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.NET Forum / Languages / C# / September 2005trim spaces before evaluating regex
What's the proper way to write a validation regex so that all leading and trailing spaces are trimmed before evaluation ? Thanks. You may just want to do a string.Trim() before passing to regex.  SignatureWilliam Stacey [MVP] > What's the proper way to write a validation regex so that all leading and > trailing spaces are trimmed before evaluation ? Thanks. It's inside a validator evaluated on the client-side. > You may just want to do a string.Trim() before passing to regex. > >> What's the proper way to write a validation regex so that all leading and >> trailing spaces are trimmed before evaluation ? Thanks. > What's the proper way to write a validation regex so that all leading > and trailing spaces are trimmed before evaluation ? Thanks. You could extend that regex, so that leading/trailing spaces are accepted. Don't forget to trim the value then before you use it. [ ]*<original RE>[ ]* Hans Kesting >> What's the proper way to write a validation regex so that all leading >> and trailing spaces are trimmed before evaluation ? Thanks. [quoted text clipped - 4 lines] > >Hans Kesting Almost a correct answer, but you forgot the symbols for matching the complete string. ^[ ]*<original RE>[ ]*$ Also, if you want to remove tabs and other whitespace characters you can also use \s. ^\s*<original RE>\s*$ -- Marcus Andrén <snip> > Almost a correct answer, but you forgot the symbols for matching the > complete string. > > ^[ ]*<original RE>[ ]*$ That's not true, if it's for use in RegularExpressionValidator (as i suppose is true for the OP). RegularExpressionValidator checks, if the whole value matches the RE, not if it contains a match. (Internally it adds ^ and $ IMHO. Christof >> ^[ ]*<original RE>[ ]*$ > [quoted text clipped - 5 lines] > >Christof Oops. That is what I get for assuming that the validator took a true regular expression instead of a deformed one. :) Marcus In this format: [ ]*<original RE>[ ]* when I move the cursor over it , I am finding a what appears to be a space character between the []. Is that how a regex works? It can identify a single space as a character to be matched ? >>> ^[ ]*<original RE>[ ]*$ >> [quoted text clipped - 12 lines] > > Marcus >In this format: > [quoted text clipped - 5 lines] >Is that how a regex works? It can identify a single space as a character >to be matched ? In addition to what Marcus said, be careful when using spaces or other whitespace in regular expressions. There's a flag in the RegexOptions enumeration that can be passed in to various methods of the Regex class, it's called IgnorePatternWhitespace. If this flag is set when the expression is used, any whitespace in the pattern is not assumed to be part of the match. Obviously, if you call the Regex method yourself, you're bound to know whether you're passing that flag or not. But when the call is made from other code than your own, or you're using a regular expression tester, it's important to know this when working with expressions that want to match whitespace. Oliver Sturm SignatureExpert programming and consulting services available See http://www.sturmnet.org (try /blog as well) Also, in the form : \s*<original RE>\s* why is it possible to not include the [] ? In other words, why is it that the form does not have to be: [\s]*<original RE>[\s]* Thanks. >>> ^[ ]*<original RE>[ ]*$ >> [quoted text clipped - 12 lines] > > Marcus >Also, in the form : > [quoted text clipped - 7 lines] > >Thanks. Actually the [] clammers aren't needed. They were probably used to clearly demonstrate taht it was a space. [] is a construct to match multiple types of characters. [abc] for example matches any of the characters a, b and c. [abc]* can be used to match any string containing only a,b and c's. Since space is a single character it could just as easily have been used without []. " *<original RE> *" would work just fine. While \s represents all whitespace characters it is still treated as a single character in the regular expressions syntax so it can also be written without []. If however you would like to match any whitespace character or the character b or c, you would have to use [bc\s]. If you want a brief description of all regular expression elements you can use the following link http://msdn.microsoft.com/library/default.asp?url=/library/en-us/cpgenref/html/c pconregularexpressionslanguageelements.asp A good and quite simple tutorial that I used once upon a time is here http://gnosis.cx/publish/programming/regular_expressions.html Finally, A minor advice. If you ever use regular expressions inside your c# code, prepend @ to your string. @ tells the compiler that the \ character shouldn't be treated as a special character. This is useful when working with strings that contain the character \, like regular expressions or file locations. For example: "c:\\program files\\myprogram\\program.exe" can also be written @"c:\program files\myprogram\program.exe" in c#. -- Marcus Andrén Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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