Finding quoted strings with escaped quotes in C# using a regular expression_问答_开发者

I'm trying to find all of the quoted text on a single line.

Example:

"Some Text"
"Some more Text"
"Even more text about \"this text\""

I need to get:

"Some Text"
"Some more Text"
"Even more text about \"this text\""

\"[^\"\r]*\" gives me everything except for the last one, because of the escaped quotes.

I have read about \"[^\"\\]*(?:\\.[^\"\\]*)*\" working, but I get an error at run time:

parsing ""[^"\]*(?:\.[^"\]*)*"" - Unt开发者_StackOverflow社区erminated [] set.

How do I fix this?

What you've got there is an example of Friedl's "unrolled loop" technique, but you seem to have some confusion about how to express it as a string literal. Here's how it should look to the regex compiler:

"[^"\\]*(?:\\.[^"\\]*)*"

The initial "[^"\\]* matches a quotation mark followed by zero or more of any characters other than quotation marks or backslashes. That part alone, along with the final ", will match a simple quoted string with no embedded escape sequences, like "this" or "".

If it does encounter a backslash, \\. consumes the backslash and whatever follows it, and [^"\\]* (again) consumes everything up to the next backslash or quotation mark. That part gets repeated as many times as necessary until an unescaped quotation mark turns up (or it reaches the end of the string and the match attempt fails).

Note that this will match "foo\"- in \"foo\"-"bar". That may seem to expose a flaw in the regex, but it doesn't; it's the input that's invalid. The goal was to match quoted strings, optionally containing backslash-escaped quotes, embedded in other text--why would there be escaped quotes outside of quoted strings? If you really need to support that, you have a much more complex problem, requiring a very different approach.

As I said, the above is how the regex should look to the regex compiler. But you're writing it in the form of a string literal, and those tend to treat certain characters specially--i.e., backslashes and quotation marks. Fortunately, C#'s verbatim strings save you the hassle of having to double-escape backslashes; you just have to escape each quotation mark with another quotation mark:

Regex r = new Regex(@"""[^""\\]*(?:\\.[^""\\]*)*""");

So the rule is double quotation marks for the C# compiler and double backslashes for the regex compiler--nice and easy. This particular regex may look a little awkward, with the three quotation marks at either end, but consider the alternative:

Regex r = new Regex("\"[^\"\\\\]*(?:\\\\.[^\"\\\\]*)*\"");

In Java, you always have to write them that way. :-(

Regex for capturing strings (with \ for character escaping), for the .NET engine:

(?>(?(STR)(?(ESC).(?<-ESC>)|\\(?<ESC>))|(?!))|(?(STR)"(?<-STR>)|"(?<STR>))|(?(STR).|(?!)))+

Here, a "friendly" version:

(?>                            | especify nonbacktracking
   (?(STR)                     | if (STRING MODE) then
         (?(ESC)               |     if (ESCAPE MODE) then
               .(?<-ESC>)      |          match any char and exits escape mode (pop ESC)
               |               |     else
               \\(?<ESC>)      |          match '\' and enters escape mode (push ESC)
         )                     |     endif
         |                     | else
         (?!)                  |     do nothing (NOP)
   )                           | endif
   |                           | -- OR
   (?(STR)                     | if (STRING MODE) then
         "(?<-STR>)            |     match '"' and exits string mode (pop STR)
         |                     | else
         "(?<STR>)             |     match '"' and enters string mode (push STR)
   )                           | endif
   |                           | -- OR
   (?(STR)                     | if (STRING MODE) then
         .                     |     matches any character
         |                     | else
         (?!)                  |     do nothing (NOP)  
   )                           | endif
)+                             | REPEATS FOR EVERY CHARACTER

Based on http://tomkaminski.com/conditional-constructs-net-regular-expressions examples. It relies in quotes balancing. I use it with great success. Use it with Singleline flag.

To play around with regexes, I recommend Rad Software Regular Expression Designer, which has a nice "Language Elements" tab with quick access to some basic instructions. It's based at .NET's regex engine.

"(\\"|\\\\|[^"\\])*"

should work. Match either an escaped quote, an escaped backslash, or any other character except a quote or backslash character. Repeat.

In C#:

StringCollection resultList = new StringCollection();
Regex regexObj = new Regex(@"""(\\""|\\\\|[^""\\])*""");
Match matchResult = regexObj.Match(subjectString);
while (matchResult.Success) {
    resultList.Add(matchResult.Value);
    matchResult = matchResult.NextMatch();
}

Edit: Added escaped backslash to the list to correctly handle "This is a test\\".

Explanation:

First match a quote character.

Then the alternatives are evaluated from left to right. The engine first tries to match an escaped quote. If that doesn't match, it tries an escaped backslash. That way, it can distinguish between "Hello \" string continues" and "String ends here \\".

If either don't match, then anything else is allowed except for a quote or backslash character. Then repeat.

Finally, match the closing quote.

I recommend getting RegexBuddy. It lets you play around with it until you make sure everything in your test set matches.

As for your problem, I would try four /'s instead of two:

\"[^\"\\\\]*(?:\\.[^\"\\\\]*)*\"

I know this isn't the cleanest method, but with your example I would check the character before the " to see if it's a \. If it is, I would ignore the quote.

The regular expression

(?<!\\)".*?(?<!\\)"

will also handle text that starts with an escaped quote:

\"Some Text\" Some Text "Some Text", and "Some more Text" an""d "Even more text about \"this text\""

Well, Alan Moore's answer is good, but I would modify it a bit to make it more compact. For the regex compiler:

"([^"\\]*(\\.)*)*"

Compare with Alan Moore's expression:

"[^"\\]*(\\.[^"\\]*)*"

The explanation is very similar to Alan Moore's one:

The first part " matches a quotation mark.

The second part [^"\\]* matches zero or more of any characters other than quotation marks or backslashes.

And the last part (\\.)* matches backslash and whatever single character follows it. Pay attention on the *, saying that this group is optional.

The parts described, along with the final " (i.e. "[^"\\]*(\\.)*"), will match: "Some Text" and "Even more Text\"", but will not match: "Even more text about \"this text\"".

To make it possible, we need the part: [^"\\]*(\\.)* gets repeated as many times as necessary until an unescaped quotation mark turns up (or it reaches the end of the string and the match attempt fails). So I wrapped that part by brackets and added an asterisk. Now it matches: "Some Text", "Even more Text\"", "Even more text about \"this text\"" and "Hello\\".

In C# code it will look like:

var r = new Regex("\"([^\"\\\\]*(\\\\.)*)*\"");

BTW, the order of the two main parts: [^"\\]* and (\\.)* does not matter. You can write:

"([^"\\]*(\\.)*)*"

"((\\.)*[^"\\]*)*"

The result will be the same.

Now we need to solve another problem: \"foo\"-"bar". The current expression will match to "foo\"-", but we want to match it to "bar". I don't know

why would there be escaped quotes outside of quoted strings

but we can implement it easily by adding the following part to the beginning:(\G|[^\\]). It says that we want the match start at the point where the previous match ended or after any character except backslash. Why do we need \G? This is for the following case, for example: "a""b".

Note that (\G|[^\\])"([^"\\]*(\\.)*)*" matches -"bar" in \"foo\"-"bar". So, to get only "bar", we need to specify the group and optionally give it a name, for example "MyGroup". Then C# code will look like:

[TestMethod]
public void RegExTest()
{
    //Regex compiler: (?:\G|[^\\])(?<MyGroup>"(?:[^"\\]*(?:\.)*)*")
    string pattern = "(?:\\G|[^\\\\])(?<MyGroup>\"(?:[^\"\\\\]*(?:\\\\.)*)*\")";
    var r = new Regex(pattern, RegexOptions.IgnoreCase);

    //Human readable form:       "Some Text"  and  "Even more Text\""     "Even more text about  \"this text\""      "Hello\\"      \"foo\"  - "bar"  "a"   "b" c "d"
    string inputWithQuotedText = "\"Some Text\" and \"Even more Text\\\"\" \"Even more text about \\\"this text\\\"\" \"Hello\\\\\" \\\"foo\\\"-\"bar\" \"a\"\"b\"c\"d\"";
    var quotedList = new List<string>();
    for (Match m = r.Match(inputWithQuotedText); m.Success; m = m.NextMatch())
        quotedList.Add(m.Groups["MyGroup"].Value);

    Assert.AreEqual(8, quotedList.Count);
    Assert.AreEqual("\"Some Text\"", quotedList[0]);
    Assert.AreEqual("\"Even more Text\\\"\"", quotedList[1]);
    Assert.AreEqual("\"Even more text about \\\"this text\\\"\"", quotedList[2]);
    Assert.AreEqual("\"Hello\\\\\"", quotedList[3]);
    Assert.AreEqual("\"bar\"", quotedList[4]);
    Assert.AreEqual("\"a\"", quotedList[5]);
    Assert.AreEqual("\"b\"", quotedList[6]);
    Assert.AreEqual("\"d\"", quotedList[7]);
}

Similar to RegexBuddy posted by @Blankasaurus, RegexMagic helps too.

A simple answer, without the use of ?, is