I would like a simple explanation about regex's escaping structure in C#. I've read the MSDN pages but it seems that i cannot write a working Regex.Escape()
Additionally, a working example of escaping "(", ")" and "." characters would be great. For example somestring = Regex.Escape("("+"(.*?))");
Thanks
As stated in the documentation:
Escapes a minimal set of characters (,\, *, +, ?, |, {, [, (,), #, ^, $, .,
and white space) by replacing them with their escape codes. This instructs the regular expression engine to interpret these characters
literally rather than as metacharacters.
Which basically means that, in regular expression language, you have some characters which are special. These characters include, operators such as ?, *, ., +, etc.
To have a regular expression threat for instance, the + as the character +, and not the one or more of the previous operator, we escape it like so: \+. This tells the parsing engine to treat the + as is.
What the escape method does is that it adds the extra backslash to these characters.
Thus, given this: Regex.Escape("("+"(.*?))");, the output string would be \(\(\.\*\?\)\), which would mean, match the given string: (.*?)).
There may some possibilities of regex meta characters present in a variable in which you're trying to use the value of that variable as a regex to search for a particular substring. In this case , we need to put the variable inside the Regex.Escape function in-order to make special characters present inside the variable to get automatically escaped.
Regex.Escape("("+"(.*?))")
Essentially any meta-character in the input gets a backslash in front of it. So:
\(\(\.\*\?\)\)
But of course, anything that shows the string as if it were in C# source code (like the VS debugger tool windows) will itself escape the backslashes, hence a display something like:
\\(\\(\\.\\*\\?\\)\\)
(Hence why verbastin strings are so useful with regexes.)
PS. Do not write your own Regex.Escape: you'll just miss some edge cases of the syntax (and there are lots). The Framework method is there to use, so use it.
Related
Can someone explain to me when using regular expressions when a double backslash or single backslash needs to be used to escape a character?
A lot of references online use a single backslash and online regex testers work with single backslashes, but in practice I often have to use a double backslash to escape a character.
For example:
"SomeString\."
Works in an online regex tester and matches "SomeString" followed by a dot.
However in practice I have to use a double escape:
if (Regex.IsMatch(myString, "SomeString\\."))
C# does not have a special syntax for construction of regular expressions, like Perl, Ruby or JavaScript do. It instead uses a constructor that takes a string. However, strings have their own escaping mechanism, because you want to be able to put quotes inside the string. Thus, there are two levels of escaping.
So, in a regular expression, w means the letter "w", while \w means a word character. However, if you make a string "\w", you are escaping the character "w", which makes no sense, since character "w" is not a quote or a backslash, so "w" == "\w". Then this string containing only "w" gets passed to the regexp constructor, and you end up matching the letter "w" instead of any word character. Thus, to pass the backslash to regexp, you need to put in two backslashes in the string literal (\\w): one will be removed when the string literal is interpreted, one will be used by the regular expression.
When working with regular expressions directly (such as on most online regexp testers, or when using verbatim strings #"..."), you don't have to worry about the interpretation of string literals, and you always write just one backslash (except when you want to match the backslash itself, but then you're espacing the backslash for the regexp, not for the string).
\ Is also an escape character for string literals in c# so the first \ is escaping the second \ being passed to the method and the second one is escaping the . in the regex.
Use:
if (Regex.IsMatch(myString, #"SomeString\."))
If you want to avoid double escaping.
I you use a verbatim symbol #(verbatim string), you don't need to escape the backslash again.
if (Regex.IsMatch(myString, #"SomeString\."))
Old post but Regex.Escape may be useful
In JavaScript you have to use double escape character: \
let m = "My numer is [56]".match("\\[(.*)\\]");
alert(m[1]);//outputs 56
In C# single \
May I know what \? means in a regular expression? For example, what is its significance in this expression.
I have used this for validating 7 digit telephone no
Any help is highly appreciated.
"\?" means "?" itself. "\" - is escape character. "?" is quantifier and "\" is used to escape it.
I have used this for validating 7 digit telephone no
"[[:number:]]\{3\}[ -]\?[[:number:]]\{4\}"
Looking at your example, it seems that you are talking about BRE, then the \ (escaping) gave ? special meaning: one or zero[ -]
If it is ERE/PCRE, the \ will take that speical meaning away from ?, that is, \? means literal question mark: ?
The properly-escaped "?" will match that exact character, the "?", as it appears in the text.
For instance, if you do
Regex re = new Regex(#"\d{3}-\?\d{4}");
, you will be able to get a positive match for 123-?1234.
If you want to get a positive match for 1231234 OR 123-1234, you can use the special character "?" without escape, like this:
Regex re = new Regex(#"\d{3}-?\d{4}");
P.S. for C# .NET, I find the best regex-testing place online is MyRegexTester. If you use it for C#, don't forget to check the appropriate "C# .NET" checkbox.
P.P.S. as per the comment, putting "\s*" into the regex will match any length white space (spaces and tabs included), "\ ?" will match an optional space, and "[ ]" will match exactly one space (no less).
"\?" escapes "?" that have a special meaning in the regex (0 or 1 match) so "\?" escapes it and identifies the literal "?"
your regex looks strange to me, it looks that all the special character are escaped (also "{" ) and doesn't appear to be valid from what i know.
i think you want to write
"\d{3}[ -]?\d{4}"
if you want to match something that respect the pattern or
"^\d{3}[ -]?\d{4}$"
if you want to have a match something that is exactly the pattern
If you have a string with special characters that you want to match with:
System.Text.RegularExpressions.Regex.Matches(theTextToCheck, myString);
It will obviously give you wrong results, if you have special characters inside myString like "%" or "\".
The idea is to convert myString and replacing all occurences of special characters like "%" to be replaced by their corresponding characters.
Does anyone know how to solve that or does someone have a RegEx for that? :)
Update:
The following characters have a special meaning, that I should turn of with adding a leading backslash: \, &, ~, ^, %, [, ], {, }, ?, +, *,(,),|,$
are there any others I should replace?
As #Kobi links to in the comments, you need to use Regex.Escape to ensure that that regular expression string is properly escaped.
Escapes a minimal set of characters (\, *, +, ?, |, {, [, (,), ^, $,., #, and white space) by replacing them with their escape codes. This instructs the regular expression engine to interpret these characters literally rather than as metacharacters.
If you want to escape all characters that carry a special meaning in regex, you could simply escape every character with a backslash (There is no harm in escaping characters that don't need to be escaped).
But if you do, why are you using Regex at all instead of string.IndexOf?
Regex.Escape will do that for you. Somewhere in msdn doc it reads:
Escape converts a string so that the regular expression engine will interpret any metacharacters that it may contain as character literals
which is much more informative that the function description.
This is left for search/replace reference.
Use this as your regex:
(\\|\&|\~|\^|\%|\[|\]|\{|\}|\?|\+|\*|\(|\)|\||\$)
gets your chars of interes in a numbered group
And this as your replacement string:
\$1
replaces the matches with backslash plus the group content
Sample code:
Regex re = new Regex(#"(\\|\&|\~|\^|\%|\[|\]|\{|\}|\?|\+|\*|\(|\)|\||\$)");
string replaced = re.Replace(#"Look for (special {characters} and scape [100%] of them)", #"\$1");
I'm currently facing a (little) blocking issue. I'd like to replace a substring by one another using regular expression. But here is the trick : I suck at regex.
Regex.Replace(contenu, "Request.ServerVariables("*"))",
"ServerVariables('test')");
Basically I'd like to replace whatever is between the " by "test". I tried ".{*}" as a pattern but it doesn't work.
Could you give me some tips, I'd appreciate it!
There are several issues you need to take care of.
You are using special characters in your regex (., parens, quotes) -- you need to escape these with a slash. And you need to escape the slashes with another slash as well because we 're in a C# string literal, unless you prefix the string with # in which case the escaping rules are different.
The expression to match "any number of whatever characters" is .*. In this case, you would want to match any number of non-quote characters, which is [^"]*.
In contrast to (1) above, the replacement string is not a regular expression so you don't want any slashes there.
You need to store the return value of the replace somewhere.
The end result is
var result = Regex.Replace(contenu,
#"Request\.ServerVariables\(""[^""]*""\)",
"Request.ServerVariables('test')");
Based purely on my knowledge of regex (and not how they are done in C#), the pattern you want is probably:
"[^"]*"
ie - match a " then match everything that's not a " then match another "
You may need to escape the double-quotes to make your regex-parser actually match on them... that's what I don't know about C#
Try to avoid where you can the '.*' in regex, you can usually find what you want to get by avoiding other characters, for example [^"]+ not quoted, or ([^)]+) not in parenthesis. So you may just want "([^"]+)" which should give you the whole thing in [0], then in [1] you'll find 'test'.
You could also just replace '"' with '' I think.
Taryn Easts regex includes the *. You should remove it, if it is just a placeholder for any value:
"[^"]"
BTW: You can test this regex with this cool editor: http://rubular.com/r/1MMtJNF3kM
Pattern is
Regex splRegExp = new System.Text.RegularExpressions.Regex(#"[\,#,+,\,?,\d,%,.,?,*,&,^,$,(,!,),#,-,_]");
All characters work except '-'. Please advise.
Use
#"[,#+\\?\d%.*&^$(!)#_-]"
No need for all those commas.
If you place a - inside a character class, it means a literal dash only if it's at the start or end of the class. Otherwise it denotes a range like A-Z. As Damien put it, the range ,-, is indeed rather small (and doesn't contain the -, of course).
'-' has to be the first charater in your regex.
Regex splRegExp = new System.Text.RegularExpressions.Regex(#"[-,\,#,+,\,?,\d,%,.,?,*,&,^,$,(,!,),#,_]");
You need to escape the -character for it to work (it's a regular expression syntax)
Try this:
"[\,#,+,\,?,\d,%,.,?,*,&,^,$,(,!,),#,\-,_]"