? [duplicate]

I find it curious that the most obvious way to create Date objects in Java has been deprecated and appears to have been "substituted" with a not so obvious to use lenient calendar.
How do you check that a date, given as a combination of day, month, and year, is a valid date?
For instance, 2008-02-31 (as in yyyy-mm-dd) would be an invalid date.
Key is df.setLenient(false);. This is more than enough for simple cases. If you are looking for a more robust (I doubt that) and/or alternate libraries like joda-time, then look at the answer by user "tardate"
final static String DATE_FORMAT = "dd-MM-yyyy";
public static boolean isDateValid(String date)
{
try {
DateFormat df = new SimpleDateFormat(DATE_FORMAT);
df.setLenient(false);
df.parse(date);
return true;
} catch (ParseException e) {
return false;
}
}
As shown by #Maglob, the basic approach is to test the conversion from string to date using SimpleDateFormat.parse. That will catch invalid day/month combinations like 2008-02-31.
However, in practice that is rarely enough since SimpleDateFormat.parse is exceedingly liberal. There are two behaviours you might be concerned with:
Invalid characters in the date string
Surprisingly, 2008-02-2x will "pass" as a valid date with locale format = "yyyy-MM-dd" for example. Even when isLenient==false.
Years: 2, 3 or 4 digits?
You may also want to enforce 4-digit years rather than allowing the default SimpleDateFormat behaviour (which will interpret "12-02-31" differently depending on whether your format was "yyyy-MM-dd" or "yy-MM-dd")
A Strict Solution with the Standard Library
So a complete string to date test could look like this: a combination of regex match, and then a forced date conversion. The trick with the regex is to make it locale-friendly.
Date parseDate(String maybeDate, String format, boolean lenient) {
Date date = null;
// test date string matches format structure using regex
// - weed out illegal characters and enforce 4-digit year
// - create the regex based on the local format string
String reFormat = Pattern.compile("d+|M+").matcher(Matcher.quoteReplacement(format)).replaceAll("\\\\d{1,2}");
reFormat = Pattern.compile("y+").matcher(reFormat).replaceAll("\\\\d{4}");
if ( Pattern.compile(reFormat).matcher(maybeDate).matches() ) {
// date string matches format structure,
// - now test it can be converted to a valid date
SimpleDateFormat sdf = (SimpleDateFormat)DateFormat.getDateInstance();
sdf.applyPattern(format);
sdf.setLenient(lenient);
try { date = sdf.parse(maybeDate); } catch (ParseException e) { }
}
return date;
}
// used like this:
Date date = parseDate( "21/5/2009", "d/M/yyyy", false);
Note that the regex assumes the format string contains only day, month, year, and separator characters. Aside from that, format can be in any locale format: "d/MM/yy", "yyyy-MM-dd", and so on. The format string for the current locale could be obtained like this:
Locale locale = Locale.getDefault();
SimpleDateFormat sdf = (SimpleDateFormat)DateFormat.getDateInstance(DateFormat.SHORT, locale );
String format = sdf.toPattern();
Joda Time - Better Alternative?
I've been hearing about joda time recently and thought I'd compare. Two points:
Seems better at being strict about invalid characters in the date string, unlike SimpleDateFormat
Can't see a way to enforce 4-digit years with it yet (but I guess you could create your own DateTimeFormatter for this purpose)
It's quite simple to use:
import org.joda.time.format.*;
import org.joda.time.DateTime;
org.joda.time.DateTime parseDate(String maybeDate, String format) {
org.joda.time.DateTime date = null;
try {
DateTimeFormatter fmt = DateTimeFormat.forPattern(format);
date = fmt.parseDateTime(maybeDate);
} catch (Exception e) { }
return date;
}
tl;dr
Use the strict mode on java.time.DateTimeFormatter to parse a LocalDate. Trap for the DateTimeParseException.
LocalDate.parse( // Represent a date-only value, without time-of-day and without time zone.
"31/02/2000" , // Input string.
DateTimeFormatter // Define a formatting pattern to match your input string.
.ofPattern ( "dd/MM/uuuu" )
.withResolverStyle ( ResolverStyle.STRICT ) // Specify leniency in tolerating questionable inputs.
)
After parsing, you might check for reasonable value. For example, a birth date within last one hundred years.
birthDate.isAfter( LocalDate.now().minusYears( 100 ) )
Avoid legacy date-time classes
Avoid using the troublesome old date-time classes shipped with the earliest versions of Java. Now supplanted by the java.time classes.
LocalDate & DateTimeFormatter & ResolverStyle
The LocalDate class represents a date-only value without time-of-day and without time zone.
String input = "31/02/2000";
DateTimeFormatter f = DateTimeFormatter.ofPattern ( "dd/MM/uuuu" );
try {
LocalDate ld = LocalDate.parse ( input , f );
System.out.println ( "ld: " + ld );
} catch ( DateTimeParseException e ) {
System.out.println ( "ERROR: " + e );
}
The java.time.DateTimeFormatter class can be set to parse strings with any of three leniency modes defined in the ResolverStyle enum. We insert a line into the above code to try each of the modes.
f = f.withResolverStyle ( ResolverStyle.LENIENT );
The results:
ResolverStyle.LENIENTld: 2000-03-02
ResolverStyle.SMARTld: 2000-02-29
ResolverStyle.STRICTERROR: java.time.format.DateTimeParseException: Text '31/02/2000' could not be parsed: Invalid date 'FEBRUARY 31'
We can see that in ResolverStyle.LENIENT mode, the invalid date is moved forward an equivalent number of days. In ResolverStyle.SMART mode (the default), a logical decision is made to keep the date within the month and going with the last possible day of the month, Feb 29 in a leap year, as there is no 31st day in that month. The ResolverStyle.STRICT mode throws an exception complaining that there is no such date.
All three of these are reasonable depending on your business problem and policies. Sounds like in your case you want the strict mode to reject the invalid date rather than adjust it.
About java.time
The java.time framework is built into Java 8 and later. These classes supplant the troublesome old legacy date-time classes such as java.util.Date, Calendar, & SimpleDateFormat.
To learn more, see the Oracle Tutorial. And search Stack Overflow for many examples and explanations. Specification is JSR 310.
The Joda-Time project, now in maintenance mode, advises migration to the java.time classes.
You may exchange java.time objects directly with your database. Use a JDBC driver compliant with JDBC 4.2 or later. No need for strings, no need for java.sql.* classes.
Where to obtain the java.time classes?
Java SE 8, Java SE 9, Java SE 10, Java SE 11, and later - Part of the standard Java API with a bundled implementation.
Java 9 adds some minor features and fixes.
Java SE 6 and Java SE 7
Most of the java.time functionality is back-ported to Java 6 & 7 in ThreeTen-Backport.
Android
Later versions of Android bundle implementations of the java.time classes.
For earlier Android (<26), the ThreeTenABP project adapts ThreeTen-Backport (mentioned above). See How to use ThreeTenABP….
The ThreeTen-Extra project extends java.time with additional classes. This project is a proving ground for possible future additions to java.time. You may find some useful classes here such as Interval, YearWeek, YearQuarter, and more.
You can use SimpleDateFormat
For example something like:
boolean isLegalDate(String s) {
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
sdf.setLenient(false);
return sdf.parse(s, new ParsePosition(0)) != null;
}
The current way is to use the calendar class. It has the setLenient method that will validate the date and throw and exception if it is out of range as in your example.
Forgot to add:
If you get a calendar instance and set the time using your date, this is how you get the validation.
Calendar cal = Calendar.getInstance();
cal.setLenient(false);
cal.setTime(yourDate);
try {
cal.getTime();
}
catch (Exception e) {
System.out.println("Invalid date");
}
java.time
With the Date and Time API (java.time classes) built into Java 8 and later, you can use the LocalDate class.
public static boolean isDateValid(int year, int month, int day) {
try {
LocalDate.of(year, month, day);
} catch (DateTimeException e) {
return false;
}
return true;
}
Building on Aravind's answer to fix the problem pointed out by ceklock in his comment, I added a method to verify that the dateString doesn't contain any invalid character.
Here is how I do:
private boolean isDateCorrect(String dateString) {
try {
Date date = mDateFormatter.parse(dateString);
Calendar calendar = Calendar.getInstance();
calendar.setTime(date);
return matchesOurDatePattern(dateString); //added my method
}
catch (ParseException e) {
return false;
}
}
/**
* This will check if the provided string matches our date format
* #param dateString
* #return true if the passed string matches format 2014-1-15 (YYYY-MM-dd)
*/
private boolean matchesDatePattern(String dateString) {
return dateString.matches("^\\d+\\-\\d+\\-\\d+");
}
An alternative strict solution using the standard library is to perform the following:
1) Create a strict SimpleDateFormat using your pattern
2) Attempt to parse the user entered value using the format object
3) If successful, reformat the Date resulting from (2) using the same date format (from (1))
4) Compare the reformatted date against the original, user-entered value. If they're equal then the value entered strictly matches your pattern.
This way, you don't need to create complex regular expressions - in my case I needed to support all of SimpleDateFormat's pattern syntax, rather than be limited to certain types like just days, months and years.
I suggest you to use org.apache.commons.validator.GenericValidator class from apache.
GenericValidator.isDate(String value, String datePattern, boolean strict);
Note: strict - Whether or not to have an exact match of the datePattern.
I think the simpliest is just to convert a string into a date object and convert it back to a string. The given date string is fine if both strings still match.
public boolean isDateValid(String dateString, String pattern)
{
try
{
SimpleDateFormat sdf = new SimpleDateFormat(pattern);
if (sdf.format(sdf.parse(dateString)).equals(dateString))
return true;
}
catch (ParseException pe) {}
return false;
}
Assuming that both of those are Strings (otherwise they'd already be valid Dates), here's one way:
package cruft;
import java.text.DateFormat;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;
public class DateValidator
{
private static final DateFormat DEFAULT_FORMATTER;
static
{
DEFAULT_FORMATTER = new SimpleDateFormat("dd-MM-yyyy");
DEFAULT_FORMATTER.setLenient(false);
}
public static void main(String[] args)
{
for (String dateString : args)
{
try
{
System.out.println("arg: " + dateString + " date: " + convertDateString(dateString));
}
catch (ParseException e)
{
System.out.println("could not parse " + dateString);
}
}
}
public static Date convertDateString(String dateString) throws ParseException
{
return DEFAULT_FORMATTER.parse(dateString);
}
}
Here's the output I get:
java cruft.DateValidator 32-11-2010 31-02-2010 04-01-2011
could not parse 32-11-2010
could not parse 31-02-2010
arg: 04-01-2011 date: Tue Jan 04 00:00:00 EST 2011
Process finished with exit code 0
As you can see, it does handle both of your cases nicely.
This is working great for me. Approach suggested above by Ben.
private static boolean isDateValid(String s) {
SimpleDateFormat sdf = new SimpleDateFormat("dd/MM/yyyy");
try {
Date d = asDate(s);
if (sdf.format(d).equals(s)) {
return true;
} else {
return false;
}
} catch (ParseException e) {
return false;
}
}
looks like SimpleDateFormat is not checking the pattern strictly even after setLenient(false); method is applied on it, so i have used below method to validate if the date inputted is valid date or not as per supplied pattern.
import java.time.format.DateTimeFormatter;
import java.time.format.DateTimeParseException;
public boolean isValidFormat(String dateString, String pattern) {
boolean valid = true;
DateTimeFormatter formatter = DateTimeFormatter.ofPattern(pattern);
try {
formatter.parse(dateString);
} catch (DateTimeParseException e) {
valid = false;
}
return valid;
}
Two comments on the use of SimpleDateFormat.
it should be declared as a static instance
if declared as static access should be synchronized as it is not thread safe
IME that is better that instantiating an instance for each parse of a date.
Above methods of date parsing are nice , i just added new check in existing methods that double check the converted date with original date using formater, so it works for almost each case as i verified. e.g. 02/29/2013 is invalid date.
Given function parse the date according to current acceptable date formats. It returns true if date is not parsed successfully.
public final boolean validateDateFormat(final String date) {
String[] formatStrings = {"MM/dd/yyyy"};
boolean isInvalidFormat = false;
Date dateObj;
for (String formatString : formatStrings) {
try {
SimpleDateFormat sdf = (SimpleDateFormat) DateFormat.getDateInstance();
sdf.applyPattern(formatString);
sdf.setLenient(false);
dateObj = sdf.parse(date);
System.out.println(dateObj);
if (date.equals(sdf.format(dateObj))) {
isInvalidFormat = false;
break;
}
} catch (ParseException e) {
isInvalidFormat = true;
}
}
return isInvalidFormat;
}
Here's what I did for Node environment using no external libraries:
Date.prototype.yyyymmdd = function() {
var yyyy = this.getFullYear().toString();
var mm = (this.getMonth()+1).toString(); // getMonth() is zero-based
var dd = this.getDate().toString();
return zeroPad([yyyy, mm, dd].join('-'));
};
function zeroPad(date_string) {
var dt = date_string.split('-');
return dt[0] + '-' + (dt[1][1]?dt[1]:"0"+dt[1][0]) + '-' + (dt[2][1]?dt[2]:"0"+dt[2][0]);
}
function isDateCorrect(in_string) {
if (!matchesDatePattern) return false;
in_string = zeroPad(in_string);
try {
var idate = new Date(in_string);
var out_string = idate.yyyymmdd();
return in_string == out_string;
} catch(err) {
return false;
}
function matchesDatePattern(date_string) {
var dateFormat = /[0-9]+-[0-9]+-[0-9]+/;
return dateFormat.test(date_string);
}
}
And here is how to use it:
isDateCorrect('2014-02-23')
true
// to return valid days of month, according to month and year
int returnDaysofMonth(int month, int year) {
int daysInMonth;
boolean leapYear;
leapYear = checkLeap(year);
if (month == 4 || month == 6 || month == 9 || month == 11)
daysInMonth = 30;
else if (month == 2)
daysInMonth = (leapYear) ? 29 : 28;
else
daysInMonth = 31;
return daysInMonth;
}
// to check a year is leap or not
private boolean checkLeap(int year) {
Calendar cal = Calendar.getInstance();
cal.set(Calendar.YEAR, year);
return cal.getActualMaximum(Calendar.DAY_OF_YEAR) > 365;
}
Here is I would check the date format:
public static boolean checkFormat(String dateTimeString) {
return dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}") || dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}\\s\\d{2}:\\d{2}:\\d{2}")
|| dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}T\\d{2}:\\d{2}:\\d{2}") || dateTimeString
.matches("^\\d{4}-\\d{2}-\\d{2}T\\d{2}:\\d{2}:\\d{2}Z") ||
dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}\\s\\d{2}:\\d{2}:\\d{2}Z");
}
public static String detectDateFormat(String inputDate, String requiredFormat) {
String tempDate = inputDate.replace("/", "").replace("-", "").replace(" ", "");
String dateFormat;
if (tempDate.matches("([0-12]{2})([0-31]{2})([0-9]{4})")) {
dateFormat = "MMddyyyy";
} else if (tempDate.matches("([0-31]{2})([0-12]{2})([0-9]{4})")) {
dateFormat = "ddMMyyyy";
} else if (tempDate.matches("([0-9]{4})([0-12]{2})([0-31]{2})")) {
dateFormat = "yyyyMMdd";
} else if (tempDate.matches("([0-9]{4})([0-31]{2})([0-12]{2})")) {
dateFormat = "yyyyddMM";
} else if (tempDate.matches("([0-31]{2})([a-z]{3})([0-9]{4})")) {
dateFormat = "ddMMMyyyy";
} else if (tempDate.matches("([a-z]{3})([0-31]{2})([0-9]{4})")) {
dateFormat = "MMMddyyyy";
} else if (tempDate.matches("([0-9]{4})([a-z]{3})([0-31]{2})")) {
dateFormat = "yyyyMMMdd";
} else if (tempDate.matches("([0-9]{4})([0-31]{2})([a-z]{3})")) {
dateFormat = "yyyyddMMM";
} else {
return "Pattern Not Added";
//add your required regex
}
try {
String formattedDate = new SimpleDateFormat(requiredFormat, Locale.ENGLISH).format(new SimpleDateFormat(dateFormat).parse(tempDate));
return formattedDate;
} catch (Exception e) {
//
return "";
}
}
setLenient to false if you like a strict validation
public boolean isThisDateValid(String dateToValidate, String dateFromat){
if(dateToValidate == null){
return false;
}
SimpleDateFormat sdf = new SimpleDateFormat(dateFromat);
sdf.setLenient(false);
try {
//if not valid, it will throw ParseException
Date date = sdf.parse(dateToValidate);
System.out.println(date);
} catch (ParseException e) {
e.printStackTrace();
return false;
}
return true;
}
With 'legacy' date format, we can format the result and compare it back to the source.
public boolean isValidFormat(String source, String pattern) {
SimpleDateFormat sd = new SimpleDateFormat(pattern);
sd.setLenient(false);
try {
Date date = sd.parse(source);
return date != null && sd.format(date).equals(source);
} catch (Exception e) {
return false;
}
}
This execerpt says 'false' to source=01.01.04 with pattern '01.01.2004'
We can use the org.apache.commons.validator.GenericValidator's method directly without adding the whole library:
public static boolean isValidDate(String value, String datePattern, boolean strict) {
if (value == null
|| datePattern == null
|| datePattern.length() <= 0) {
return false;
}
SimpleDateFormat formatter = new SimpleDateFormat(datePattern, Locale.ENGLISH);
formatter.setLenient(false);
try {
formatter.parse(value);
} catch(ParseException e) {
return false;
}
if (strict && (datePattern.length() != value.length())) {
return false;
}
return true;
}
A simple and elegant way for Android developers (Java 8 not required):
// month value is 1-based. e.g., 1 for January.
public static boolean isDateValid(int year, int month, int day) {
Calendar calendar = Calendar.getInstance();
try {
calendar.setLenient(false);
calendar.set(year, month-1, day);
calendar.getTime();
return true;
} catch (Exception e) {
return false;
}
}
Below code works with dd/MM/yyyy format and can be used to check NotNull,NotEmpty as well.
public static boolean validateJavaDate(String strDate) {
if (strDate != null && !strDate.isEmpty() && !strDate.equalsIgnoreCase(" ")) {
{
SimpleDateFormat date = new SimpleDateFormat("dd/MM/yyyy");
date.setLenient(false);
try {
Date javaDate = date.parse(strDate);
System.out.println(strDate + " Valid Date format");
}
catch (ParseException e) {
System.out.println(strDate + " Invalid Date format");
return false;
}
return true;
}
} else {
System.out.println(strDate + "----> Date is Null/Empty");
return false;
}
}

Possible to create a Java.time.LocalDate object from a String?

I have been trying to use the date.format(DateTimeFormatter formatter) method to format a list of date strings, where 'date' is a java.time.LocalDate object, for example. The problem is, I cannot find a straight-forward way to create a Year object from a string. For instance, if I have the string yearString = "90". I would like to create a Year object that is equal to this value, and then use the format method to output yearStringNew = "1990". The closest I see to a public constructor is the now() function which returns the current time.
I have also looked into creating a Calendar object and then creating a format-able date object there, but I can only create a Java.util.Date object – as opposed to an object in the Java.time package which could then ideally be formatted by the format function. Am I missing something here?
FYI I'm referencing the Java 8 SDK javadoc https://docs.oracle.com/javase/8/docs/api/
Thank you for your time.
EDIT: Per the request of another user, I have posted my code below; this is the closest I have come to accomplishing what I'm looking for:
//Module 3:
//Format Date Segments
package challenge245E;
import java.text.DateFormat;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Arrays;
import java.util.Date;
public class TestClass3 {
public static void main(String[] args) throws ParseException {
DateFormatter dateFormatter = new DateFormatter();
String myGroupedSlices [][] =
{
{"1990", "12", "06"},
{"12","6", "90"}
};
dateFormatter.formatDates(myGroupedSlices);
}
}
class DateFormatter {
public Date[][] formatDates(String[][] groupedDates) throws ParseException {
Date[][] formattedDates = new Date[groupedDates.length][3];
DateFormat yearFormat = new SimpleDateFormat("YYYY");
DateFormat monthFormat = new SimpleDateFormat("MM");
DateFormat dayFormat = new SimpleDateFormat("dd");
//iterate through each groupedSlices array
for (int i=0; i<groupedDates.length;i++) {
//Conditions
if (groupedDates[i][0].length()<3) {
//MDDYY format: if date[0].length < 3
//Re-arrange into YDM order
String m = groupedDates[i][0];
String y = groupedDates[i][2];
groupedDates[i][0] = y;
groupedDates[i][2] = m;
//convert dates to correct format
formattedDates[i][0] = yearFormat.parse(groupedDates[i][0]);
formattedDates[i][1] = monthFormat.parse(groupedDates[i][1]);
formattedDates[i][2] = dayFormat.parse(groupedDates[i][2]);
//testing if block
System.out.println("MDY Order: " + Arrays.toString(formattedDates[i]));
}
if (groupedDates[i][0].length()>3) {
//YYYYMMDD format: if date[0].length > 3
//convert dates to correct format
formattedDates[i][0] = yearFormat.parse(groupedDates[i][0]);
formattedDates[i][1] = monthFormat.parse(groupedDates[i][1]);
formattedDates[i][2] = dayFormat.parse(groupedDates[i][2]);
//testing if block
System.out.println("YMD Order: " + Arrays.toString(formattedDates[i]));
}
}
return formattedDates;
}
}
If I understand your requirement correctly, have a look at the LocalDate.parse() methods.
Example:
LocalDate date = LocalDate.parse("1990-01-01", DateTimeFormatter.ofPattern("yyyy-MM-dd"));
int year = date.getYear(); // 1990
Parse Each Number Separately
It’s good to see you using the java.time framework rather than the troublesome old date-time classes. The old java.util.Date/.Calendar classes have been supplanted by the new framework.
The DateTimeFormatter class parses any two digit year as being in the 2000s. From class doc:
If the count of letters is two… will parse using the base value of 2000, resulting in a year within the range 2000 to 2099 inclusive.
To override this behavior, see this Answer by assylias. But that issue may be moot in your case. You already have the individual year, month, and date values isolated. So they need not be parsed together.
I suggest you convert each string into a integer. For the year, if less than 100 then add 1900.
String inputYear = "90";
String inputMonth = "12";
String inputDayOfMonth = "6";
int year = Integer.parseInt( inputYear );
int month = Integer.parseInt( inputMonth );
int dayOfMonth = Integer.parseInt( inputDayOfMonth );
if( year < 100 ) { // If two-digit year, assume the 1900s century. Even better: Never generate two-digit year text!
year = ( year + 1900 );
}
LocalDate localDate = LocalDate.of( year , month , dayOfMonth );
Create an Instance of class GregorianCalendar, set your date in that calendar and then use the method toZonedDateTime(). This will give you ZonedDateTime instance. form it you can use method LocalDate.from(TemporalAccessor temporal) method to get your LocalDate. Here how it might look:
//....
GregorianCalendar calendar = new GregorianCalendar();
// Set the deasired date in your calendar
LocalDate localDate = LocalDate.from(calendar.toZonedDateTime());
//...
import java.time.LocalDate;
public class DaysTilNextMonth {
public static void main(String[] args) {
//create an object for LocalDate class
LocalDate date = LocalDate.now();
//get today's day
int today = date.getDayOfMonth();
//get number of days in the current month
int numOfDaysInMonth = date.lengthOfMonth();
//compute the days left for next month
int dayForNextMonth = numOfDaysInMonth - today;
//display the result
System.out.println("The next month is: " + date.plusMonths(7).getMonth());
System.out.println("We have " + dayForNextMonth + " days left until first day of next month.");
}
}

Java time Error converting data

I have a little problem with the translation of this data 19/12/2005
17:30:45 to mileseconds. I don't know why, I am getting a wrong translation to Jun 28 17:30:45 CEST 1995 Time :804353445798
The code I used is this :
private static long ConvertTimeToTimeStamp(String time) {
Integer[] data = new Integer[6];
String [] tokens = time.split(" ");
System.out.println(tokens[0]);
System.out.println(tokens[1]);
String[] d_m_y = tokens[0].split("/");
String[] hh_mm_ss = tokens[1].split(":");
for (int i = 0; i < d_m_y.length; i++) {
data[i]=Integer.parseInt(d_m_y[i]);
// System.out.println(d_m_y[i]);
}
for (int i = 0; i < hh_mm_ss.length; i++) {
data[i+3]=Integer.parseInt(hh_mm_ss[i]);
// System.out.println(hh_mm_ss[i]);
}
//Calendar calendar = Calendar.getInstance();
GregorianCalendar calendar = new GregorianCalendar();
calendar.set(data[0]+1970, data[1], data[2], data[3],data[4],data[5]);
System.out.println(calendar.getTime().toString());
return calendar.getTimeInMillis();
}
You should not parse the date manually - use a date format instead:
String s = "19/12/2005 17:30:45";
Date d = new SimpleDateFormat("dd/MM/yyyy HH:mm:ss").parse(s);
Treat this object from the outset as a date. That will be much more reliable.
Use SimpleDateFormat.parse() to parse your string into a date/time. Otherwise you're reduced to string parsing/splitting/recombining etc with no bounds-checking, type-checking etc. Your solution is stringly-typed, not strongly-typed.
I've referenced the standard Java libraries, but Joda-Time is a better bet for doing date work in Java and I would advise adopting this for a more intuitive and safer API.
Forget about it. Use a SimpleDataFormat object with pattern dd/MM/yyyy HH:mm:ss to parse() the String into a date object.
try...
private static long ConvertTimeToTimeStamp(String time) {
//19/12/2005 17:30:45
SimpleDateFormat sdf = new SimpleDateFormat("dd/MM/yyyy hh:mm:ss");
java.util.Date d = sdf.parse(time);
return d.getTime();
}
You are mixing the order of year, month and day. The set method is expecting them in the following order: year, month, day and you are providing them as day + 1970, month, year.
Having said that, you might actually be much more happy with a library like joda-time.

Checking the validity of a date [duplicate]

I find it curious that the most obvious way to create Date objects in Java has been deprecated and appears to have been "substituted" with a not so obvious to use lenient calendar.
How do you check that a date, given as a combination of day, month, and year, is a valid date?
For instance, 2008-02-31 (as in yyyy-mm-dd) would be an invalid date.
Key is df.setLenient(false);. This is more than enough for simple cases. If you are looking for a more robust (I doubt that) and/or alternate libraries like joda-time, then look at the answer by user "tardate"
final static String DATE_FORMAT = "dd-MM-yyyy";
public static boolean isDateValid(String date)
{
try {
DateFormat df = new SimpleDateFormat(DATE_FORMAT);
df.setLenient(false);
df.parse(date);
return true;
} catch (ParseException e) {
return false;
}
}
As shown by #Maglob, the basic approach is to test the conversion from string to date using SimpleDateFormat.parse. That will catch invalid day/month combinations like 2008-02-31.
However, in practice that is rarely enough since SimpleDateFormat.parse is exceedingly liberal. There are two behaviours you might be concerned with:
Invalid characters in the date string
Surprisingly, 2008-02-2x will "pass" as a valid date with locale format = "yyyy-MM-dd" for example. Even when isLenient==false.
Years: 2, 3 or 4 digits?
You may also want to enforce 4-digit years rather than allowing the default SimpleDateFormat behaviour (which will interpret "12-02-31" differently depending on whether your format was "yyyy-MM-dd" or "yy-MM-dd")
A Strict Solution with the Standard Library
So a complete string to date test could look like this: a combination of regex match, and then a forced date conversion. The trick with the regex is to make it locale-friendly.
Date parseDate(String maybeDate, String format, boolean lenient) {
Date date = null;
// test date string matches format structure using regex
// - weed out illegal characters and enforce 4-digit year
// - create the regex based on the local format string
String reFormat = Pattern.compile("d+|M+").matcher(Matcher.quoteReplacement(format)).replaceAll("\\\\d{1,2}");
reFormat = Pattern.compile("y+").matcher(reFormat).replaceAll("\\\\d{4}");
if ( Pattern.compile(reFormat).matcher(maybeDate).matches() ) {
// date string matches format structure,
// - now test it can be converted to a valid date
SimpleDateFormat sdf = (SimpleDateFormat)DateFormat.getDateInstance();
sdf.applyPattern(format);
sdf.setLenient(lenient);
try { date = sdf.parse(maybeDate); } catch (ParseException e) { }
}
return date;
}
// used like this:
Date date = parseDate( "21/5/2009", "d/M/yyyy", false);
Note that the regex assumes the format string contains only day, month, year, and separator characters. Aside from that, format can be in any locale format: "d/MM/yy", "yyyy-MM-dd", and so on. The format string for the current locale could be obtained like this:
Locale locale = Locale.getDefault();
SimpleDateFormat sdf = (SimpleDateFormat)DateFormat.getDateInstance(DateFormat.SHORT, locale );
String format = sdf.toPattern();
Joda Time - Better Alternative?
I've been hearing about joda time recently and thought I'd compare. Two points:
Seems better at being strict about invalid characters in the date string, unlike SimpleDateFormat
Can't see a way to enforce 4-digit years with it yet (but I guess you could create your own DateTimeFormatter for this purpose)
It's quite simple to use:
import org.joda.time.format.*;
import org.joda.time.DateTime;
org.joda.time.DateTime parseDate(String maybeDate, String format) {
org.joda.time.DateTime date = null;
try {
DateTimeFormatter fmt = DateTimeFormat.forPattern(format);
date = fmt.parseDateTime(maybeDate);
} catch (Exception e) { }
return date;
}
tl;dr
Use the strict mode on java.time.DateTimeFormatter to parse a LocalDate. Trap for the DateTimeParseException.
LocalDate.parse( // Represent a date-only value, without time-of-day and without time zone.
"31/02/2000" , // Input string.
DateTimeFormatter // Define a formatting pattern to match your input string.
.ofPattern ( "dd/MM/uuuu" )
.withResolverStyle ( ResolverStyle.STRICT ) // Specify leniency in tolerating questionable inputs.
)
After parsing, you might check for reasonable value. For example, a birth date within last one hundred years.
birthDate.isAfter( LocalDate.now().minusYears( 100 ) )
Avoid legacy date-time classes
Avoid using the troublesome old date-time classes shipped with the earliest versions of Java. Now supplanted by the java.time classes.
LocalDate & DateTimeFormatter & ResolverStyle
The LocalDate class represents a date-only value without time-of-day and without time zone.
String input = "31/02/2000";
DateTimeFormatter f = DateTimeFormatter.ofPattern ( "dd/MM/uuuu" );
try {
LocalDate ld = LocalDate.parse ( input , f );
System.out.println ( "ld: " + ld );
} catch ( DateTimeParseException e ) {
System.out.println ( "ERROR: " + e );
}
The java.time.DateTimeFormatter class can be set to parse strings with any of three leniency modes defined in the ResolverStyle enum. We insert a line into the above code to try each of the modes.
f = f.withResolverStyle ( ResolverStyle.LENIENT );
The results:
ResolverStyle.LENIENTld: 2000-03-02
ResolverStyle.SMARTld: 2000-02-29
ResolverStyle.STRICTERROR: java.time.format.DateTimeParseException: Text '31/02/2000' could not be parsed: Invalid date 'FEBRUARY 31'
We can see that in ResolverStyle.LENIENT mode, the invalid date is moved forward an equivalent number of days. In ResolverStyle.SMART mode (the default), a logical decision is made to keep the date within the month and going with the last possible day of the month, Feb 29 in a leap year, as there is no 31st day in that month. The ResolverStyle.STRICT mode throws an exception complaining that there is no such date.
All three of these are reasonable depending on your business problem and policies. Sounds like in your case you want the strict mode to reject the invalid date rather than adjust it.
About java.time
The java.time framework is built into Java 8 and later. These classes supplant the troublesome old legacy date-time classes such as java.util.Date, Calendar, & SimpleDateFormat.
To learn more, see the Oracle Tutorial. And search Stack Overflow for many examples and explanations. Specification is JSR 310.
The Joda-Time project, now in maintenance mode, advises migration to the java.time classes.
You may exchange java.time objects directly with your database. Use a JDBC driver compliant with JDBC 4.2 or later. No need for strings, no need for java.sql.* classes.
Where to obtain the java.time classes?
Java SE 8, Java SE 9, Java SE 10, Java SE 11, and later - Part of the standard Java API with a bundled implementation.
Java 9 adds some minor features and fixes.
Java SE 6 and Java SE 7
Most of the java.time functionality is back-ported to Java 6 & 7 in ThreeTen-Backport.
Android
Later versions of Android bundle implementations of the java.time classes.
For earlier Android (<26), the ThreeTenABP project adapts ThreeTen-Backport (mentioned above). See How to use ThreeTenABP….
The ThreeTen-Extra project extends java.time with additional classes. This project is a proving ground for possible future additions to java.time. You may find some useful classes here such as Interval, YearWeek, YearQuarter, and more.
You can use SimpleDateFormat
For example something like:
boolean isLegalDate(String s) {
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
sdf.setLenient(false);
return sdf.parse(s, new ParsePosition(0)) != null;
}
The current way is to use the calendar class. It has the setLenient method that will validate the date and throw and exception if it is out of range as in your example.
Forgot to add:
If you get a calendar instance and set the time using your date, this is how you get the validation.
Calendar cal = Calendar.getInstance();
cal.setLenient(false);
cal.setTime(yourDate);
try {
cal.getTime();
}
catch (Exception e) {
System.out.println("Invalid date");
}
java.time
With the Date and Time API (java.time classes) built into Java 8 and later, you can use the LocalDate class.
public static boolean isDateValid(int year, int month, int day) {
try {
LocalDate.of(year, month, day);
} catch (DateTimeException e) {
return false;
}
return true;
}
Building on Aravind's answer to fix the problem pointed out by ceklock in his comment, I added a method to verify that the dateString doesn't contain any invalid character.
Here is how I do:
private boolean isDateCorrect(String dateString) {
try {
Date date = mDateFormatter.parse(dateString);
Calendar calendar = Calendar.getInstance();
calendar.setTime(date);
return matchesOurDatePattern(dateString); //added my method
}
catch (ParseException e) {
return false;
}
}
/**
* This will check if the provided string matches our date format
* #param dateString
* #return true if the passed string matches format 2014-1-15 (YYYY-MM-dd)
*/
private boolean matchesDatePattern(String dateString) {
return dateString.matches("^\\d+\\-\\d+\\-\\d+");
}
An alternative strict solution using the standard library is to perform the following:
1) Create a strict SimpleDateFormat using your pattern
2) Attempt to parse the user entered value using the format object
3) If successful, reformat the Date resulting from (2) using the same date format (from (1))
4) Compare the reformatted date against the original, user-entered value. If they're equal then the value entered strictly matches your pattern.
This way, you don't need to create complex regular expressions - in my case I needed to support all of SimpleDateFormat's pattern syntax, rather than be limited to certain types like just days, months and years.
I suggest you to use org.apache.commons.validator.GenericValidator class from apache.
GenericValidator.isDate(String value, String datePattern, boolean strict);
Note: strict - Whether or not to have an exact match of the datePattern.
I think the simpliest is just to convert a string into a date object and convert it back to a string. The given date string is fine if both strings still match.
public boolean isDateValid(String dateString, String pattern)
{
try
{
SimpleDateFormat sdf = new SimpleDateFormat(pattern);
if (sdf.format(sdf.parse(dateString)).equals(dateString))
return true;
}
catch (ParseException pe) {}
return false;
}
Assuming that both of those are Strings (otherwise they'd already be valid Dates), here's one way:
package cruft;
import java.text.DateFormat;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;
public class DateValidator
{
private static final DateFormat DEFAULT_FORMATTER;
static
{
DEFAULT_FORMATTER = new SimpleDateFormat("dd-MM-yyyy");
DEFAULT_FORMATTER.setLenient(false);
}
public static void main(String[] args)
{
for (String dateString : args)
{
try
{
System.out.println("arg: " + dateString + " date: " + convertDateString(dateString));
}
catch (ParseException e)
{
System.out.println("could not parse " + dateString);
}
}
}
public static Date convertDateString(String dateString) throws ParseException
{
return DEFAULT_FORMATTER.parse(dateString);
}
}
Here's the output I get:
java cruft.DateValidator 32-11-2010 31-02-2010 04-01-2011
could not parse 32-11-2010
could not parse 31-02-2010
arg: 04-01-2011 date: Tue Jan 04 00:00:00 EST 2011
Process finished with exit code 0
As you can see, it does handle both of your cases nicely.
This is working great for me. Approach suggested above by Ben.
private static boolean isDateValid(String s) {
SimpleDateFormat sdf = new SimpleDateFormat("dd/MM/yyyy");
try {
Date d = asDate(s);
if (sdf.format(d).equals(s)) {
return true;
} else {
return false;
}
} catch (ParseException e) {
return false;
}
}
looks like SimpleDateFormat is not checking the pattern strictly even after setLenient(false); method is applied on it, so i have used below method to validate if the date inputted is valid date or not as per supplied pattern.
import java.time.format.DateTimeFormatter;
import java.time.format.DateTimeParseException;
public boolean isValidFormat(String dateString, String pattern) {
boolean valid = true;
DateTimeFormatter formatter = DateTimeFormatter.ofPattern(pattern);
try {
formatter.parse(dateString);
} catch (DateTimeParseException e) {
valid = false;
}
return valid;
}
Two comments on the use of SimpleDateFormat.
it should be declared as a static instance
if declared as static access should be synchronized as it is not thread safe
IME that is better that instantiating an instance for each parse of a date.
Above methods of date parsing are nice , i just added new check in existing methods that double check the converted date with original date using formater, so it works for almost each case as i verified. e.g. 02/29/2013 is invalid date.
Given function parse the date according to current acceptable date formats. It returns true if date is not parsed successfully.
public final boolean validateDateFormat(final String date) {
String[] formatStrings = {"MM/dd/yyyy"};
boolean isInvalidFormat = false;
Date dateObj;
for (String formatString : formatStrings) {
try {
SimpleDateFormat sdf = (SimpleDateFormat) DateFormat.getDateInstance();
sdf.applyPattern(formatString);
sdf.setLenient(false);
dateObj = sdf.parse(date);
System.out.println(dateObj);
if (date.equals(sdf.format(dateObj))) {
isInvalidFormat = false;
break;
}
} catch (ParseException e) {
isInvalidFormat = true;
}
}
return isInvalidFormat;
}
Here's what I did for Node environment using no external libraries:
Date.prototype.yyyymmdd = function() {
var yyyy = this.getFullYear().toString();
var mm = (this.getMonth()+1).toString(); // getMonth() is zero-based
var dd = this.getDate().toString();
return zeroPad([yyyy, mm, dd].join('-'));
};
function zeroPad(date_string) {
var dt = date_string.split('-');
return dt[0] + '-' + (dt[1][1]?dt[1]:"0"+dt[1][0]) + '-' + (dt[2][1]?dt[2]:"0"+dt[2][0]);
}
function isDateCorrect(in_string) {
if (!matchesDatePattern) return false;
in_string = zeroPad(in_string);
try {
var idate = new Date(in_string);
var out_string = idate.yyyymmdd();
return in_string == out_string;
} catch(err) {
return false;
}
function matchesDatePattern(date_string) {
var dateFormat = /[0-9]+-[0-9]+-[0-9]+/;
return dateFormat.test(date_string);
}
}
And here is how to use it:
isDateCorrect('2014-02-23')
true
// to return valid days of month, according to month and year
int returnDaysofMonth(int month, int year) {
int daysInMonth;
boolean leapYear;
leapYear = checkLeap(year);
if (month == 4 || month == 6 || month == 9 || month == 11)
daysInMonth = 30;
else if (month == 2)
daysInMonth = (leapYear) ? 29 : 28;
else
daysInMonth = 31;
return daysInMonth;
}
// to check a year is leap or not
private boolean checkLeap(int year) {
Calendar cal = Calendar.getInstance();
cal.set(Calendar.YEAR, year);
return cal.getActualMaximum(Calendar.DAY_OF_YEAR) > 365;
}
Here is I would check the date format:
public static boolean checkFormat(String dateTimeString) {
return dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}") || dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}\\s\\d{2}:\\d{2}:\\d{2}")
|| dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}T\\d{2}:\\d{2}:\\d{2}") || dateTimeString
.matches("^\\d{4}-\\d{2}-\\d{2}T\\d{2}:\\d{2}:\\d{2}Z") ||
dateTimeString.matches("^\\d{4}-\\d{2}-\\d{2}\\s\\d{2}:\\d{2}:\\d{2}Z");
}
public static String detectDateFormat(String inputDate, String requiredFormat) {
String tempDate = inputDate.replace("/", "").replace("-", "").replace(" ", "");
String dateFormat;
if (tempDate.matches("([0-12]{2})([0-31]{2})([0-9]{4})")) {
dateFormat = "MMddyyyy";
} else if (tempDate.matches("([0-31]{2})([0-12]{2})([0-9]{4})")) {
dateFormat = "ddMMyyyy";
} else if (tempDate.matches("([0-9]{4})([0-12]{2})([0-31]{2})")) {
dateFormat = "yyyyMMdd";
} else if (tempDate.matches("([0-9]{4})([0-31]{2})([0-12]{2})")) {
dateFormat = "yyyyddMM";
} else if (tempDate.matches("([0-31]{2})([a-z]{3})([0-9]{4})")) {
dateFormat = "ddMMMyyyy";
} else if (tempDate.matches("([a-z]{3})([0-31]{2})([0-9]{4})")) {
dateFormat = "MMMddyyyy";
} else if (tempDate.matches("([0-9]{4})([a-z]{3})([0-31]{2})")) {
dateFormat = "yyyyMMMdd";
} else if (tempDate.matches("([0-9]{4})([0-31]{2})([a-z]{3})")) {
dateFormat = "yyyyddMMM";
} else {
return "Pattern Not Added";
//add your required regex
}
try {
String formattedDate = new SimpleDateFormat(requiredFormat, Locale.ENGLISH).format(new SimpleDateFormat(dateFormat).parse(tempDate));
return formattedDate;
} catch (Exception e) {
//
return "";
}
}
setLenient to false if you like a strict validation
public boolean isThisDateValid(String dateToValidate, String dateFromat){
if(dateToValidate == null){
return false;
}
SimpleDateFormat sdf = new SimpleDateFormat(dateFromat);
sdf.setLenient(false);
try {
//if not valid, it will throw ParseException
Date date = sdf.parse(dateToValidate);
System.out.println(date);
} catch (ParseException e) {
e.printStackTrace();
return false;
}
return true;
}
With 'legacy' date format, we can format the result and compare it back to the source.
public boolean isValidFormat(String source, String pattern) {
SimpleDateFormat sd = new SimpleDateFormat(pattern);
sd.setLenient(false);
try {
Date date = sd.parse(source);
return date != null && sd.format(date).equals(source);
} catch (Exception e) {
return false;
}
}
This execerpt says 'false' to source=01.01.04 with pattern '01.01.2004'
We can use the org.apache.commons.validator.GenericValidator's method directly without adding the whole library:
public static boolean isValidDate(String value, String datePattern, boolean strict) {
if (value == null
|| datePattern == null
|| datePattern.length() <= 0) {
return false;
}
SimpleDateFormat formatter = new SimpleDateFormat(datePattern, Locale.ENGLISH);
formatter.setLenient(false);
try {
formatter.parse(value);
} catch(ParseException e) {
return false;
}
if (strict && (datePattern.length() != value.length())) {
return false;
}
return true;
}
A simple and elegant way for Android developers (Java 8 not required):
// month value is 1-based. e.g., 1 for January.
public static boolean isDateValid(int year, int month, int day) {
Calendar calendar = Calendar.getInstance();
try {
calendar.setLenient(false);
calendar.set(year, month-1, day);
calendar.getTime();
return true;
} catch (Exception e) {
return false;
}
}
Below code works with dd/MM/yyyy format and can be used to check NotNull,NotEmpty as well.
public static boolean validateJavaDate(String strDate) {
if (strDate != null && !strDate.isEmpty() && !strDate.equalsIgnoreCase(" ")) {
{
SimpleDateFormat date = new SimpleDateFormat("dd/MM/yyyy");
date.setLenient(false);
try {
Date javaDate = date.parse(strDate);
System.out.println(strDate + " Valid Date format");
}
catch (ParseException e) {
System.out.println(strDate + " Invalid Date format");
return false;
}
return true;
}
} else {
System.out.println(strDate + "----> Date is Null/Empty");
return false;
}
}

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