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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.

java regex minimum character not working

^[a-zA-Z1-9][a-zA-Z1-9_\\.-]{2,64}[^\\.-]$
this is the regex that should match the following conditions
should start only with alphabets and numbers ,
contains alphabets numbers ,dot and hyphen
should not end with hyphen
it works for all conditions but when i try with three character like
vu6
111
aaa
after four characters validation is working properly did i miss anything

Reason why your Regex doesn't work:
Hope breaking it into smaller pieces will help:
^[a-zA-Z1-9][a-zA-Z1-9_\\.-]{2,64}[^\\.-]$
[a-zA-Z1-9]: Will match a single alphanumeric character ( except for _ )
[a-zA-Z1-9_\\.-]{2,64}: Will match alphanumeric character + "." + -
[^\\.-]: Will expect exactly 1 character which should not be "." or "-"
Solution:
You can use 2 simple regex:
This answer assumes that the length of the string you want to match lies between [3-65] (both inclusive)
First, that will actually validate the string
[a-zA-Z1-9][a-zA-Z1-9_\\.-]{2,64}
Second, that will check the char doesn't end with ".|-"
[^\\.-]$
In Java
Pattern pattern1 = Pattern.compile("^[a-zA-Z1-9][a-zA-Z1-9_\\.-]{2,64}$");
Pattern pattern2 = Pattern.compile("[^\\.-]$");
Matcher m1 = pattern1.matcher(input);
Matcher m2 = pattern1.matcher(input);
if(m1.find() && m2.find()) {
System.out.println("found");
}

java regex char sequence

I have a string with multiple "message" inside it. "message" starts with certain char sequence. I've tried:
String str = 'ab message1ab message2ab message3'
Pattern pattern = Pattern.compile('(?<record>ab\\p{ASCII}+(?!ab))');
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
handleMessage(matcher.group('record'))
}
but \p{ASCII}+ greedy eat everything.
Symbols a, b can be inside message only their sequence mean start of next message

p{ASCII}+ is the greedy regex for one or more ASCII characters, meaning that it will use the longest possible match. But you can use the reluctant quantifier if you want the shortest possible match: p{ASCII}+?. In that case, you should use a positive lookahead assertion.
The regex could become:
Pattern pattern = Pattern.compile("(?<record>ab\\p{ASCII}+?)(?=(ab)|\\z)");
Please note the (ab)|\z to match the last message...

Regular expression for a string starting with some string

I have some string, that has this type: (notice)Any_other_string (notes that : () has in this string`.
So, I want to separate this string to 2 part : (notice) and the rest. I do as follow :
private static final Pattern p1 = Pattern.compile("(^\\(notice\\))([a-z_A-Z1-9])+");
String content = "(notice)Stack Over_Flow 123";
Matcher m = p1.matcher(content);
System.out.println("Printing");
if (m.find()) {
System.out.println(m.group(0));
System.out.println(m.group(1));
}
I hope the result will be (notice) and Stack Over_Flow 123, but instead, the result is : (notice)Stack and (notice)
I cannot explain this result. Which regex is suitable for my purpose?

Issue 1: group(0) will always return the entire match - this is specified in the javadoc - and the actual capturing groups start from index 1. Simply replace it with the following:
System.out.println(m.group(1));
System.out.println(m.group(2));
Issue 2: You do not take spaces and other characters, such as underscores, into account (not even the digit 0). I suggest using the dot, ., for matching unknown characters. Or include \\s (whitespace) and _ into your regex. Either of the following regexes should work:
(^\\(notice\\))(.+)
(^\\(notice\\))([A-Za-z0-9_\\s]+)
Note that you need the + inside the capturing group, or it will only find the last character of the second part.

regex pattern - extract a string only if separated by a hyphen

I've looked at other questions, but they didn't lead me to an answer.
I've got this code:
Pattern p = Pattern.compile("exp_(\\d{1}-\\d)-(\\d+)");
The string I want to be matched is: exp_5-22-718
I would like to extract 5-22 and 718. I'm not too sure why it's not working What am I missing? Many thanks

Try this one:
Pattern p = Pattern.compile("exp_(\\d-\\d+)-(\\d+)");
In your original pattern you specified that second number should contain exactly one digit, so I put \d+ to match as more digits as we can.
Also I removed {1} from the first number definition as it does not add value to regexp.

If the string is always prefixed with exp_ I wouldn't use a regular expression.
I would:
replaceFirst() exp_
split() the resulting string on -
Note: This answer is based on the assumptions. I offer it as a more robust if you have multiple hyphens. However, if you need to validate the format of the digits then a regular expression may be better.

In your regexp you missed required quantifier for second digit \\d. This quantifier is + or {2}.
String yourString = "exp_5-22-718";
Matcher matcher = Pattern.compile("exp_(\\d-\\d+)-(\\d+)").matcher(yourString);
if (matcher.find()) {
System.out.println(matcher.group(1)); //prints 5-22
System.out.println(matcher.group(2)); //prints 718
}

You can use the string.split methods to do this. Check the following code.
I assume that your strings starts with "exp_".
String str = "exp_5-22-718";
if (str.contains("-")){
String newStr = str.substring(4, str.length());
String[] strings = newStr.split("-");
for (String string : strings) {
System.out.println(string);
}
}