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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)));

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);

Splitting a comma-separated string but ignoring commas in whole word [closed]

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i have a string like:
String test = "firstName:a,lastName:b,addressOne:line 1,line 2,city:other";
i am trying to do test.replaceAll(",","\",\"").I mean replace , with ",". I want to this only for the whole strings. For eg addressOne is a single string with comma seperated i dont want to replace that. Is there any regex or someother way i can do this?
i should get a string like
"firstName":"a","lastName":"b","addressOne":"line 1 , line 2","city":"other"
after replace , but i am getting
"firstName":"a","lastName":"b","addressOne":"line 1 "," line 2","city":"other".

You can split your string with commas that are followed by a word which has been followed by :. And for this aim you can use a positive look-ahead and for matching the leading word use a negated character class [^:,]+ which will match any string except : and ,:
test.split(",(?=[^:,]+:)")

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("\\,",""));
}
}