Develop Reference

Regular Expression WildCard matching split with java split method

I know there's similar questions like this asked before, but i want to do a custom operation and i don't know how to go about it.
I want to split a string of data with a regular expression like, but this time like i know the starting character and the ending character like:
String myString="Google is a great search engine<as:...s>";
The <as: and s> is the beginning and closing characters
the ... is dynamic which i cant predict its value
I want to be able to split the string from the beginning <as: to the end s>
with the dynamic string in it.
Like:
myString.split("<as:/*s>");
Something like that. I also want to get all the occurrence of the <as:..s> in the string.
i know this can be done with regex, but I've never done it before. I need a simple and neat way to do this.
Thanks in advance

Rather than using a .split(), I would just extract using Pattern and Matcher. This approach finds everything between <as: and s> and extracts it to a capture group. Group 1 then has the text you would like.
public static void main(String[] args)
{
final String myString="Google is a great search engine<as:Some stuff heres>";
Pattern pat = Pattern.compile("^[^<]+<as:(.*)s>$");
Matcher m = pat.matcher(myString);
if (m.matches()) {
System.out.println(m.group(1));
}
}
Output:
Some stuff here
If you need the text at the beginning, you can put it in a capture group as well.
Edit: If there are more than one <as...s> in the input, then the following will gather all of them.
Edit 2: increased the logic. Added checks for emptiness.
public static List<String> multiEntry(final String myString)
{
String[] parts = myString.split("<as:");
List<String> col = new ArrayList<>();
if (! parts[0].trim().isEmpty()) {
col.add(parts[0]);
}
Pattern pat = Pattern.compile("^(.*?)s>(.*)?");
for (int i = 1; i < parts.length; ++i) {
Matcher m = pat.matcher(parts[i]);
if (m.matches()) {
for (int j = 1; j <= m.groupCount(); ++j) {
String s = m.group(j).trim();
if (! s.isEmpty()) {
col.add(s);
}
}
}
}
return col;
}
Output:
[Google is a great search engine, Some stuff heress, Here is Facebook, More Stuff, Something else at the end]
Edit 3: This approach uses find and looping to do the parsing. It uses optional capture groups as well.
public static void looping()
{
final String myString="Google is a great search engine"
+ "<as:Some stuff heresss>Here is Facebook<as:More Stuffs>"
+ "Something else at the end" +
"<as:Stuffs>" +
"<as:Yet More Stuffs>";
Pattern pat = Pattern.compile("([^<]+)?(<as:(.*?)s>)?");
Matcher m = pat.matcher(myString);
List<String> col = new ArrayList<>();
while (m.find()) {
String prefix = m.group(1);
String contents = m.group(3);
if (prefix != null) { col.add(prefix); }
if (contents != null) { col.add(contents); }
}
System.out.println(col);
}
Output:
[Google is a great search engine, Some stuff heress, Here is Facebook, More Stuff, Something else at the end, Stuff, Yet More Stuff]
Additional Edit: wrote some quick test cases (with super hacked helper class) to help validate. These all pass (updated) multiEntry:
public static void main(String[] args)
{
Input[] inputs = {
new Input("Google is a great search engine<as:Some stuff heres>", 2),
new Input("Google is a great search engine"
+ "<as:Some stuff heresss>Here is Facebook<as:More Stuffs>"
+ "Something else at the end" +
"<as:Stuffs>" +
"<as:Yet More Stuffs>" +
"ending", 8),
new Input("Google is a great search engine"
+ "<as:Some stuff heresss>Here is Facebook<as:More Stuffs>"
+ "Something else at the end" +
"<as:Stuffs>" +
"<as:Yet More Stuffs>", 7),
new Input("No as here", 1),
new Input("Here is angle < input", 1),
new Input("Angle < plus <as:Stuff in as:s><as:Other stuff in as:s>", 3),
new Input("Angle < plus <as:Stuff in as:s><as:Other stuff in as:s>blah", 4),
new Input("<as:To start with anglass>Some ending", 2),
};
List<String> res;
for (Input inp : inputs) {
res = multiEntry(inp.inp);
if (res.size() != inp.cnt) {
System.err.println("FAIL: " + res.size()
+ " did not match exp of " + inp.cnt
+ " on " + inp.inp);
System.err.println(res);
continue;
}
System.out.println(res);
}
}

JDBCTemplate queryForMap compare and replace value in extracted lines

Can you help me to understand why my code doesn't work, please?
I am trying to get values from 2 columns from my database and store them in a hashmap where K_PARAM is my key and L_PARAM is my value. Then I would like to compare 2 characters from a line that I am extracting and see if these 2 characters are equals to my key or not. In case they are equals, I replace key with value.
Thanks in advance. This is the code :
if (action.equals("RP")) {
if (marqueCarte = null) {
jdbcTemplate.query(" select K_PARAM, L_PARAM from DLCOA.DLC_ADM_PARAMS where K_CHX_PARAM = '50'", new ResultSetExtractor<Map>(){
#Override
public Map extractData(ResultSet rs) throws SQLException,DataAccessException {
 HashMap<String,String> marqueCarte = new HashMap<String,String>();
while (rs.next()) {
marqueCarte.put(rs.getString("K_PARAM"),rs.getString("L_PARAM"));
if (line.contains("blocE")) {
if (line.substring(line.indexOf("blocE") + 15, line.indexOf("blocE") + 15 + (line.substring(line.indexOf("blocE")+15)).indexOf("#")).equals(rs.getString("K_PARAM"))){
line = line.replace(line.substring(line.indexOf("blocE") + 15, line.indexOf("blocE") + 15 + (line.substring(line.indexOf("blocE")+15)).indexOf("#")),rs.getString("L_PARAM") );
}
}
   }
return marqueCarte;
}
}
}
}

I got a more readable and modifiable solution for your second problem.
(I'm still not sure what's your first one)
Using regex and patterns you can achieve the replacement you want.
Let's assume that you are searching for the text "blocE" followed by 15 characters, followed at the same time by the text contained in rs.getString("K_PARAM") plus an "#"
We can model what you search as a pattern like this
"(blocE)(.{15})(" + key + "#)"
Parenthesis allow us to establish different groups in the regex.
Group 1 - blocE
Group 2 - 15 characters
Group 3 - key + #
Being group 0 the complete matching expression.
Knowing this you can do the replacement applying the following code
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class TestRegex {
public static void main(String[] args) {
String key = "KEY"; // rs.getString("K_PARAM")
String value = "VALUE"; // rs.getString("L_PARAM")
Pattern pattern = Pattern.compile("(blocE)(.{15})(" + key + "#)");
String input ="helloworldblocE111111111111111KEY#blocE111111111111111KEY";
Matcher m = pattern.matcher(input);
if (m.find()) {
String text2replace = m.group(0);
String replacement = m.group(1) + m.group(2) + value;
System.out.println(input.replaceFirst(text2replace, replacement));
}
}
}
If your pattern changes, you only have to change one line and you do not have to worry about such quantity of indexOf.

phone Number validation in java

I want to validate a phone number in such Way :-
The field should allow the user to enter characters and should auto-correct. So an entry of "+1-908-528-5656" would not create an error for the user, it would just change to "19085285656".
I also want to number range between 9 to 11.
I also tried with the below code but not concluded to the final solution:
final String PHONE_REGEX = "^\\+([0-9\\-]?){9,11}[0-9]$";
final Pattern pattern = Pattern.compile(PHONE_REGEX);
String phone = "+1-908-528-5656";
phone=phone.replaceAll("[\\-\\+]", "");
System.out.println(phone);
final Matcher matcher = pattern.matcher(phone);
System.out.println(matcher.matches());

You can use simple String.matches(regex) to test any string against a regex pattern instead of using Pattern and Matcher classes.
Sample:
boolean isValid = phoneString.matches(regexPattern);
Find more examples
Here is the regex pattern as per your input string:
\+\d(-\d{3}){2}-\d{4}
Online demo
Better use Spring validation annotation for validation.
Example

// The Regex not validate mobile number, which is in internation format.
// The Following code work for me.
// I have use libphonenumber library to validate Number from below link.
// http://repo1.maven.org/maven2/com/googlecode/libphonenumber/libphonenumber/8.0.1/
// https://github.com/googlei18n/libphonenumber
// Here, is my source code.
public boolean isMobileNumberValid(String phoneNumber)
{
boolean isValid = false;
// Use the libphonenumber library to validate Number
PhoneNumberUtil phoneUtil = PhoneNumberUtil.getInstance();
Phonenumber.PhoneNumber swissNumberProto =null ;
try {
swissNumberProto = phoneUtil.parse(phoneNumber, "CH");
} catch (NumberParseException e) {
System.err.println("NumberParseException was thrown: " + e.toString());
}
if(phoneUtil.isValidNumber(swissNumberProto))
{
isValid = true;
}
// The Library failed to validate number if it contains - sign
// thus use regex to validate Mobile Number.
String regex = "[0-9*#+() -]*";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(phoneNumber);
if (matcher.matches()) {
isValid = true;
}
return isValid;
}

Assuming your input field take any kind of character and you just want the digits.
String phone = "+1-908-528-5656";
phone=phone.replaceAll("[\\D]","");
if(phone.length()>=9 || phone.length()<=11)
System.out.println(phone);

We can use String.matches(String regex)1 to validate phone numbers using java.
Sample code snippet
package regex;
public class Phone {
private static boolean isValid(String s) {
String regex = "\\d{3}-\\d{3}-\\d{4}"; // XXX-XXX-XXXX
return s.matches(regex);
}
public static void main(String[] args) {
System.out.println(isValid("123-456-7890"));
}
}
P.S. The regex pattern we use extra '\' for escaping when we use in java string. (Try to use "\d{3}-\d{3}-\d{4}" in java program, you will get an error.

Assuming you want an optimization (which is what your comment suggests).
How bout this? (the "0" is to exclude if they give complete garbage without even a single digit).
int parse(String phone){
int num = Integer.parseInt("0"+phone.replaceAll("[^0-9]",""));
return 100000000<=num&&num<100000000000?num:-1;
}

I am not sure but removing the garbage characters parenthesis, spaces and hyphens, if you match with ^((\+[1-9]?[0-9])|0)?[7-9][0-9]{9}$ , you may validate a mobile number
private static final String PHONE_NUMBER_GARBAGE_REGEX = "[()\\s-]+";
private static final String PHONE_NUMBER_REGEX = "^((\\+[1-9]?[0-9])|0)?[7-9][0-9]{9}$";
private static final Pattern PHONE_NUMBER_PATTERN = Pattern.compile(PHONE_NUMBER_REGEX);
public static boolean validatePhoneNumber(String phoneNumber) {
return phoneNumber != null && PHONE_NUMBER_PATTERN.matcher(phoneNumber.replaceAll(PHONE_NUMBER_GARBAGE_REGEX, "")).matches();
}

One easy and simple to use java phone validation regex:
public static final String PHONE_VERIFICATION = "^[+0-9-\\(\\)\\s]*{6,14}$";
private static Pattern p;
private static Matcher m;
public static void main(String[] args)
{
//Phone validation
p = Pattern.compile(PHONE_VERIFICATION);
m = p.matcher("+1 212-788-8609");
boolean isPhoneValid = m.matches();
if(!isPhoneValid)
{
System.out.println("The Phone number is NOT valid!");
return;
}
System.out.println("The Phone number is valid!");
}

i have done testing one regex for this combination of phone numbers
(294) 784-4554
(247) 784 4554
(124)-784 4783
(124)-784-4783
(124) 784-4783
+1(202)555-0138
THIS REGEX SURELY WILL BE WORKING FOR ALL THE US NUMBERS
\d{10}|(?:\d{3}-){2}\d{4}|\(\d{3}\)\d{3}-?\d{4}|\(\d{3}\)-\d{3}-?\d{4}|\(\d{3}\) \d{3} ?\d{4}|\(\d{3}\)-\d{3} ?\d{4}|\(\d{3}\) \d{3}-?\d{4}

Building on #k_g's answers, but for US numbers.
static boolean isValidTelephoneNumber(String number) {
long num = Long.parseLong("0" + number.replaceAll("[^0-9]", ""));
return 2000000000L <= num && num < 19999999999L;
}
public static void main(String[] args) {
var numbers = List.of("+1 212-788-8609", "212-788-8609", "1 212-788-8609", "12127888609", "2127888609",
"7143788", "736103355");
numbers.forEach(number -> {
boolean isPhoneValid = isValidTelephoneNumber(number);
log.debug(number + " matches = " + isPhoneValid);
});
}

How can i match particular format in input using java.util.regex in java?

INPUT
Input can be in any of the form shown below with following mandatory content TXT{Any comma separated strings in any format}
String loginURL = "http://ip:port/path?username=abcd&location={LOCATION}&TXT{UE-IP,UE-Username,UE-Password}&password={PASS}";
String loginURL1 = "http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}&TXT{UE-IP,UE-Username,UE-Password}";
String loginURL2 = "http://ip:port/path?TXT{UE-IP,UE-Username,UE-Password}&username=abcd&location={LOCATION}&password={PASS}";
String loginURL3 = "http://ip:port/path?TXT{UE-IP,UE-Username,UE-Password}";
String loginURL4 = "http://ip:port/path?username=abcd&password={PASS}";
Required Output
1. OutputURL corresponding to loginURL.
String outputURL = "http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}";
String outputURL1 = "http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}";
String outputURL2 = "http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}";
String outputURL3 = "http://ip:port/path?";
String outputURL4 = "http://ip:port/path?username=abcd&password={PASS}";
2. Deleted pattern(if any)
String deletedPatteren = TXT{UE-IP,UE-Username,UE-Password}
My Attempts
String loginURLPattern = TXT+"\\{([\\w-,]*)\\}&*";
System.out.println("1. ");
getListOfTemplates(loginURL, loginURLPattern);
System.out.println();
System.out.println("2. ");
getListOfTemplates(loginURL1, loginURLPattern);
System.out.println();
private static void getListOfTemplates(String inputSequence,String pattern){
System.out.println("Input URL : " + inputSequence);
Matcher templateMatcher = Pattern.compile(pattern).matcher(inputSequence);
if (templateMatcher.find() && templateMatcher.group(1).length() > 0) {
System.out.println(templateMatcher.group(1));
System.out.println("OutputURL : " + templateMatcher.replaceAll(""));
}
}
OUTPUT obtained
1.
Input URL : http://ip:port/path?username=abcd&location={LOCATION}&TXT{UE-IP,UE-Username,UE-Password}&password={PASS}
UE-IP,UE-Username,UE-Password}&password={PASS
OutputURL : http://ip:port/path?username=abcd&location={LOCATION}&
2.
Input URL : http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}&TXT{UE-IP,UE-Username,UE-Password}
UE-IP,UE-Username,UE-Password
OutputURL : http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}&
DRAWBACK OF ABOVE PATTERN
If i add any String containing character like #,%,# in between TXT{} then my code breaks.
How can i achieve it using java.util.regex library so that user can input any comma separated String between TXT{Any Comma Separated Strings}.

I would recommend using Matcher.appendReplacement:
public static void main(final String[] args) throws Exception {
final String[] loginURLs = {
"http://ip:port/path?username=abcd&location={LOCATION}&TXT{UE-IP,UE-Username,UE-Password}&password={PASS}",
"http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}&TXT{UE-IP,UE-Username,UE-Password}",
"http://ip:port/path?TXT{UE-IP,UE-Username,UE-Password}&username=abcd&location={LOCATION}&password={PASS}",
"http://ip:port/path?TXT{UE-IP,UE-Username,UE-Password}",
"http://ip:port/path?username=abcd&password={PASS}"};
final Pattern patt = Pattern.compile("(\\?)?&?(TXT\\{[^}]++})(&)?");
for (final String loginURL : loginURLs) {
System.out.printf("%1$-10s %2$s%n", "Processing", loginURL);
final StringBuffer sb = new StringBuffer();
final Matcher matcher = patt.matcher(loginURL);
while (matcher.find()) {
final String found = matcher.group(2);
System.out.printf("%1$-10s %2$s%n", "Found", found);
if (matcher.group(1) != null && matcher.group(3) != null) {
matcher.appendReplacement(sb, "$1");
} else {
matcher.appendReplacement(sb, "$3");
}
}
matcher.appendTail(sb);
System.out.printf("%1$-10s %2$s%n%n", "Processed", sb.toString());
}
}
Output:
Processing http://ip:port/path?username=abcd&location={LOCATION}&TXT{UE-IP,UE-Username,UE-Password}&password={PASS}
Found TXT{UE-IP,UE-Username,UE-Password}
Processed http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}
Processing http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}&TXT{UE-IP,UE-Username,UE-Password}
Found TXT{UE-IP,UE-Username,UE-Password}
Processed http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}
Processing http://ip:port/path?TXT{UE-IP,UE-Username,UE-Password}&username=abcd&location={LOCATION}&password={PASS}
Found TXT{UE-IP,UE-Username,UE-Password}
Processed http://ip:port/path?username=abcd&location={LOCATION}&password={PASS}
Processing http://ip:port/path?TXT{UE-IP,UE-Username,UE-Password}
Found TXT{UE-IP,UE-Username,UE-Password}
Processed http://ip:port/path
Processing http://ip:port/path?username=abcd&password={PASS}
Processed http://ip:port/path?username=abcd&password={PASS}
As you rightly point out, there are 3 possible cases:
"?{TEXT}&" -> "?"
"&{TEXT}&" -> "&"
"?{TEXT}" -> ""
So what we need to do is test for those cases in the regex. Here is the pattern:
(\\?)?&?(TXT\\{[^}]++})(&)?
Explanation:
(\\?)? optionally matches and captures a ?
&? optionally captures an &
(TXT\\{[^}]++}) matches and captures TXT, followed by {, followed by one or most not } (possessively), followed by } (closing brackets don't need to be escaped
(&)? optionally matches and captures a &
We have 3 groups:
potentially a ?
the required text
potentially an &
Now when we find a match we need to replace with the appropriate capture of case 1..3
if (matcher.group(1) != null && matcher.group(3) != null) {
matcher.appendReplacement(sb, "$1");
} else {
matcher.appendReplacement(sb, "$3");
}
If groups 1 and 3 are both present:
We must be in case 1; we must replace with "?" which is in group 1 so $1.
Otherwise we are in case 2 or 3:
In case 2 we need to replace with "&" and in 3 with "".
In case 2 group 3 will hold "&" and in case 3 it will hold "" so we can replace with $3 in both these cases.
Here I only capture the TXT{...} part using a match group. This means that although the leading ? or & is replaced it is not in the String found. I you only want the bit between {} then just move the parenthesis.
Note that I reuse the Pattern - you can also reuse the Matcher if performance is a concern. You should always reuse the Pattern as it is (very) expensive to create. Store it in a static final if you can - it's threadsafe, matchers are not. The usual way to do it is to store the Pattern in a static final and then reuse the Matcher in the context of a method.
Also, the use of Matcher.appendReplacement is much more efficient than your current approach as it only needs to process the input once. Your approach parses the string twice.

Use RegEx to extract number from coordinates

I am a beginner of Java Programming language.
When I input (1,2) into the console (brackets included), how can I write the code to extract the first and the second number using RegEx?
If there is no such expression to extract the first/second number within the brackets, I will have to change the way of inputing coordinates to x,y without the brackets and that should be a lot easier to extract numbers to be used.

Try this code:
public static void main(String[] args) {
String searchString = "(7,32)";
Pattern compile1 = Pattern.compile("\\(\\d+,");
Pattern compile2 = Pattern.compile(",\\d+\\)");
Matcher matcher1 = compile1.matcher(searchString);
Matcher matcher2 = compile2.matcher(searchString);
while (matcher1.find() && matcher2.find()) {
String group1 = matcher1.group();
String group2 = matcher2.group();
System.out.println("value 1: " + group1.substring(1, group1.length() - 1 ) + " value 2: " + group2.substring(1, group2.length() - 1 ));
}
}
Not that I think regex is the best to use here. If you know the input will be in the form of: (number, number), I would first get rid of brackets:
stringWithoutBrackets = searchString.substring(1, searchString.length()-1)
and than tokenize it with split
String[] coordiantes = stringWithoutBrackets.split(",");
Looked through Regex API and you can also do something like this:
public static void main(String[] args) {
String searchString = "(7,32)";
Pattern compile1 = Pattern.compile("(?<=\\()\\d+(?=,)");
Pattern compile2 = Pattern.compile("(?<=,)\\d+(?=\\))");
Matcher matcher1 = compile1.matcher(searchString);
Matcher matcher2 = compile2.matcher(searchString);
while (matcher1.find() && matcher2.find()) {
String group1 = matcher1.group();
String group2 = matcher2.group();
System.out.println("value 1: " + group1 + " value 2: " + group2);
}
}
The main change is that I used (?<==\)), (?=,), (?<=,), (?=\)), to search for brackets and commas but not caputre them. But I really think its an overkill for this task.