This question already has answers here:
Why does Java's java.time.format.DateTimeFormatter#format(LocalDateTime) add a year?
(2 answers)
Using DateTimeFormatter on january first cause an invalid year value
(1 answer)
Closed 3 years ago.
I am trying to convert a date of the format 2019-12-30 to a date with format 30-12-2019 and for this I thought of using DateTimeFormatter and I have the following code for this:
LocalDate date = LocalDate.parse("2019-12-30");
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("dd-MM-YYYY");
System.out.println(formatter.format(date));
However, to my surprise, this returns an output:
30-12-2020
instead of the expected 30-12-2019. If the date is set to 2019-11-30, it correctly returns 30-11-2019. I am doing something terribly wrong but I am not able to figure out what exactly. Can someone please help?
From the DateTimeFormatter documentation:, indicating the symbol, meaning and examples:
Y week-based-year year 1996; 96
So you're formatting the week-based-year, not the regular year. December 30th 2019 belongs to the first week of 2020, hence your output.
Use yyyy (year-of-era) or uuuu (year) instead of YYYY and you'll get 2019 instead.
Basically, YYYY should usually be used with w (week-of-week-based-year) and E (day-of-week).
Use "dd-MM-yyyy" instead of "dd-MM-YYYY"
Try the following code snippet to get expected result -
LocalDate date = LocalDate.parse("2019-12-30");
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("dd-MM-yyyy");
System.out.println("date " + formatter.format(date)); // date 30-12-2019
You are using DateTimeFormatter.ofPattern("dd-MM-YYYY"); and notice that Y stands for week-based-year, which you probably don't want. Perhaps you want yout pattern to be "dd-MM-uuuu". For more information on patterns, refer to the official Oracle documentation: Patterns for Formatting and Parsing
Related
This question already has answers here:
Unable to parse DateTime-string with AM/PM marker
(6 answers)
Closed 1 year ago.
Below is my code snippet and it is failing in my local but works in other DEV's machine, can someone give pointers on what am I missing?
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss a z");
ZonedDateTime releaseDate = ZonedDateTime
.parse("2021-03-31 10:15:30 AM +05:30", formatter);
Exception
java.time.format.DateTimeParseException: Text '2021-03-31 10:15:30 AM +05:30' could not be parsed at index 20
at java.base/java.time.format.DateTimeFormatter.parseResolved0(DateTimeFormatter.java:2046)
at java.base/java.time.format.DateTimeFormatter.parse(DateTimeFormatter.java:1948)
I note two problems with your pattern:
You include HH (24-hour hour) and a (AM/PM). You probably want hh.
You use z (time-zone name) but then show +05:30. Perhaps you want either Z or X.
This question already has answers here:
How can I convert a date in YYYYMMDDHHMMSS format to epoch or unix time?
(2 answers)
Changing String date format
(4 answers)
Change date format in a Java string
(22 answers)
Closed 3 years ago.
I am currently facing a problem when it comes to some Java methods that weren't explained to me in lectures very well. I need to write a program that accepts user-inputted strings (particularly a date) in yyyymmddhhss format, which should then convert to hh:mm Month day, year.
E.g. 201901151500 outputs: "03:00 PM January 15, 2019".
Currently, in my program, I have accepted the user's input and implemented a method that returns an error message if the inputted format is invalid.
Any tips on where to go from here? Advice is greatly appreciated-- thank you!
If you are using Java 8 you can use java.time API like so :
String input = "201901151500";
LocalDateTime dt = LocalDateTime.parse(input, DateTimeFormatter.ofPattern("uuuuMMddHHmm"));
String output = dt.format(DateTimeFormatter.ofPattern("hh:mm a MMMM dd, uuuu"));
>> output = 03:00 PM janvier 15, 2019
Use the DateTimeFormatter as defined here: https://docs.oracle.com/javase/8/docs/api/java/time/format/DateTimeFormatter.html
Pay attention to the parse method. You can define a formatter that takes in a string and then returns it in a certain way, almost any way you choose.
Here is the LocalDataTime class:
https://docs.oracle.com/javase/8/docs/api/java/time/LocalDateTime.html
Example code:
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy MM dd");
LocalDateTime local = LocalDateTime.parse("2004 12 25", formatter);
This question already has answers here:
How do I simply parse a date without a year specified?
(5 answers)
Closed 4 years ago.
I want to parse a date which only consists of date and month information, without a year. The resulting date should be set to the current year.
The input for example will be something like "11. November".
My idea was to use something like this:
DateTimeFormatter.ofPattern("EEEE dd. MMMM");
But in the resulting date the year is set to 1970 which makes sense.
Is there a way of setting the omitted year to a certain value within DateTimeFormatter?
Parse into a MonthDay and add the year. Change the Locale as appropriate.
Parse the input string.
final DateTimeFormatter formatter = DateTimeFormatter.ofPattern("dd. MMMM", Locale.UK);
final MonthDay monthDay = MonthDay.parse("11. November", formatter) ;
Apply the desired year.
final int currentYear = LocalDate.now().getYear();
final LocalDate date = monthDay.atYear(currentYear);
Worth noting, if you have something like 29th February and you add the year 2018 (not a leap year) then it will get pushed back to the 28th because the 29/02/2018 is not a valid date.
This question already has answers here:
SimpleDateFormat producing wrong date time when parsing "YYYY-MM-dd HH:mm"
(5 answers)
Closed 4 years ago.
I want to convert 2018-02-21 15:47:35 UTC to epoch UTC form. How do we do it? I am currently in PST.
SimpleDateFormat df = new SimpleDateFormat("YYYY-MM-DD HH:MM:SS");
df.setTimeZone(TimeZone.getTimeZone("UTC"));
date = df.parse(dateString).getTime();
The code above should return the number of milliseconds since January 1, 1970, 00:00:00 GMT, but I'm getting an incorrect value.
The only problem with your code is DateFormat
please check.
https://docs.oracle.com/javase/7/docs/api/java/text/SimpleDateFormat.html
String dateString = "2018-02-21 15:47:35";
SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
df.setTimeZone(TimeZone.getTimeZone("UTC"));
Date date = df.parse(dateString);
long time = date.getTime();
System.out.println(time);
System.out.println(new Date(time));
System.out.println(date);
I'm in PKT so output would differ...
1519228055000
Wed Feb 21 20:47:35 PKT 2018
Wed Feb 21 20:47:35 PKT 2018
Expected: 2018-02-21 15:47:35 UTC is equivalent to 1 519 228 055 000 milliseconds since the epoch of January 1, 1970 at 0:00 UTC.
Observed: Your code in the question gives 1 514 818 800 035. So it’s 4 409 254 965 milliseconds off, a little over 51 days.
The solution:
DateTimeFormatter dtf = DateTimeFormatter.ofPattern("uuuu-MM-dd HH:mm:ss");
date = LocalDateTime.parse("2018-02-21 15:47:35", dtf)
.atOffset(ZoneOffset.UTC)
.toInstant()
.toEpochMilli();
This gives the correct 1 519 228 055 000.
What went wrong?
One of the many troublesome traits of SimpleDateFormat is that with its default settings, if you specify an incorrect format pattern string, it will very often give you an incorrect result and pretend all is well. The modern Java date and time API that I am using in my snippet, is trying somewhat harder to figure out when the pattern doesn’t make sense and tell you it’s wrong somehow. As an example, let’s try your format pattern with the modern DateTimeFormatter:
final DateTimeFormatter dtf = DateTimeFormatter.ofPattern("YYYY-MM-DD HH:MM:SS");
LocalDateTime.parse(dateString, dtf);
This will throw a java.time.format.DateTimeParseException: Text '2018-02-21 15:47:35' could not be parsed at index 14. Index 14 is where 47 is in the string, it was supposed to be the minutes. Apparently 47 doesn’t match MM in the format. If you haven’t figured out yet, check the documentation. It says that uppercase M is for “month-of-year”. So what the formatter is trying to tell you is there are not 47 months in a year. In the documentation you will also find lowercase m for “minute-of-hour”. As you correct the case of the letters in the format pattern string, you will receive other exceptions until you end up with either yyyy-MM-dd HH:mm:ss or uuuu-MM-dd HH:mm:ss (lowercase yyyy is year or era while uuuu is a signed year, both work for years after year 0).
Links
Oracle tutorial: Date Time explaining how to use java.time.
DateTimeFormatter documentation spelling out the uppercase and lowercase letters of format pattern strings.
Also can be done via java8 time library:
String dateString = "2018-02-21 15:47:35";
DateTimeFormatter dateTimeFormatter = DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss", Locale.ENGLISH);
dateTimeFormatter.withZone(ZoneId.of("UTC"));
LocalDateTime parsedDateTime = LocalDateTime.from(dateTimeFormatter.parse(dateString));
ZonedDateTime timeAtYourZone = parsedDateTime.atZone(ZoneId.systemDefault());
System.out.println(timeAtYourZone.toInstant().toEpochMilli());
System.out.println(timeAtYourZone);
Your pattern must be yyyy-MM-dd HH:mm:ss, as the other answers told you:
SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
I just want to add some more details.
First of all, take a look at the patterns description in the javadoc: https://docs.oracle.com/javase/7/docs/api/java/text/SimpleDateFormat.html
Note that a lowercase y is not the same as the uppercase Y (lowercase is the year, while uppercase is the week year - 2 different fields with completely different definitions)
Also note that uppercase D is the day of the year, while the day of the month (which is what you want) is the lowercase d. And uppercase M is the month, while lowercase m is the minute of hour.
And uppercase S is the milliseconds field, while the seconds are represented by lowercase s.
And SimpleDateFormat's design doesn't help: the class simply tries to parse the string, even if the month field (MM) appears twice in your pattern, while the minutes field doesn't appear (and it's set to a default value of zero - all behind the scenes, without any warning, no indication of error at all).
Conclusion: always read the docs :-)
For Java 8 or higher, consider using the new date API, which is much better because it doesn't have all these behind-the-scenes stuff:
DateTimeFormatter fmt = DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss")
.withZone(ZoneOffset.UTC);
long epochMilli = Instant.from(fmt.parse("2018-02-21 15:47:35")).toEpochMilli();
This API will also throw an exception if you use a pattern like YYYY-MM-DD HH:MM:SS, because it will try to parse the minutes value 47 as a month (because uppercase MM will be in the respective position), and 47 is not a valid month.
This question already has answers here:
How to include milliseconds in a formatted date string?
(3 answers)
Closed 5 years ago.
I have a date that I want to change into a different format.
Current format: Fri Apr 07 08:21:19 MDT 2017
Desired format: 2017-04-07T13:28:41.00MDT. I want to have the desired format output as a string.
I am currently doing this:
DateTimeFormatter sourceFormat = DateTimeFormatter.ofPattern("EEE MMM dd HH:mm:ss z uuuu");
ZonedDateTime dateParsed = ZonedDateTime.parse(dateStr, sourceFormat);
String fullDateTimeStr = fullDateTime.format(dateParsed);
The formatting is correct, but the main issue I'm having is preserving the fractional seconds (the two numbers right before the time zone.) I know that the date I am receiving has the ms/fractional seconds preserved because when I print out long epoch = ((Date) date).getTime(); I get the epoch time and I can convert it (using an online conversion tool) and it shows the correct date/time with ms.
What's going on?
DateTimeFormatter targetFormat = DateTimeFormatter.ofPattern("uuuu-MM-dd'T'HH:mm:ss.SSSz");
ZonedDateTime zdt = ZonedDateTime.now();
System.out.println(zdt.format(targetFormat));
Sample output:
2017-04-14T21:19:58.409CEST
You need SSS in the format pattern for the milliseconds.
Please consider whether you need the three or four letter time zone abbreviations that are often ambiguous. I would prefer to use DateTimeFormatter.ISO_ZONED_DATE_TIME for an output like 2017-04-14T21:22:49.344+02:00[Europe/Berlin]. This also saves you from building the pattern string yourself.