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use regex to map TY_111.22-L007-C010 from a text

I want to get TY_111.22-L007-C010,Tzo11-L010-C100 and Tff-L010-C110 from this string with regex
"12.5*MAX(\"TY_111.22-L007-C010\";\"Tzo11-L010-C100\";\"Tff-L010-C110\")
I tested this T.*-L\d*-C\d* but it don't give the result I want :
My code java for test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "T.*-L\\d*-C\\d*";
final String string = "\"12.5*MAX(\\\"TY_111.22-L007-C010\\\";\\\"Tzo11-L010-C100\\\";\\\"Tff-L010-C110\\\"";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}

You need to use this regex T.*?\-L\d*?\-C\d*
final String regex = "T.*?\\-L\\d*?\\-C\\d*";
Note: you need to escape the hyphens \- and use non-greedy quantifier .*? instead of .*, also you can use only matcher.group() instead of matcher.group(0), in your regex you don't have any groups, so the 0 is useless.
Outputs
Full match: TC_24.00-L010-C090
Full match: TC_24.00-L010-C100
Full match: TC_24.00-L010-C110

Why use a verbose regex pattern matcher when you can handle the problem with one line of code:
String input = "12.5*MAX(\"Txxxx-L007-C010\";\"Txxxx-L010-C100\";\"Txxxx-L010-C110\")";
String[] matches = input.replaceAll("^.*?\"|\"[^\"]*$", "")
.split("\";\"");
System.out.println(Arrays.toString(matches));
This prints:
[Txxxx-L007-C010, Txxxx-L010-C100, Txxxx-L010-C110]
OK...I used three lines of code, but the first and third are just for setting up the data and printing it.

Regex Pattern required in java for matching string starts with '{{' and ends with "}}"

Hi,
I need to create a regex pattern that will pick the matching string starts with '{{' and ends with
"}}" from a given string.
The pattern I have created is working same with the strings starting with '{{{' and '{{', Similarly with ending with '}}}' and
'}}'
Output of above code:
matches = {{phone2}}
matches = {{phone3}}
matches = {{phone5}}
**Expected Output**:
matches = {{phone5}}
I need only Strings which follows two consecutive pattern of '{' and '}' not three.
Sharing the code below
package com.test;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexTest {
public static void main(String[] args) {
String text = "<test>{{#phone1}}{{{phone3}}}{{/phone4}} {{phone5}}></test>";
//String pattern = "\\{\\{\\s*?(\\w*?)\\s*?(?!.*\\}\\}\\}$)";
String pattern = "\\{\\{\\s*?(\\w*?)\\s*?}}";
Pattern placeholderPattern = Pattern.compile(pattern);
Matcher placeholderMatcher = placeholderPattern.matcher(text);
while (placeholderMatcher.find()) {
System.out.println("matches = " + placeholderMatcher.group());
}
}
}

You may use
String pattern = "(?<!\\{)\\{{2}\\s*(\\w*)\\s*\\}{2}(?!\\})";
Or, if empty or blank {{...}} are not expected, use
String pattern = "(?<!\\{)\\{{2}\\s*(\\w+)\\s*\\}{2}(?!\\})";
See the regex demo.
Details
(?<!\{) - a negative lookbehind failing the match if there is a { char immediately to the left of the current location
\{{2} - {{ substring
\s* - 0+ whitespaces
(\w*) - Group 1: one or more word chars (1 or more if + quantifier is used)
\s* - 0+ whitespaces
\}{2} - }} string
(?!\}) - a negative lookahead that fails the match if there is a } char immediately to the right of the current location.
See the Java demo:
String text = "<test>{{#phone1}}{{{phone3}}}{{/phone4}} {{phone5}}></test>";
String pattern = "(?<!\\{)\\{{2}\\s*(\\w*)\\s*\\}{2}(?!\\})";
Pattern placeholderPattern = Pattern.compile(pattern);
Matcher placeholderMatcher = placeholderPattern.matcher(text);
while (placeholderMatcher.find()) {
System.out.println("Match: " + placeholderMatcher.group());
System.out.println("Group 1: " + placeholderMatcher.group(1));
}
Output:
Match: {{phone5}}
Group 1: phone5

Get text in the URL with dynamic date - Regex Java

I need to get the text between the URL which has a date in Java
Input 1:
/test1/raw/2019-06-11/testcustomer/usr/pqr/DATA/mn/export/
Output: testcustomer
Only /raw/ remains, date will change and testcustomer will change
Input 2:
/test3/raw/2018-09-01/newcustomer/usr/pqr/DATA/mn/export/
Output: newcustomer
String url = "/test3/raw/2018-09-01/newcustomer/usr/pqr/DATA/mn/export/";
String customer = getCustomer(url);
public String getCustomer (String _url){
String source = "default";
String regex = basePath + "/raw/\\d{4}-\\d{2}-\\d{2}/usr*";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(_url);
if (m.find()) {
source = m.group(1);
} else {
logger.error("Cant get customer with regex " + regex);
}
return source;
}
It's returning 'default' :(

Your regex /raw/\\d{4}-\\d{2}-\\d{2}/usr* is missing the part for the value you want, you need a regex that find the date, and keep what's next :
/\w*/raw/[0-9-]+/(\w+)/.* or (?<=raw\/\d{4}-\d{2}-\d{2}\/)(\w+) will be good
Pattern p = Pattern.compile("/\\w*/raw/[0-9-]+/(\\w+)/.*");
Matcher m = p.matcher(str);
if (m.find()) {
String value = m.group(1);
System.out.println(value);
}
Or if it's always the 4th part, use split()
String value = str.split("/")[4];
System.out.println(value);
And here a >> code demo

Here, we can likely use raw followed by the date as a left boundary, then we would collect our desired output in a capturing group, we would add an slash and consume the rest of our string, with an expression similar to:
.+raw\/[0-9]{4}-[0-9]{2}-[0-9]{2}\/(.+?)\/.+
Demo
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = ".+raw\\/[0-9]{4}-[0-9]{2}-[0-9]{2}\\/(.+?)\\/.+";
final String string = "/test1/raw/2019-06-11/testcustomer/usr/pqr/DATA/mn/export/\n"
+ "/test3/raw/2018-09-01/newcustomer/usr/pqr/DATA/mn/export/";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
RegEx
If this expression wasn't desired or you wish to modify it, please visit regex101.com.
RegEx Circuit
jex.im visualizes regular expressions:

Regex designing for special string

Unable to implement regex for string in line starts with ? and ends with ;
appended with single quotes or double quotes.
for example:
?abcdef;
'?abcdef;'
"?abcdef;"
I tried a lot, like this "^\\?([^;]+)\\;$" but it did not work.
Test code snippet:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexTest {
public static void main(final String[] args) {
final String text = "This is param-start ?abcdef; param-end";
final String patternString = "(['\"]?)\\?.*;\\1";
final Pattern pattern = Pattern.compile(patternString, Pattern.CASE_INSENSITIVE);
final Matcher matcher = pattern.matcher(text);
System.out.println("lookingAt = " + matcher.lookingAt());
System.out.println("matches = " + matcher.matches());
}
}

You may use this regex:
(["']?)\?.*;\1
For Java use:
final String pattern = "(['\"]?)\\?.*;\\1";
RegEx Demo
(["']?) matches an optional ' or " and captures in group #1
\1 is back-reference for the same value as in group #1

You may use
final String text = "This is param-start ?abcdef; param-end";
final String patternString = "(['\"]?)\\?.*?;\\1";
final Pattern pattern = Pattern.compile(patternString, Pattern.DOTALL);
final Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
System.out.println("Match found: " + matcher.group());
}
// => Match found: = ?abcdef;
See the Java demo and the regex demo. Regulex graph:
Notes:
while (matcher.find()) - iterates through all potential matches in the string
matcher.group() - accesses the match value.
Regex
(['"]?) - Capturing group 1: either ' or "
\? - a ? char
.*? - any 0+ chars, as few as possible (the pattern is compiled with Pattern.DOTALL, so it will match across line breaks, too)
; - a semi-colon
\1 - the same value as captured in Group 1.

Your description stated "starts with ? and ends with ;", but your sample string does not end with a semicolon - it's ending with wtih a double quote.
If you want to allow the string to start or end with either a single quote or double quote, your match should be:
^['"]?\?[^;]+;\['"]?$

How to use REGEX to validate EditText?

I want to validate a username against these requirements:
Just accept character or digital
At least one character
I tried with
public boolean validateFormat(String input){
return Pattern.compile("^[A-Za-z0-9]+$").matcher(input).matches();
}
How can I do this one?

Try with this regex:
^(\w|\d)+$
^ indicates the start of the string
$ indicates the end of the string
\w means any word character
\d means any digit
| is the logical OR operator
Anyway, i suggest you to use an online regex tester like regex101.com .It is very helpful to quickly test regular expressions.
Hope it can help!
== UPDATE ==
In Java code:
final String regex = "^(\\w|\\d)+$";
final String string = "myCoolUsername12";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
if(matcher.matches()) {
// if you are interested only in matching the full regex
}
// Otherwise, you can iterate over the matched groups (including the full match)
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}

/^[A-Za-z0-9]+(?:[ _-][A-Za-z0-9]+)*$/