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

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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);
}

Related

Find last letter in String using Regex [closed]

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I'm trying to find the regex to get the last letter in a string:
String str = "A76B62Z**F**63";
Finding last letter via regex should return 'F'.

You can greedily match any sequence of characters before a letter:
String s = "A76B62ZF63";
Matcher m = Pattern.compile(".*([A-Za-z])").matcher(s);
if(m.find()) System.out.println(m.group(1));
With Java 9+:
String s = "A76B62ZF63";
Pattern.compile(".*([A-Za-z])").matcher(s).results()
.findFirst().ifPresent(r -> System.out.println(r.group(1)));

Regular Expression to match first character and after a character [closed]

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I have the following string:
___abcd-metadata.json
I am trying to do a regex to get ___ and everything after -. So far I have this Regex:
(\-.*?\.json)
To find the last part and (___) to find the first part, but I have not been able to figure how to combine both regexes to make this happen.
My desired result would be ___ -metadata.json = true

"(\-)(.*)" is the pattern, which will give you 2 matches if the string contains them.

Take your string and use String.split('-') to break it into parts.
EDIT
var parts = ('_abcd-metadata.json').split('-'),
start = parts[0].substr(0,3),
rest = parts[1];

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");

What would be regular expression of finding all strings starts with $ in java [closed]

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I am having one string containing "This is a time to get involve into $FO, $RTP, $DFG and $RG"

Use following regular expression:
"\\$\\w+"
$ should be escaped.
\w match digits, alphabet, _.
If you need only match alphabets, use [a-zA-Z] instead.

This will work too
String str="This is a time to get involve into $FO, $RTP, $DFG and $RG" ;
String[] arr=str.split(" ");
for (String i:arr){
if(i.indexOf("$")==0){
System.out.println(i.replaceAll("\\,",""));
}
}

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