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Use regular expression in C# to select a specific occurrence from a string by limiting input

Using C#, i am stuck while trying to extract a specific string while limiting the string to be matched. Here is my input string:
NPS_CNTY01_10112018_Adult_Submittal.txt
I would like to extract 01 after CNTY and ingnore anything after 01.
So far i have the regex to be:
(?!NPS_CNTY)\d{2}
But the above regex gets many other digit matches from the input string. One approach i was thinking was to limit the input to 9 characters to eventually get 01. But somehow not able to achieve that. Any help is appreciated.
I would like to add that the only variable data in this input string is:
NPS_CNTY[two digit county code excluding this bracket]_[date in MMDDYYYY format excluding the brackets]_Adult_Submittal.txt.
Also please limit solutions to regex's.

The (?!NPS_CNTY)\d{2} pattern matches a location that is not immediately followed with NPS_CNTY and then matches 2 digits. The lookahead always returns true since two digits cannot start a NPS_CNTY char sequence, it is redundant.
You may use a positive lookbehind like this to get 01:
var m = Regex.Match(s, #"(?<=NPS_CNTY)\d+");
var result = "";
if (m.Success)
{
result = m.Value;
}
See the .NET regex demo
Here, (?<=NPS_CNTY), a positive lookbehind, matches a location that is immediately preceded with NPS_CNTY and then \d+ matches 1 or more digits.
An equivalent solution using capturing mechanism is
var m = Regex.Match(s, #"NPS_CNTY(\d+)");
var result = "";
if (m.Success)
{
result = m.Groups[1].Value;
}

If the string always start with NPS_CNTY and you have to extract 2 digits then you don't need a regular expression. Just use Substring() method:
string text = #"NPS_CNTY01_01141980_Adult_Submittal.txt";
string digits = text.Substring(8, 2);
EDIT:
In case you need to match N digits after NPS_CNTY you can use the following code:
string text = #"NPS_CNTY012_01141980_Adult_Submittal.txt";
string digits = text.Replace("NPS_CNTY", string.Empty)
.Split("_", StringSplitOptions.RemoveEmptyEntries)
.FirstOrDefault();

Regex first digits occurrence

My task is extract the first digits in the following string:
GLB=VSCA|34|speed|1|
My pattern is the following:
(?x:VSCA(\|){1}(\d.))
Basically I need to extract "34", the first digits occurrence after the "VSCA". With my pattern I obtain a group but would be possibile to get only the number? this is my c# snippet:
string regex = #"(?x:VSCA(\|){1}(\d.))";
Regex rx = new Regex(regex);
string s = "GLB=VSCA|34|speed|1|";
if (rx.Match(s).Success)
{
var test = rx.Match(s).Groups[1].ToString();
}

You could match 34 (the first digits after VSCA) using a positive lookbehind (?<=VSCA\D*) to assert that what is on the left side is VSCA followed by zero or times not a digit \D* and then match one or more digits \d+:
(?<=VSCA\D*)\d+
If you need the pipe to be after VSCA the you could include that in the lookbehind:
(?<=VSCA\|)\d+
Demo

This regex pattern: (?<=VSCA\|)\d+?(?=\|) will match only the number. (If your number can be negative / have decimal places you may want to use (?<=VSCA\|).+?(?=\|) instead)

You don't need Regex for this, you can simply split on the '|' character:
string s = "GLB=VSCA|34|speed|1|";
string[] parts = s.Split('|');
if(parts.Length >= 2)
{
Console.WriteLine(parts[1]); //prints 34
}
The benefit here is that you can access all parts of the original string based on the index:
[0] - "GLB=VSCA"
[1] - "34"
[2] - "speed"
[3] - "1"
Fiddle here

While the other answers work really well, if you really must use a regular expression, or are interested in knowing how to get to that straight away you can use a named group for the number. Consider the following code:
string regex = #"(?x:VSCA(\|){1}(?<number>\d.?))";
Regex rx = new Regex(regex);
string s = "GLB:VSCA|34|speed|1|";
var match = rx.Match(s);
if(match.Success) Console.WriteLine(match.Groups["number"]);

How about (?<=VSCA\|)[0-9]+?
Try it out here

Regex to text before set of numbers

I have text like this
Inc12345_Month
Ted12345_Month
J8T12345_Month
What I need to do is extract the 12345 and also remove everything before it. This will be done in C#
.+?(?=\d_Monthly) was working in a regex tester online but when I put it in my code it only returned 5_Month.
Edit: the 12345 could be a variable length so I cannot [0-9] multiple times.
Edit2: Code this was just to try and remove everything before the 12345
string text = /* the above text pulled in from a file */;
Regex reg = new Regex(#".+?(?=\d+_Monthly)");
text = reg.Replace(string, "");

You can use this function to strip it:
private static Regex getNumberAndBeyondRegex = new Regex(^.{2}\D+(\d.*)$", RegexOptions.Compiled);
public static string GetNumberAndBeyond(string input)
{
var match = getNumberAndBeyondRegex.Match(input);
if (!match.Success) throw new ArgumentException("String isn't in the correct format.", "input");
return match.Groups[1].Value;
}
The regex at work is ^.{2}\D+(\d.*)$
It works by grabbing anything that's a number, after at least one character that isn't a number. It'll not only match _Month but also other endings.
The regex exists out of a few parts:
^ matches the beginning of the string
.{2} matches any two characters, to prevent a digit from matching if it's the first or 2nd character, you can increase this number to be equal to the minimum prefix length - 1
\D+ matches at least one character that isn't a number
( starts capturing a group
\d.* matches at least one number and any values beyond that
) closes the capturing group
$ matches the end of the string
There are a lot of different regex flavors, many of them have slight differences in terms of escaping, capturing, replacing and quite surely some others.
For testing .NET regexes online I use the free version of the tool RegexHero, it has an popup every now and then, but it makes up for that time by showing you live results, capture groups, and instant replacing. Next to having quite a lot of features.
If you want to match anywhere within the string, you can use the regex \d+_Month, it is very similiar to your original regex. In code:
new Regex("\d+_Month").Match(input).Value
Edit:
Based on the format you supplied in the comment I've created a regex and function to parse the entire file name:
private static Regex parseFileNameRegex = new Regex(#"^.*\D(\d+)_Month_([a-zA-Z]+)\.(\w+)$", RegexOptions.Compiled);
public static bool TryParseFileName(string fileName, out int id, out string month, out string fileExtension)
{
id = 0; month = null; fileExtension = null;
if (fileName == null) return false;
var match = parseFileNameRegex.Match(fileName);
if (!match.Success) return false;
if (!int.TryParse(match.Groups[1].Value, out id) || id < 1) return false; // Convert the ID into a number
month = match.Groups[2].Value;
fileExtension = match.Groups[3].Value;
return true;
}
In the parse function it requires the ID to be at least 1, 0 isn't accepted (and negative numbers won't match the regex), if you don't want this restriction, simply remove || id < 1 from the function.
Using the function would look like:
int id; string month, fileExtension;
if (!TryParseFileName("CompanyName_ClientName12345_Month_Nov.pdf", out id, out month, out fileExtension))
throw new FormatException("File name is incorrectly formatted."); // Do whatever you want when you get an invalid filename
// Use id, month and fileExtension here :)
The regex ^.*\D(\d+)_Month_([a-zA-Z]+)\.(\w+)$ works like:
^ matches the beginning of the string
.*\D matches at least one non-numeric character
(\d+) captures at least 1 number, this is the ID
_Month_ is the literal text in between
([a-zA-Z]+) matches and captures at least 1 letter, this is the month
\. matches a . character
(\w+) matches and captures any alphanumeric (letters and numbers), this is the file extension
$ matches the end of the string

Using :
Regex reg = new Regex(#"\D+(?=(\d+)_Monthly)");
is more explicit, the result is in Groups[1].

Part by part:
.+?
Match anything, maybe. This doesn't make any sense to me. It would be equivalent to ".*", which may or may not be what you meant.
(?=
start a group
\d
Match exactly 1 decimal, which explains what you are seeing, and the rest of the number is matched by .+? which is outside the group
_Monthly
match the literal text
)
end group
I think what you want is:
.*(?=\d+_Monthly)

I guess you are missing the + sign after \d
.+?(?=\d+_Monthly)
This should ask for one or more digits.

If you don't need anything before the number, this should work:
(\d+_Month)
I use Derek Slager's regex tester when I'm working with C# regex.
Better dotnet regular expression tester

Replace all characters and first 0's (zeroes)

I am trying to replace all characters inside a Regular Expression expect the number, but the number should not start with 0
How can I achieve this using Regular Expression?
I have tried multiple things like #"^([1-9]+)(0+)(\d*)"and "(?<=[1-9])0+", but those does not work
Some examples of the text could be hej:\\\\0.0.0.22, hej:22, hej:\\\\?022 and hej:\\\\?22, and the result should in all places be 22

Rather than replace, try and match against [1-9][0-9]*$ on your string. Grab the matched text.
Note that as .NET regexes match Unicode number characters if you use \d, here the regex restricts what is matched to a simple character class instead.
(note: regex assumes matches at end of line only)

According to one of your comments hej:\\\\0.011.0.022 should yield 110022. First select the relevant string part from the first non zero digit up to the last number not being zero:
([1-9].*[1-9]\d*)|[1-9]
[1-9] is the first non zero digit
.* are any number of any characters
[1-9]\d* are numbers, starting at the first non-zero digit
|[1-9] includes cases consisting of only one single non zero digit
Then remove all non digits (\D)
Match match = Regex.Match(input, #"([1-9].*[1-9]\d*)|[1-9]");
if (match.Success) {
result = Regex.Replace(match.Value, "\D", "");
} else {
result = "";
}

Use following
[1-9][0-9]*$
You don't need to do any recursion, just match that.

Here is something that you can try The87Boy you can play around with or add to the pattern as you like.
string strTargetString = #"hej:\\\\*?0222\";
string pattern = "[\\\\hej:0.?*]";
string replacement = " ";
Regex regEx = new Regex(pattern);
string newRegStr = Regex.Replace(regEx.Replace(strTargetString, replacement), #"\s+", " ");
Result from the about Example = 22

Remove alphabets from a string

I want to remove alphabets from a string. What is the best way to do it. To be more precise, i have MAC address of a system, and I want to extract only the numbers from it. I have found this article or stackoverflow. link text
I want to know, if using the regex is the best way or there are other ways to do it (maybe using LINQ).

To get the digits, you can use this regex:
var digits = Regex.Replace(text, #"\D", "");
\D matches anything that is not a digit, so removing those will give you the remaining digits.

The LINQ approach would be as follows:
string input = "12-34-56-78-9A-BC";
string result = new String(input.Where(Char.IsDigit).ToArray());
Non-LINQ / 2.0 approach:
string result = new String(Array.FindAll(input.ToCharArray(),
delegate(char c) { return Char.IsDigit(c); }));

This will replace anything that's not a number and leave you with just numbers:
string text = "abc123abc:13sdf2";
string numbers = Regex.Replace(text, #"[^\d]+", "");
Console.WriteLine(numbers);