Match a string until it meets a '('

Question

I've managed to get everything (well, all letters) up to a whitespace using the following:

@"^.*([A-Z][a-z].*)]\s" 

However, I want to to match to a ( instead of a whitespace... how can I manage this?

Without having the '(' in the match

Answer 1

If what you want is to match any character up until the ( character, then this should work:

@"^.*?(?=\()"

If you want all letters, then this should do the trick:

@"^[a-zA-Z]*(?=\()"

Explanation:

^           Matches the beginning of the string

.*?         One or more of any character. The trailing ? means 'non-greedy', 
            which means the minimum characters that match, rather than the maximum

(?=         This means 'zero-width positive lookahead assertion'. That means that the 
            containing expression won't be included in the match.

\(          Escapes the ( character (since it has special meaning in regular 
            expressions)

)           Closes off the lookahead

[a-zA-Z]*?  Zero or more of any character from a to z, or from A to Z

Reference: Regular Expression Language - Quick Reference (MSDN)

EDIT: Actually, instead of using .*?, as Casimir has noted in his answer it's probably easier to use [^\)]*. The ^ used inside a character class (a character class is the [...] construct) inverts the meaning, so instead of "any of these characters", it means "any except these characters". So the expression using that construct would be:

@"^[^\(]*(?=\()"

Answer 2

Using a constraining character class is the best way

@"^[^(]*" 

[^(] means all characters but (

Note that you don't need a capture group since that you want is the whole pattern.