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
Related
So i have the following RegEx for the purpose of finding and adding whitespace:
(\S)(\()
So for a string like "SomeText(Somemoretext)" I want to update this to "SomeText (Somemoretext)" it matches "t(" and so my replace eliminates the "t" from the string which is not good. I also do not know what the character could be, I'm merely trying to find the non-existence of whitespace.
Is there a better expression to use or is there a way to exclude the found character from the match returned so that I can safely replace without catching characters i do not want to replace?
Thanks
I find lookarounds hard to read and would prefer using substitutions in the replacement string instead:
var s = Regex.Replace("test1() test2()", #"(\S)\(", "$1 (");
Debug.Assert(s == "test1 () test2 ()");
$1 inserts the first capture group from the regex into the replacement string which is the non-space character before the opening parenthesis (.
If you need to detect the absence of space before a specific character (such as bracket) after a word, how about the following?
\b(?=[^\s])\(
This will detect words ( [a-zA-z0-9_] that are followed by a bracket, without a space).
(if I got your problem correctly) you can replace the full match with ( and get exactly what you need.
In case you need to look for absence spaces before a symbol (like a bracket) in any kind of text (as in the text may be non-word, such as punctuation) you might want to use the following instead.
^(?:\S*)(\()(?:\S*)$
When using this, your result will be in group 1, instead of just full match (which now contains the whole line, if a line is matched).
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
I'm trying to resolve tokens in a string.
What I would like is given input like this:
string input = "asdf %(text) %(123) %(a\)a) asdf";
That I could run that through regex.Replace() and have it replace on "%(text)", "%(123)" and "%(a\)a)".
That is, that it would match everything between a starting "%(" and a closing ")" unless the closing ")" was escaped. (But of course, then you could escape the slash with another slash, which would prevent it from escaping the end paren...)
I'm pretty sure standard regular expressions can't do this, but I'm wondering if any of the various fancy expanded capabilities of the C# regular expression library could, rather than just iterating across the string totally manually? Or some other method that could do this? I feel like it's a common enough program that there has to be some way to solve it without implementing the solution from scratch, given the immensity of the .net framework? If I do have to implement iterating through the string and replacing with string.Replace(), I will, but it just seems so inelegant.
How about
var regex = new Regex(#"%\(.*?(?<!\\)(?:\\\\)*\)");
var result = regex.Replace(source,"");
%\( match literal %(
.*? match anything non-greedy
(?<!\\) preceding character to next match must not be \
(?:\\\\)* match zero or more literal \\ (i.e. match escaped \
\) match literal )
This is working for me :
String something = "\"asdf %(text) %(123) %(a\\)a) asdf\";";
String change = something.replaceAll("%\\(.*\\)", "");
System.out.println(change);
The output
"asdf asdf";
Hi I need to do like this.
Actually **ctu** is a good university but **ctu's** is not. There are many **,ctus,** present.
What I want to do is, I want to replace ctu in the string like this.
Actually **<s>ctu<e>** is a good university but **<s>ctu's<e>** is not. There are many **,<s>ctus<e>,** present.
But with the following pattern
**\\bctu*(?:['\\\\|""\\\\]*)\\w+\\b**
I'm getting the out put as:
A**<s>ctu<e>**ally **<s>ctu<e>** is a good university but **<s>ctu's<e>** is not. There are many **,ctus,** present.
I dont want to replace ctu inside words Actually. and also I need to replace " ,ctus, " with " ,<s>ctus<e>, "
How do I achieve this using regex. I need this in c#. csharp.
Thanks in advance.
The following regex matches all the cases listed in your example:
#"(\bctu(?:'\w+)?\w*\b)"
Then just replace the match with #"<s>\1<e>" where \1 is the backreference to the match above.
Are you looking for #"\bctu\b" ("ctu" with word boundaries on both sides, so it matches ctu but not Actually, ctu's, or ,ctus,) for the first search pattern and ",ctus," (exactly the string ,ctus,, regardless of where it might fall in a word) as the second search pattern? To search for both of these at once, you could use #"(\bctu\b|,ctus,)".
As a slight aside, in C# you can write regex literals much easier by using the #"" notation (verbatim strings) instead of "". E.g. to get regex to understand a word boundary, it must see \b, which can be represented as #"\b" or "\\b", and a literal \ is "\\\\" or #"\\". The first is easier to read, especially in more complex cases.
If this doesn't answer your question, please give a clear example of expected input/output.
This is the sample
"abc","abcsds","adbc,ds","abc"
Output should be
abc
abcsds
adbc,ds
abc
Try this:
"(.*?)"
if you need to put this regex inside a literal, don't forget to escape it:
Regex re = new Regex("\"(.*?)\"");
This is a tougher job than you realize -- not only can there be commas inside the quotes, but there can also be quotes inside the quotes. Two consecutive quotes inside of a quoted string does not signal the end of the string. Instead, it signals a quote embedded in the string, so for example:
"x", "y,""z"""
should be parsed as:
x
y,"z"
So, the basic sequence is something like this:
Find the first non-white-space character.
If it was a quote, read up to the next quote. Then read the next character.
Repeat until that next character is not also a quote.
If the next (non-whitespace) character is not a comma, input is malformed.
If it was not a quote, read up to the next comma.
Skip the comma, repeat the whole process for the next field.
Note that despite the tag, I'm not providing a regex -- I'm not at all sure I've seen a regex that can really handle this properly.
This answer has a C# solution for dealing with CSV.
In particular, the line
private static Regex rexCsvSplitter = new Regex( #",(?=(?:[^""]*""[^""]*"")*(?![^""]*""))" );
contains the Regex used to split properly, i.e., taking quoting and escaping into consideration.
Basically what it says is, match any comma that is followed by an even number of quote marks (including zero). This effectively prevents matching a comma that is part of a quoted string, since the quote character is escaped by doubling it.
Keep in mind that the quotes in the above line are doubled for the sake of the string literal. It might be easier to think of the expression as
,(?=(?:[^"]*"[^"]*")*(?![^"]*"))
If you can be sure there are no inner, escaped quotes, then I guess it's ok to use a regular expression for this. However, most modern languages already have proper CSV parsers.
Use a proper parser is the correct answer to this. Text::CSV for Perl, for example.
However, if you're dead set on using regular expressions, I'd suggest you "borrow" from some sort of module, like this one:
http://metacpan.org/pod/Regexp::Common::balanced