Develop Reference

Regex C# - optional group in the middle

I have such source text with an optional group in the middle:
GH22-O0-TFS-SFSD 00-1-006.19135
GH22-O0-TFS-SFSD 00-1-006.1.19135
Desired value in the first case will be '19135' and in the second '1.19135'.
Regex has to match the whole string and select all characters after first "." - which is my Group 1. I tried to make subgroups and mark Group 3 as optional but it is not working.
Regex:
.*\.0*(([0-9])(\.0*([0-9]+)))
How it should be changed to capture desired values?

This should work for you:
.*?\.(.*)
This will match the whole string and include everything after the first period in capture group 1 regardless of character type.

You can use
^(.*?)\.0*(\d+)(?:\.0*(\d+))?$
See the regex demo. Details:
^ - start of string
(.*?) - Group 1: any zero or more chars other than an LF char as few as possoble (as *? is a lazy quantifier)
\. - a dot
0* - zero or more zeros
(\d+) - Group 2: any one or more digits
(?:\.0*(\d+))? - an optional occurrence of ., zero or more zeros, then Group 3 capturing one or more digits
$ - end of string.

I hope I understand your goals and this should work:
.*?\.([\d.]+)
.*?\. - loosely capture everything leading up to the first period
([\d.]+) - capture the remaining digits and periods into capture group #1
https://regex101.com/r/0t9Ijy/1

Regex to match 7 same digits in a number regardless of position

I want to match an 8 digit number. Currently, I have the following regex but It is failing in some cases.
(\d+)\1{6}
It matches only when a number is different at the end such as 44444445 or 54444444. However, I am looking to match cases where at least 7 digits are the same regardless of their position.
It is failing in cases like
44454444
44544444
44444544
What modification is needed here?

It's probably a bad idea to use this in a performance-sensitive location, but you can use a capture reference to achieve this.
The Regex you need is as follows:
(\d)(?:.*?\1){6}
Breaking it down:
(\d) Capture group of any single digit
.*? means match any character, zero or more times, lazily
\1 means match the first capture group
We enclose that in a non-capturing group {?:
And add a quantifier {6} to match six times

You can sort the digits before matching
string input = "44444445 54444444 44454444 44544444 44444544";
string[] numbers = input.Split(' ');
foreach (var number in numbers)
{
number = String.Concat(str.OrderBy(c => c));
if (Regex.IsMatch(number, #"(\d+)\1{6}"))
// do something
}
Still not a good idea to use regex for this though

The pattern that you tried (\d+)\1{6} matches 6 of the same digits in a row. If you want to stretch the match over multiple same digits, you have to match optional digits in between.
Note that in .NET \d matches more digits than 0-9 only.
If you want to match only digits 0-9 using C# without matching other characters in between the digits:
([0-9])(?:[0-9]*?\1){6}
The pattern matches:
([0-9]) Capture group 1
(?: Non capture group
[0-9]*?\1 Match optional digits 0-9 and a backreference to group 1
){6} Close non capture group and repeat 6 times
See a .NET Regex demo
If you want to match only 8 digits, you can use a positive lookahead (?= to assert 8 digits and word boundaries \b
\b(?=\d{8}\b)[0-9]*([0-9])(?:[0-9]*?\1){6}\d*\b
See another .NET Regex demo

C# regex string that is not another string

I want to match an at least 3 letter word, preceded by any character from class [-_ :] any amount of times, that is not this specific 3 letter word string2.
Ex:
if string2="VER"
in
" ODO VER7"
matched " ODO"
or
"_::ATTPQ VER7"
matched "_::ATTPQ"
but if
" VER7"
it shoudn't match " VER"
so I thought about
Regex.Match(inputString, #"[-_:]*[A-Z]{3,}[^(VER)]", RegexOptions.IgnoreCase);
where
[-_:]* checks for any character in class, appearing 0 or more times
[A-Z] the range of letters that could form the word
{3,} the minimum amount of letters to form the word
[^(VER)] the grouping construct that shouldn't appear
I believe however that [A-Z]{3,} results in any letter at least 3 times (not what i want)
and [^(VER)] not sure what it's doing

Using [^(VER)] means a negated character class where you would match any character except ( ) V E or R
For you example data, you could match 0+ spaces or tabs (or use \s to also match a newline).
Then use a negative lookahead before matching 3 or more times A-Z to assert what is on the right is not VER.
If that is the case, match 3 or more times A-Z followed by a space and VER itself.
^[ \t]*[-_:]*(?!VER)[A-Z]{3,} VER
Regex demo

^\s*[-_:]*(?!VER)[A-Z]{3,}
This regex asserts that between the start and end of the string, there's zero or more of your characters, followed by at least 3 letters. It uses a negative lookahead to make sure that VER (or whatever you want) is not present.
Demo

This would match the preceding class characters [-_ :] of 3 or more letters/numbers
that do not start with VER (as in the samples given) :
[-_ :]+(?!VER)[^\W_]{3,}
https://regex101.com/r/wLw23I/1

Regex to take first set after Space and want to remove $ with same regex

My input string:-
" $440,765.12 12-108(e)\n3 "
Output String i want as:-
"440,765.12"
I have tried with below regex and it's working but I am not able to remove $ with the same regex so anyone knows how to do the same task with below regex?
Regex rx = new Regex(#"^(.*? .*?) ");
var match = rx.Match(" $440,765.12 12-108(e)\n3 ");
var text = match.Groups[1].Value;
output after using above regex:-
$440,765.12
I know I can do the same task using string.replace function but I want to do the same with regex only.

You may use
var result = Regex.Match(s, #"\$(\d[\d,.]*)")?.Groups[1].Value;
See the regex demo:
Details
\$ - matches a $ char
(\d[\d,.]*) - captures into Group 1 ($1) a digit and then any 0 or more digits, , or . chars.
If you want a more "precise" pattern (just in case the match may appear within some irrelevant dots or commas), you may use
\$(\d{1,3}(?:,\d{3})*(?:\.\d+)?)
See this regex demo. Here, \d{1,3}(?:,\d{3})*(?:\.\d+)? matches 1, 2 or 3 digits followed with 0 or more repetitions of , and 3 digits, followed with an optional sequence of a . char and 1 or more digits.
Also, if there can be any currency symol other than $ replace \$ with \p{Sc} Unicode category that matches any currency symbol:
\p{Sc}(\d{1,3}(?:,\d{3})*(?:\.\d+)?)
See yet another regex demo.

Regex matching recurring patterns

I want to validate input in a C# TextBox by using regular expressions. The expected input is in this format:
CCCCC-CCCCC-CCCCC-CCCCC-CCCCC-CCCCC-C
So I've got six elements of five separated characters and one separated character at the end.
Now my regex matches any character between five and 255 chars: .{5,255}
How do I need to modify it in order to match the format mentioned above?

Update: -
If you want to match any character, then you can use: -
^(?:[a-zA-Z0-9]{5}-){6}[a-zA-Z0-9]$
Explanation: -
(?: // Non-capturing group
[a-zA-Z0-9]{5} // Match any character or digit of length 5
- // Followed by a `-`
){6} // Match the pattern 6 times (ABCD4-) -> 6 times
[a-zA-Z0-9] // At the end match any character or digit.
Note: - The below regex will only match pattern like you posted: -
CCCCC-CCCCC-CCCCC-CCCCC-CCCCC-CCCCC-C
You can try this regex: -
^(?:([a-zA-Z0-9])\1{4}-){6}\1$
Explanation: -
(?: // Non-capturing group
( // First capture group
[a-zA-Z0-9] // Match any character or digit, and capture in group 1
)
\1{4} // Match the same character as in group 1 - 4 times
- // Followed by a `-`
){6} // Match the pattern 6 times (CCCCC-) -> 6 times
\1 // At the end match a single character.

Untested, but I think this will work:
([A-Za-z0-9]{5}-){6}[A-Za-z0-9]

For your example, in general replace C to the character class you want:
^(C{5}-){6}C$
^([a-z]{5}-){6}[a-z]$ # Just letter, use case insensitive modifier
^([a-z0-9]{5}-){6}[a-z0-9]$ # Letters and digits..

Try this:
^(C{5}-){6}C$
The ^ and $ denote the begiining and end of the string repectively and make sure that no additional characters are entered.