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Regex - Find all the digits that occur to a certain character [closed]

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I would like to ask you about regex expression - I need to get all numbers that occur to a certain character. For example:
"$z4~min.~00~s" -> 4
"$z12~min.~00~s" -> 12
I simply need first number in the string, I don't need numbers after dot in the string.
I am using Java for this project.
Do you have any suggestions? Thanks a lot.

java.util.regex.Pattern pattern = java.util.regex.Pattern.compile("^\\D*(\\d+)");
java.util.regex.Matcher matcher = pattern.matcher("$z12~min.~00~s");
if (matcher.find()) {
String firstNumber = matcher.group(1);
System.out.println(firstNumber);
}

RegEx to extract/group the all values which comes after colon from below string in Java [closed]

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I've got a string like this:
1454974419:1234;1454974448:3255,2255,66789
I would like to extract/group these values by using regular expression in java.
1234
3255225566789

You can use this lookbehind and negation based regex:
(?<=[:,])[^;,]+
RegEx Demo
Breakup:
(?<=[:,]) # lookbehind to assert if previous char is : or ,
[^;,]+ # match 1 or more of anything that is not a ; or ,

Try this
String yourString= "1454974419:1234;1454974448:3255,2255,66789";
Pattern myPattern = Pattern.compile("[^a-zA-Z0-9]");
Matcher myMatcher = myPattern.matcher(yourString);
while(myMatcher.find())
{
String temp= myMatcher.group();
yourString=yourString.replaceAll("\\"+temp, "");
}
System.out.println(yourString);

Is there a way to use \p{Punct}\p{Lower}\p{Upper} and in a regex(java), but without the "." character? [closed]

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I need to take out all the characters of a string that are not numbers.

You could use a Character Class Intersection like [\p{Punct}\p{Lower}\p{Upper}&&[^.]]
But why not just use
[^\d.]+
As Java String "[^\\d.]+"
This would match one or more characters, that are not \d a digit or the . period.

I'd suggest using \\d+ then (it's consecutive digits), and a capture group. Something like
String str = "";
str = str.replaceAll("(\\d+\\.\\d+)", "$1");

java regexp for reading number after certain symbol [closed]

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I have the following string
String names= D:45454546544654 A:45454545454 C:454545474
I need Output to be
String data[]=[45454546544654,45454545454,454545474]

First replace all the uppercase letter plus the following colon : with an empty string and then split the resultant string according to the spaces.
String names = "D:45454546544654 A:45454545454 C:454545474";
String parts[] = names.replaceAll("[A-Z]:", "").split("\\s+");
System.out.println(Arrays.toString(parts));
Output:
[45454546544654, 45454545454, 454545474]

Rather than split you can just match:
(?<=:)\d+(?!\d)
Using this Pattern:
Pattern p = Pattern.compile("(?<=:)\\d+(?!\\d)");
Then you can use Matcher.find API to get all the matches.
RegEx Demo

Pattern matching for validating the input [closed]

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I have to validate the input as
Asterisks are permitted in positions 2-5.
Position One should be alphabetic (except for the ~)
-No characters should accept numbers.
Other than the exceptions mentioned above, no special characters are allowed
I am trying to build as this.
final Pattern pattern =
Pattern.compile("^[a-zA-Z~][a-zA-Z*]*$", Pattern.CASE_INSENSITIVE);
final Matcher matcher = pattern.matcher(this.mainStaOrgBO.getStaOrgCode());
final boolean specialCharCheck = matcher.find();
if (specialCharCheck) {
}

How about:
^[a-zA-Z~][a-zA-Z*]{1,4}[a-zA-Z]*$
Explanation:
^ : start of string
[a-zA-Z~] : First char can be letter or ~
[a-zA-Z*]{1,4} : char 2 to 5 can be letter or *
[a-zA-Z]* : rest of string only letter
$ : end of string.

This should work
[a-zA-Z~]\*[a-zA-Z~]{2}\*[a-zA-Z~]*
If the * are optional
[a-zA-Z~][a-zA-Z~\*][a-zA-Z~]{2}[a-zA-Z~\*][a-zA-Z~]*
Test is here Online Java Regex Test