Develop Reference

Java Regex .* usage

.*[A-Z].*
Here A-Z means any letter from Capital A to Z.
What is the purpose of using .* starting and beginning.
Please Explain above regex code

. means match any character in regular expressions. * means zero or more occurrences of the single regex preceding it.

Regex Match Reset \K Equalent In Java

I have come up with a regex pattern to match a part of a Json value. But only PRCE engine is supporting this. I want to know the Java equalent of this regex.
Simplified version
cif:\K.*(?=(.+?){4})
Matches part of the value, leaving the last 4 characters.
cif:test1234
Matched value will be test
https://regex101.com/r/xV4ZNa/1
Note: I can only define the regex and the replace text. I don't have access to the Java code since it's handle by a propriotery log masking framework.

You can write simplify the pattern to:
(?<=cif:).*(?=....)
Explanation
(?<=cif:) Positive lookbehind, assert cif: to the left
.* Match 0+ times any character without newlines
(?=....) Positive lookahead, assert 4 characters (which can include spaces)
See a regex demo.
If you don't want to match empty strings, then you can use .+ instead
(?<=cif:).+(?=....)

You can use a lookbehind assertion instead:
(?<=cif:).*(?=(.+?){4})
Demo: https://regex101.com/r/xV4ZNa/3

Expression to capture only 1 occurrence for a single character but multiple for others

I am trying to use the following regex to capture following values. This is for use in Java.
(\$|£|$|£)([ 0-9.]+)
Example values which I do want to be captured via above regex which works.
$100
$100.5
$100
$100.6
£200
£200.6
But the following as gets captured which is wrong. I only want to capture values when thereis only 1 dot in the text. Not multiples.
£200.15.
£200.6.6.6.6
Is there a way to select such that multiple periods doesn't count?
I can't do something like following cos that would affect the numbers too. Please advice.
(\$|£|$|£)([ 0-9.]{1})

You can use
(\$|£|$|£)(\d+(?:\.\d+)?)\b(?!\.)
See the regex demo.
In this regex, (\d+(?:\.\d+)?)\b(?!\.) matches
(\d+(?:\.\d+)?) - Group 1: one or more digits, then an optional occurrence of . and one or more digits
\b - a word boundary
(?!\.) - not immediately followed with a . char.
Another solution for Java (where the regex engine supports possessive quantifiers) will be
(\$|£|$|£)(\d++(?:\.\d+)?+)(?!\.)
See this regex demo. \d++ and (?:\.\d+)?+ contain ++ and ?+ possessive quantifiers that prevent backtracking into the quantified subpatterns.
In Java, do not forget to double the backslashes in the string literals:
String regex = "(\\$|£|$|£)(\\d++(?:\\.\\d+)?+)(?!\\.)";

You could try this
(\$|£|$|£)([0-9]+(?:\.[0-9]+)?)$
one or more digits followed by an optional dot and some digits and then the end of the string.
EDIT: some typos fixed
And it's not ok to delete the whole sentence obove, due to one word against my self. :(

Regex with Whitespace

I am try to write a regex to match the following:
act=MATCHME
act=Match me too
I have the following regex to match either one but not both. Here is my effort:
matches MATCHME: act=(\w+)
matches Match me too: (\w+\s\w+\s\w+)
Is there anyway to can combine the two with OR, or may I be looking at this wrong?
I am using the JAVA regex engine.

You may use an optional non-capturing group:
act=(\w+(?:\s+\w+\s+\w+)?)
^^^^^^^^^^^^^^^^^
See the regex demo
The ? matches 1 or 0 occurrences of the quantified subpattern. When it is applied to a grouping construct, the quantification is applied to the whole pattern sequence, so (?:\s+\w+\s+\w+)? matches 1 or 0 sequences of 1+ whitespaces, 1+ word chars, 1+ whitespaces and again 1+ word chars.
You may further subsegment the pattern if you need to capture 2-word substrings after act=.

Surely you know how to compose regular expressions by alternation.

This regular expression may help you
^[a-zA-Z ]*$

Java String validation only one alphanumeric with Regex

I want to do validation for a String which can only contains alphanumeric and only one special character. I tried with (\\W).{1,1}(\\w+).
But it is true only when I start with a special character. But I can have one special character at any place in String.

Use the ^ and $ anchors to instruct the regex engine to start matching from the beginning of the string and stop matching at the end of the string, so taking your regex:
^(\\W).{1,1}(\\w+)$
Please take a look at this Oracle (Java) tutorial on regular expressions.

Try this regexp: \w*\W?\w* (Java string: "\\w*\\W?\\w*")
This expression has a drawback of matching zero-length strings. If your input must have exactly one special character, remove the question mark ? from the expression.

use matcher.find() and not matcher.match() and search for \\w and remove plus (+) because it will match all alphanumeric characters sequence in your string.If your string contains only them, your regex will match whole string.

if I understand your regex correctly, this could solve your problem:
([\w]+)([^\w])([\w]+)