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Use regex to get 2 specific groups of substring

String s = #Section250342,Main,First/HS/12345/Jack/M,2000 10.00,
#Section250322,Main,First/HS/12345/Aaron/N,2000 17.00,
#Section250399,Main,First/HS/12345/Jimmy/N,2000 12.00,
#Section251234,Main,First/HS/12345/Jack/M,2000 11.00
Wherever there is the word /Jack/M in the3 string, I want to pull the section numbers(250342,251234) and the values(10.00,11.00) associated with it using regex each time.
I tried something like this https://regex101.com/r/4te0Lg/1 but it is still messed.
.Section(\d+(?:\.\d+)?).*/Jack/M

If the only parts of each section that change are the section number, the name of the person and the last value (like in your example) then you can make a pattern very easily by using one of the sections where Jack appears and replacing the numbers you want by capturing groups.
Example:
#Section250342,Main,First/HS/12345/Jack/M,2000 10.00
becomes,
#Section(\d+),Main,First/HS/12345/Jack/M,2000 (\d+.\d{2})
If the section substring keeps the format but the other parts of it may change then just replace the rest like this:
#Section(\d+),\w+,(?:\w+/)*Jack/M,\d+ (\d+.\d{2})
I'm assuming that "Main" is a class, "First/HS/..." is a path and that the last value always has 2 and only 2 decimal places.
\d - A digit: [0-9]
\w - A word character: [a-zA-Z_0-9]
+ - one or more times
* - zero or more times
{2} - exactly 2 times
() - a capturing group
(?:) - a non-capturing group
For reference see: https://docs.oracle.com/en/java/javase/18/docs/api/java.base/java/util/regex/Pattern.html
Simple Java example on how to get the values from the capturing groups using java.util.regex.Pattern and java.util.regex.Matcher
import java.util.regex.*;
public class GetMatch {
public static void main(String[] args) {
String s = "#Section250342,Main,First/HS/12345/Jack/M,2000 10.00,#Section250322,Main,First/HS/12345/Aaron/N,2000 17.00,#Section250399,Main,First/HS/12345/Jimmy/N,2000 12.00,#Section251234,Main,First/HS/12345/Jack/M,2000 11.00";
Pattern p = Pattern.compile("#Section(\\d+),\\w+,(?:\\w+/)*Jack/M,\\d+ (\\d+.\\d{2})");
Matcher m;
String[] tokens = s.split(",(?=#)"); //split the sections into different strings
for(String t : tokens) //checks every string that we got with the split
{
m = p.matcher(t);
if(m.matches()) //if the string matches the pattern then print the capturing groups
System.out.printf("Section: %s, Value: %s\n", m.group(1), m.group(2));
}
}
}

You could use 2 capture groups, and use a tempered greedy token approach to not cross #Section followed by a digit.
#Section(\d+)(?:(?!#Section\d).)*\bJack/M,\d+\h+(\d+(?:\.\d+)?)\b
Explanation
#Section(\d+) Match #Section and capture 1+ digits in group 1
(?:(?!#Section\d).)* Match any character if not directly followed by #Section and a digit
\bJack/M, Match the word Jack and /M,
\d+\h+ Match 1+ digits and 1+ spaces
(\d+(?:\.\d+)?) Capture group 2, match 1+ digits and an optional decimal part
\b A word boundary
Regex demo
In Java:
String regex = "#Section(\\d+)(?:(?!#Section\\d).)*\\bJack/M,\\d+\\h+(\\d+(?:\\.\\d+)?)\\b";

Regex to mask multiple phone numbers (~) separated except last 4 digiits

I am trying to find a regex which masks phone numbers except last 4 digits.
example: phone=9988998888~7654321908~6789054321
Desired output : phone=******8888~******1908~*****4321
I tried below regex but it is masking only starting number
phone=******8888~7654321908~6789054321
^(phone)=(\d(?=\d{4}))*

Use replaceAll​(Function<MatchResult,​String> replacer) to replace each digit in MatchResult with "*".
public class PhoneNumberMask {
public static void main(String[] args) {
String target = "phone=9988998888~7654321908~6789054321";
Pattern pattern = Pattern.compile("(\\d+(?=\\d{4}))");
Matcher matcher = pattern.matcher(target);
String result = matcher.replaceAll((matchResult) -> matchResult.group(1).replaceAll("\\d", "*"));
System.out.println(result);
}
}

You could use:
\d(?=\d{4})
See this online demo
\d - Any single digit.
(?=\d{4}) - Positive lookahead for 4 digits.
Replace with *.
See a Java demo

Assuming you only want to mask all numbers in a string that starts with phone= separated with ~, you can use a plain regex solution without a lambda in the replacement with
String masked = text.replaceAll("(\\G(?!^)(?:\\d{4}~)?|^phone=)\\d(?=\\d{4})", "$1*");
See the regex demo. Details:
(\G(?!^)(?:\d{4}~)?|^phone=) - Group 1: end of the previous successful match and then an optional sequence of four digits and a ~ or start of string and phone=
\d - a digit
(?=\d{4}) - followed with any four digits.

REGEX extract two double number separated from hypen

I have strings like:
some foo text
some foo
1-2
1.00-2.00
3.21-1.23
2.12-2.12
I have to check if the string format contains two numbers separated by hyphen.
How can I do it?
Thanks

Regex for float is: ^[1-9]\d*\.\d+$ if decimals are optional : ^[1-9]\d*(?:\.\d+)?$
Repeat it twice with hyphen in between:
`^[1-9]\d*(?:\.\d+)?-[1-9]\d*(?:\.\d+)?$`

You can use the regex:
^\d+(\.\d+)?-\d+(\.\d+)?$
Explanation can be found here.
Using java you can create a method that checks whether your desired pattern exists or not:
public static boolean returnMatch(String input) {
Pattern p1 = Pattern.compile("^\\d+(\\.\\d+)?-\\d+(\\.\\d+)?$");
Matcher m1 = p1.matcher(input);
return m1.find() ? true : false;
}
Now call it using:
System.out.println(returnMatch("some foo text")); // false
System.out.println(returnMatch("1.00-2.00")); // true
System.out.println(returnMatch("2.12-2.12")); // true
System.out.println(returnMatch("10-20")); // true

Use a simple Regex:
(\d+(?:\.\d+)?)-(\d+(?:\.\d+)?)
This solution assumes there is always a decimal part present (at least one digit). Demo at Regex101.
\d is a digit
\d+ is at least one digit
\. matches a dot (.) literally
() is a capturing group
(?:\.\d+)? is a non-capturing group which optionally matches the decimal part
Don't forget the proper escaping in Java String regex = "(\\d+(?:\\.\\d+)?)-(\\d+(?:\\.\\d+)?)";
In case one or more spaced or blank characters appear between the dash and numbers, use:
(\d+(?:\.\d+)?)\s*-\s*(\d+(?:\.\d+)?)

Mask mobile number in Java [duplicate]

I would like to mask the last 4 digits of the identity number (hkid)
A123456(7) -> A123***(*)
I can do this by below:
hkid.replaceAll("\\d{3}\\(\\d\\)", "***(*)")
However, can my regular expression really can match the last 4 digit and replace by "*"?
hkid.replaceAll(regex, "*")
Please help, thanks.
Jessie

Personally, I wouldn't do it with regular expressions:
char[] cs = hkid.toCharArray();
for (int i = cs.length - 1, d = 0; i >= 0 && d < 4; --i) {
if (Character.isDigit(cs[i])) {
cs[i] = '*';
++d;
}
}
String masked = new String(cs);
This goes from the end of the string, looking for digit characters, which it replaces with a *. Once it's found 4 (or reaches the start of the string), it stops iterating, and builds a new string.

While I agree that a non-regex solution is probably the simplest and fastest, here's a regex to catch the last 4 digits independent if there is a grouping ot not: \d(?=(?:\D*\d){0,3}\D*$)
This expression is meant to match any digit that is followed by 0 to 3 digits before hitting the end of the input.
A short breakdown of the expression:
\d matches a single digit
\D matches a single non-digit
(?=...) is a positive look-ahead that contributes to the match but isn't consumed
(?:...){0,3} is a non-capturing group with a quantity of 0 to 3 occurences given.
$ matches the end of the input
So you could read the expression as follows: "match a single digit if it is followed by a sequence of 0 to 3 times any number of non-digits which are followed by a single digit and that sequence is followed by any number of non-digits and the end of the input" (sounds complicated, no?).
Some results when using input.replaceAll( "\\d(?=(?:\\D*\\d){0,3}\\D*$)", "*" ):
input = "A1234567" -> output = "A123****"
input = "A123456(7)" -> output = "A123***(*)"
input = "A12345(67)" -> output = "A123**(**)"
input = "A1(234567)" -> output = "A1(23****)"
input = "A1234B567" -> output = "A123*B***"
As you can see in the last example the expression will match digits only. If you want to match letters as well either replace \d and \D with \w and \W (note that \w matches underscores as well) or use custom character classes, e.g. [02468] and [^02468] to match even digits only.

Writing regex for string containing no only numbers

I need to write a regex containing not only digits [0-9]. How can I do that without explicitly specifying all possible charaters in a group. Is it possible to do through lookahead/lookbehind? Examples:
034987694 - doesn't match
23984576s9879 - match
rtfsdbhkjdfg - match
=-0io[-09uhidkbf - match
9347659837564983467 - doesn't match

^(?!\\d+$).*$
This should do it for you.See demo.
https://regex101.com/r/fM9lY3/1
The negative will lookahead will check if the string doesnt have integers from start to end.You need $ to make sure the check is till end or else it will just check at the start.

If you just need to detect whether the string is not numbers-only, then you can simply test for /\D/ - "succeed if there is a non-digit anywhere".

Why not check if it only contains digits, if not it matches
String[] strings = {"034987694", "23984576s9879",
"rtfsdbhkjdfg",
"=-0io[-09uhidkbf",
"9347659837564983467"};
for (String s : strings) {
System.out.printf("%s = %s%n", s, !s.matches("\\d*"));
}
output
034987694 = false
23984576s9879 = true
rtfsdbhkjdfg = true
=-0io[-09uhidkbf = true
9347659837564983467 = false

You may try the below,
string.matches(".*\\D.*");
This expects atleast 1 non-digit character.