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Regex for allowing semi colon

I have a regex for validating a string but it doesn't accept semicolons? Is it because I have to use some escape sequences? I tested my regex here and it passes i.e allows semi-colon but doesn't allow in my c# app.
EDITED I have following regex
^[A-Za-z0-9]{1}[A-Za-z.&0-9\s\\-]{0,21}$
And tried validating sar232 trading inc;

The & entity hints at the fact you have this regular expression inside some XML attribute, and that this & gets parsed as a single & symbol when the pattern is sent to the regex engine.
That means, your pattern lacks the semi-colon inside the second character class, and that is why your regex does not match the string you provided.
The solution is simple: add the semi-colon to the 2nd character class:
someattr="^[A-Za-z0-9][;A-Za-z.&0-9\s\\-]{0,21}$"
^
See the regex demo
Please also note that the {1} limiting quantifier is redundant since a [A-Za-z0-9] already matches only 1 symbol from the indicated ranges.

Extract string from a pattern preceded by any length

I'm looking for a regular expression to extract a string from a file name
eg if filename format is "anythingatallanylength_123_TESTNAME.docx", I'm interested in extracting "TESTNAME" ... probably fixed length of 8. (btw, 123 can be any three digit number)
I think I can use regex match ...
".*_[0-9][0-9][0-9]_[A-Z][A-Z][A-Z][A-Z][A-Z][A-Z][A-Z][A-Z].docx$"
However this matches the whole thing. How can I just get "TESTNAME"?
Thanks

Use parenthesis to match a specific piece of the whole regex.
You can also use the curly braces to specify counts of matching characters, and \d for [0-9].
In C#:
var myRegex = new Regex(#"*._\d{3}_([A-Za-z]{8})\.docx$");
Now "TESTNAME" or whatever your 8 letter piece is will be found in the captures collection of your regex after using it.
Also note, there will be a performance overhead for look-ahead and look-behind, as presented in some other solutions.

You can use a look-behind and a look-ahead to check parts without matching them:
(?<=_[0-9]{3}_)[A-Z]{8}(?=\.docx$)
Note that this is case-sensitive, you may want to use other character classes and/or quantifiers to fit your exact pattern.

In your file name format "anythingatallanylength_123_TESTNAME.docx", the pattern you are trying to match is a string before .docx and the underscore _. Keeping the thing in mind that any _ before doesn't get matched I came up with following solution.
Regex: (?<=_)[A-Za-z]*(?=\.docx$)
Flags used:
g global search
m multi-line search.
Explanation:
(?<=_) checks if there is an underscore before the file name.
(?=\.docx$) checks for extension at the end.
[A-Za-z]* checks the required match.
Regex101 Demo

Thanks to #Lucero #noob #JamesFaix I came up with ...
#"(?<=.*[0-9]{3})[A-Z]{8}(?=.docx$)"
So a look behind (in brackets, starting with ?<=) for anything (ie zero or more any char (denoted by "." ) followed by an underscore, followed by thee numerics, followed by underscore. Thats the end of the look behind. Now to match what I need (eight letters). Finally, the look ahead (in brackets, starting with ?=), which is the .docx
Nice work, fellas. Thunderbirds are go.

C# Regular Expression for String matching

I am looking for a regular expression that returns success only if the input string contains following characters:
a-zA-Z0-9~!#$^ ()_-+’:.?
Is this regular expression correct?
^[a-zA-Z0-9~!#$^ ()_-+’:.?]+$
I have understood what ^ means here but not sure about +$. Also are there any alternatives to this? By the way the above regular expression also includes a space character between ^ and (

it only contains the characters listed above
bool invalidCharsExist =
Regex.Replace(input, #"[a-zA-Z0-9~!#\$\^\ \(\)_\-\+’:\.\?]", "").Length != 0;
BTW: This is not fully equivalent to your regex (It will also include non-ascii letters and digits) but I think it is a better way to check
var specialChars = new HashSet<char>("~!#$^ ()_-+’:.?");
var allValid = input.All(c => char.IsLetterOrDigit(c) || specialChars.Contains(c));

Close, but get rid of that dash in the middle of your character class and put it at the beginning:
^[-a-zA-Z0-9~!#$^ ()_+’:.?]+$
And make sure when you put it in a string that you use the proper string qualifier (I forget what it's called):
#"^[-a-zA-Z0-9~!#$^ ()_+’:.?]+$"
As to whether or not you can do it in other ways, sure, for example a negative look-ahead that doesn't actually match anything. I don't think a proper regex optimizer would leave one better than the other, it's just a matter of preference. Do you want something that looks to succeed (selects the entire string if valid), or something that looks to fail (negative look-ahead).
Honestly if performance is at all important, you should write a good old for and loop over the characters (or the equivalent LINQ implementation). Regex won't even be in the ballpark.

the regular expression would be: ^[a-zA-Z0-9~!#$^ ()_\-+’:.?]+$
I personally recommend using https://regex101.com to check regex expressions - note that they don't have C# support, but in general javascript's RegExp has similar syntax to C#, but what it does give you a particularly useful explaination of what your expression is doing, here is this epression's explaination from there:
^ assert position at start of the string
[a-zA-Z0-9~!#$^ ()_\-\+’:.?]+ match a single character present in the list below
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
a-z a single character in the range between a and z (case sensitive)
A-Z a single character in the range between A and Z (case sensitive)
0-9 a single character in the range between 0 and 9
~!#$^ ()_ a single character in the list ~!#$^ ()_ literally
\- matches the character - literally
+’:.? a single character in the list ’:.? literally
$ assert position at end of the string
the issue with what you put in the OP was literally only forgetting to escape the - as it is reserved in the regular expression pattern to be used for special purposes (i.e in the [] notation the - is reserved to declare a character range like a-z)

how to create regular expression based on some condition

i want to create a regular expression to find and replace uppercase character based on some condition.
find the starting uppercase for a group of uppercase character in a string and replace it lowercase and * before the starting uppercase.
If there is any lowercase following the uppercase,replace the uppercase with lowercase and * before the starting uppercase.
input string : stackOVERFlow
expected output : stack*over*flow
i tried but could not get it working perfectly.
Any idea on how to create a regular expression ?
Thanks

Well the expected inputs and outputs are slightly illogical: you're lower-casing the "f" in "flow" but not including it in the asterisk.
Anyway, the regex you want is pretty simple: #"[A-Z]+?". This matches a string of one or more uppercase alpha characters, nongreedily (don't think it makes a difference either way as the matched character class is relatively narrow).
Now, to do the find/replace, you would do something like the following:
Regex.Replace(inputString, #"([A-Z]+?)", "*$1*").ToLower();
This simply finds all occurrences of one or more uppercase alpha characters, and wherever it finds a match it replaces it with itself surrounded by asterisks. This does the surrounding but not the lowercasing; .NET Regex doesn't provide for that kind of string modification. However, since the end result of the operation should be a string with all lowercase chars, just do exactly that with a ToLower() and you'll get the expected result.

KeithS's solution can be simplified a bit
Regex.Replace("stackOVERFlow","[A-Z]+","*$0*").ToLower()
However, this will yield stack*overf*low including the f between the stars. If you want to exclude the last upper case letter, use the following expression
Regex.Replace("stackOVERFlow","[A-Z]+(?=[A-Z])","*$0*").ToLower()
It will yield stack*over*flow
This uses the pattern find(?=suffix), which finds a position before a suffix.

Does this regex expression allow "*"?

I really know very little about regex's.
I'm trying to test a password validation.
Here's the regex that describes it (I didn't write it, and don't know what it means):
private static string passwordField = "[^A-Za-z0-9_.\\-!##$%^&*()=+;:'\"|~`<>?\\/{}]";
I've tried a password like "dfgbrk*", and my code, using the above regex, allowed it.
Is this consistent with what the regex defines as acceptable, or is it a problem with my code?
Can you give me an example of a string that validation using the above regex isn't suppose to allow?
Added: Here's how the original code uses this regex (and it works there):
public static bool ValidateTextExp(string regexp, string sText)
{
if ( sText == null)
{
Log.WriteWarning("ValidateTextExp got null text to validate against regExp {0} . returning false",regexp);
return false;
}
return (!Regex.IsMatch(sText, regexp));
}
It seems I'm doing something wrong..
Thanks.

Your regex matches a value that contains any single character which is not in that list.
Your test value matches because it has spaces in it, which do not appear to be in your expression.
The reason it's not is because your character class starts with ^. The reason it matches any value that contains any single character that is not that is because you did not specify the beginning or end of the string, or any quantifiers.
The above assumes I'm not missing the importance of any of the characters in the middle of the character soup :)
This answer is also dependent on how you actually use the Regex in code.
If your intention was for that Regex string to represent the only characters that are actually allowed in a password, you would change the regex like so:
string pattern = "^[A-Z0-9...etc...]+$";
The important parts there are:
The ^ has been removed from inside the bracket, to outside; where it signifies the start of the whole string.
The $ has been added to the end, where it signifies the end of the whole string.
Those are needed because otherwise, your pattern will match anything that contains the valid values anywhere inside - even if invalid values are also present.
finally, I've added the + quantifier, which means you want to find any one of those valid characters, one or more times. (this regex would not permit a 0-length password)
If you wanted to permit the ^ character also as part of the password, you would add it back in between the brackets, but just *not as the first thing right after the opening bracket [. So for example:
string pattern = "^[A-Z0-9^...etc...]+$";
The ^ has special meaning in different places at different times in Regexes.

[^A-Za-z0-9_.\-!##$%^&*()=+;:'\"|~`?\/{}]
----------------------^
Looks fine to me, at least in regards to your question title. I'm not clear yet on why the spaces in your sample don't trip it up.
Note that I'm assuming the purpose of this expression is to find invalid characters. Thus, if the expression is a positive match, you have a bad password that you must reject. Since there appears to be some confusion about this, perhaps I can clear it up with a little psuedo-code:
bool isGoodPassword = !Regex.IsMatch(#"[^A-Za-z0-9_.\-!...]", requestedPassword);
You could re-write this for a positive match (without the negation) like so:
bool isGoodPassword = Regex.IsMatch(#"^[A-Za-z0-9_.\-!...]+$", requestedPassword);
The new expression matches a string that from the beginning of the string is filled with one or more of any of the characters in the list all the way the way to end. Any character not in the list would cause the match to fail.

You regular expression is just an inverted character class and describes just one single character (but that can’t be *). So it depends on how you use that character class.

Depends on how you apply it. It describes exactly one character, however, the ^ in the beginning buggs me a little, as it prohibits every other character, so there is probably something terribly fishy there.
Edit: as pointed out in other answers, the reason for your string to match is the space, not the explanation that was replaced by this line.