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C# Regex match only if one group of numbers is found in string

I'm trying to come up with a regex to deal with a general scenario to capture a number from a string where the number may have one or more non-numeric characters pre/post-fixed to it.
The number can contain zero or one decimal or comma.
If the string contains multiple "sets" of consecutive digits separated by non-digits I would like the regex to fail ("sets" is probably not the correct terminology).
As an example the following inputs would succeed with a match:
abc12.00xyz would match 12.00
0.1$ would be valid and match 0.1
.01 would be valid and match .01
123abc would be valid and match 123
abc123 would be valid and match 123
These inputs would fail to match:
abc12.00xyz322 would fail due to the second "set" of digits, 322 in this example
12t2 would fail due to having two separate "sets" of digits
I've tried many permutations and I'm not making much headway. This is the closest I've come to so far. It matches on the numbers correctly, excluding the non-digits from the match, but it includes all "sets" of numbers in the string.
([\d]*[.,])?[\d]+
Any suggestions would be appreciated.

You can use a capture group:
^[^0-9\r\n]*?([0-9]*\.?[0-9]+)[^0-9\r\n]*$
^ Start of string
[^0-9\r\n]* Optionally match any char except a digit or a newline, as few as possible
([0-9]*\.?[0-9]+) Capture group 1, match optional digits, optional comma and 1+ digits
[^0-9\r\n]* Optionally match any char except a digit or a newline
$ End of string
See a .NET regex demo (Click on the Table tab to see the capture group values)

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 match certain amount of character and allow space inbetween

I am currently working on a regex which needs to match exactly 8 digits. But sometimes it occurs that there are spaces or dots between those numbers. This is the regex that i am currently using.
([0-9\ ?.?]{7,16})
It works fine most of the time, but the problem I am having is that it sometimes matches number with a lot of spaces tailing it so you will get something like 1234/s/s/s/s (/s stands for space) Or sometimes it is only matching spaces.
What i want is a regex that always matches at least 8 digits and also allows spaces and dots without detecting less then 8 digits. I know it may be stupid question, but I couldn't find anything I need elswhere.

Your ([0-9\ ?.?]{7,16}) expression matches 7 to 16 occurrences of any character that is either a digit, or a space, or a ?, or .. Yes, the ? inside [...] is a literal ?, not a quantifier.
You need to use an expression that will match a digit ([0-9]) and then exactly 7 sequences of a space or period ([ .]) followed with 1 digit, and to make sure you are not matching the digits in 123.156.78.146 you may use special boundaries:
(?<!\d[ .]?)\d(?:[. ]?\d){7}(?![ .]?\d)
if the space or . can only be 0 to 1 in between digits; or - if the space/dot can appear 0 or more times,
(?<!\d[ .]*)\d(?:[. ]*\d){7}(?![ .]*\d)
See the regex demo
The (?<!\d[ .]*) is a negative lookbehind that will fail any match if it starts with a digit that is followed with .(s) or space(s), and the (?![ .]*\d) negative lookahead will fail the match if the 7 digits you need are followed with .(s) or space(s) and a digit.

To solve this, describe the problem to yourself. You want to match one digit followed by seven repetitions of space-or-dot followed by a digit. This leads to a regular expression like \d([ .]?\d){7}. To avoid collecting the seven captures add a :? after the (. To capture the whole string, enclose it in brackets. Adding both changes gives the expression (\d(:?[ .]?\d){7}). If more than one space or dot is allowed between the digits then change the ? to a *.
To get just the eight digits out of the string I suggest using the string captured above and replacing any spaces or dots with nothing.

Get a number with exactly x digits from string

im looking for a regex pattern, which matches a number with a length of exactly x (say x is 2-4) and nothing else.
Examples:
"foo.bar 123 456789", "foo.bar 456789 123", " 123", "foo.bar123 " has to match only "123"
So. Only digits, no spaces, letters or other stuff.
How do I have to do it?
EDIT: I want to use the Regex.Matches() function in c# to extract this 2-4 digit number and use it in additional code.

Any pattern followed by a {m,n} allows the pattern to occur m to n times. So in your case \d{m,n} for required values of m and n. If it has to be exactly an integer, use\d{m}
If you want to match 123 in x123y and not in 1234, use \d{3}(?=\D|$)(?<=(\D|^)\d{3})
It has a look ahead to ensure the character following the 3 digits is a non-digitornothing at all and a look behind to ensure that the character before the 3 digits is a non-digit or nothing at all.

You can achieve this with basic RegEx:
\b(\d\d\d)\b or \b(\d{3})\b - for matching a number with exactly 3 digits
If you want variable digits: \b(\d{2,4})\b (explained demo here)
If you want to capture matches next to words: \D(\d{2,4})\D (explained demo here)
\b is a word boundary (does not match anything, it's a zero-match character)
\d matches only digits
\D matches any character that is NOT a digit
() everything in round brackets will capture a match