Related
I have a value like this:
"Foo Bar" "Another Value" something else
What regex will return the values enclosed in the quotation marks (e.g. Foo Bar and Another Value)?
In general, the following regular expression fragment is what you are looking for:
"(.*?)"
This uses the non-greedy *? operator to capture everything up to but not including the next double quote. Then, you use a language-specific mechanism to extract the matched text.
In Python, you could do:
>>> import re
>>> string = '"Foo Bar" "Another Value"'
>>> print re.findall(r'"(.*?)"', string)
['Foo Bar', 'Another Value']
I've been using the following with great success:
(["'])(?:(?=(\\?))\2.)*?\1
It supports nested quotes as well.
For those who want a deeper explanation of how this works, here's an explanation from user ephemient:
([""']) match a quote; ((?=(\\?))\2.) if backslash exists, gobble it, and whether or not that happens, match a character; *? match many times (non-greedily, as to not eat the closing quote); \1 match the same quote that was use for opening.
I would go for:
"([^"]*)"
The [^"] is regex for any character except '"'
The reason I use this over the non greedy many operator is that I have to keep looking that up just to make sure I get it correct.
Lets see two efficient ways that deal with escaped quotes. These patterns are not designed to be concise nor aesthetic, but to be efficient.
These ways use the first character discrimination to quickly find quotes in the string without the cost of an alternation. (The idea is to discard quickly characters that are not quotes without to test the two branches of the alternation.)
Content between quotes is described with an unrolled loop (instead of a repeated alternation) to be more efficient too: [^"\\]*(?:\\.[^"\\]*)*
Obviously to deal with strings that haven't balanced quotes, you can use possessive quantifiers instead: [^"\\]*+(?:\\.[^"\\]*)*+ or a workaround to emulate them, to prevent too much backtracking. You can choose too that a quoted part can be an opening quote until the next (non-escaped) quote or the end of the string. In this case there is no need to use possessive quantifiers, you only need to make the last quote optional.
Notice: sometimes quotes are not escaped with a backslash but by repeating the quote. In this case the content subpattern looks like this: [^"]*(?:""[^"]*)*
The patterns avoid the use of a capture group and a backreference (I mean something like (["']).....\1) and use a simple alternation but with ["'] at the beginning, in factor.
Perl like:
["'](?:(?<=")[^"\\]*(?s:\\.[^"\\]*)*"|(?<=')[^'\\]*(?s:\\.[^'\\]*)*')
(note that (?s:...) is a syntactic sugar to switch on the dotall/singleline mode inside the non-capturing group. If this syntax is not supported you can easily switch this mode on for all the pattern or replace the dot with [\s\S])
(The way this pattern is written is totally "hand-driven" and doesn't take account of eventual engine internal optimizations)
ECMA script:
(?=["'])(?:"[^"\\]*(?:\\[\s\S][^"\\]*)*"|'[^'\\]*(?:\\[\s\S][^'\\]*)*')
POSIX extended:
"[^"\\]*(\\(.|\n)[^"\\]*)*"|'[^'\\]*(\\(.|\n)[^'\\]*)*'
or simply:
"([^"\\]|\\.|\\\n)*"|'([^'\\]|\\.|\\\n)*'
Peculiarly, none of these answers produce a regex where the returned match is the text inside the quotes, which is what is asked for. MA-Madden tries but only gets the inside match as a captured group rather than the whole match. One way to actually do it would be :
(?<=(["']\b))(?:(?=(\\?))\2.)*?(?=\1)
Examples for this can be seen in this demo https://regex101.com/r/Hbj8aP/1
The key here is the the positive lookbehind at the start (the ?<= ) and the positive lookahead at the end (the ?=). The lookbehind is looking behind the current character to check for a quote, if found then start from there and then the lookahead is checking the character ahead for a quote and if found stop on that character. The lookbehind group (the ["']) is wrapped in brackets to create a group for whichever quote was found at the start, this is then used at the end lookahead (?=\1) to make sure it only stops when it finds the corresponding quote.
The only other complication is that because the lookahead doesn't actually consume the end quote, it will be found again by the starting lookbehind which causes text between ending and starting quotes on the same line to be matched. Putting a word boundary on the opening quote (["']\b) helps with this, though ideally I'd like to move past the lookahead but I don't think that is possible. The bit allowing escaped characters in the middle I've taken directly from Adam's answer.
The RegEx of accepted answer returns the values including their sourrounding quotation marks: "Foo Bar" and "Another Value" as matches.
Here are RegEx which return only the values between quotation marks (as the questioner was asking for):
Double quotes only (use value of capture group #1):
"(.*?[^\\])"
Single quotes only (use value of capture group #1):
'(.*?[^\\])'
Both (use value of capture group #2):
(["'])(.*?[^\\])\1
-
All support escaped and nested quotes.
I liked Eugen Mihailescu's solution to match the content between quotes whilst allowing to escape quotes. However, I discovered some problems with escaping and came up with the following regex to fix them:
(['"])(?:(?!\1|\\).|\\.)*\1
It does the trick and is still pretty simple and easy to maintain.
Demo (with some more test-cases; feel free to use it and expand on it).
PS: If you just want the content between quotes in the full match ($0), and are not afraid of the performance penalty use:
(?<=(['"])\b)(?:(?!\1|\\).|\\.)*(?=\1)
Unfortunately, without the quotes as anchors, I had to add a boundary \b which does not play well with spaces and non-word boundary characters after the starting quote.
Alternatively, modify the initial version by simply adding a group and extract the string form $2:
(['"])((?:(?!\1|\\).|\\.)*)\1
PPS: If your focus is solely on efficiency, go with Casimir et Hippolyte's solution; it's a good one.
A very late answer, but like to answer
(\"[\w\s]+\")
http://regex101.com/r/cB0kB8/1
The pattern (["'])(?:(?=(\\?))\2.)*?\1 above does the job but I am concerned of its performances (it's not bad but could be better). Mine below it's ~20% faster.
The pattern "(.*?)" is just incomplete. My advice for everyone reading this is just DON'T USE IT!!!
For instance it cannot capture many strings (if needed I can provide an exhaustive test-case) like the one below:
$string = 'How are you? I\'m fine, thank you';
The rest of them are just as "good" as the one above.
If you really care both about performance and precision then start with the one below:
/(['"])((\\\1|.)*?)\1/gm
In my tests it covered every string I met but if you find something that doesn't work I would gladly update it for you.
Check my pattern in an online regex tester.
This version
accounts for escaped quotes
controls backtracking
/(["'])((?:(?!\1)[^\\]|(?:\\\\)*\\[^\\])*)\1/
MORE ANSWERS! Here is the solution i used
\"([^\"]*?icon[^\"]*?)\"
TLDR;
replace the word icon with what your looking for in said quotes and voila!
The way this works is it looks for the keyword and doesn't care what else in between the quotes.
EG:
id="fb-icon"
id="icon-close"
id="large-icon-close"
the regex looks for a quote mark "
then it looks for any possible group of letters thats not "
until it finds icon
and any possible group of letters that is not "
it then looks for a closing "
I liked Axeman's more expansive version, but had some trouble with it (it didn't match for example
foo "string \\ string" bar
or
foo "string1" bar "string2"
correctly, so I tried to fix it:
# opening quote
(["'])
(
# repeat (non-greedy, so we don't span multiple strings)
(?:
# anything, except not the opening quote, and not
# a backslash, which are handled separately.
(?!\1)[^\\]
|
# consume any double backslash (unnecessary?)
(?:\\\\)*
|
# Allow backslash to escape characters
\\.
)*?
)
# same character as opening quote
\1
string = "\" foo bar\" \"loloo\""
print re.findall(r'"(.*?)"',string)
just try this out , works like a charm !!!
\ indicates skip character
Unlike Adam's answer, I have a simple but worked one:
(["'])(?:\\\1|.)*?\1
And just add parenthesis if you want to get content in quotes like this:
(["'])((?:\\\1|.)*?)\1
Then $1 matches quote char and $2 matches content string.
All the answer above are good.... except they DOES NOT support all the unicode characters! at ECMA Script (Javascript)
If you are a Node users, you might want the the modified version of accepted answer that support all unicode characters :
/(?<=((?<=[\s,.:;"']|^)["']))(?:(?=(\\?))\2.)*?(?=\1)/gmu
Try here.
My solution to this is below
(["']).*\1(?![^\s])
Demo link : https://regex101.com/r/jlhQhV/1
Explanation:
(["'])-> Matches to either ' or " and store it in the backreference \1 once the match found
.* -> Greedy approach to continue matching everything zero or more times until it encounters ' or " at end of the string. After encountering such state, regex engine backtrack to previous matching character and here regex is over and will move to next regex.
\1 -> Matches to the character or string that have been matched earlier with the first capture group.
(?![^\s]) -> Negative lookahead to ensure there should not any non space character after the previous match
echo 'junk "Foo Bar" not empty one "" this "but this" and this neither' | sed 's/[^\"]*\"\([^\"]*\)\"[^\"]*/>\1</g'
This will result in: >Foo Bar<><>but this<
Here I showed the result string between ><'s for clarity, also using the non-greedy version with this sed command we first throw out the junk before and after that ""'s and then replace this with the part between the ""'s and surround this by ><'s.
From Greg H. I was able to create this regex to suit my needs.
I needed to match a specific value that was qualified by being inside quotes. It must be a full match, no partial matching could should trigger a hit
e.g. "test" could not match for "test2".
reg = r"""(['"])(%s)\1"""
if re.search(reg%(needle), haystack, re.IGNORECASE):
print "winning..."
Hunter
If you're trying to find strings that only have a certain suffix, such as dot syntax, you can try this:
\"([^\"]*?[^\"]*?)\".localized
Where .localized is the suffix.
Example:
print("this is something I need to return".localized + "so is this".localized + "but this is not")
It will capture "this is something I need to return".localized and "so is this".localized but not "but this is not".
A supplementary answer for the subset of Microsoft VBA coders only one uses the library Microsoft VBScript Regular Expressions 5.5 and this gives the following code
Sub TestRegularExpression()
Dim oRE As VBScript_RegExp_55.RegExp '* Tools->References: Microsoft VBScript Regular Expressions 5.5
Set oRE = New VBScript_RegExp_55.RegExp
oRE.Pattern = """([^""]*)"""
oRE.Global = True
Dim sTest As String
sTest = """Foo Bar"" ""Another Value"" something else"
Debug.Assert oRE.test(sTest)
Dim oMatchCol As VBScript_RegExp_55.MatchCollection
Set oMatchCol = oRE.Execute(sTest)
Debug.Assert oMatchCol.Count = 2
Dim oMatch As Match
For Each oMatch In oMatchCol
Debug.Print oMatch.SubMatches(0)
Next oMatch
End Sub
Below is a regex statement I have been working on for quite sometime:
Match parsedRequestData = Regex.Match(requestData, #"^.*\[(.*)\]$");
What this is supposed to be doing is taking the email out of the email below:
2.3|[0246303#up.com]
For clarification, this email comes from a table in SQL Server. There are many emails that are formatted like this in there and the regex is supposed to be getting all of that from inside the brackets. However, it is matching the entirety of this line instead of whats inside of it. So my question is, is there something wrong with my regex statement or do I have something in my code I need to add?
Your regex is storing the email address in capture group 1. Try referencing group 1 like this:
parsedRequestData.Groups[1];
Code Sample:
string requestData = "2.3|[0246303#up.com]";
Match parsedRequestData = Regex.Match(requestData, #"^.*\[(.*)\]$");
if (parsedRequestData.Success)
{
Console.WriteLine(parsedRequestData.Groups[1]);
}
Results:
0246303#up.com
Your regex is OK. All you need is to use the Group[1]
var email = Regex.Match("2.3|[0246303#up.com]", #"^.*\[(.*)\]$").Groups[1].Value;
However, it is matching the entirety of this line instead of whats inside of it.
Unless one uses named match captures, the match capture groups are indexed.
Match.Groups[0].Value is the whole match; it shows all the match captures and all the grouped matched text.
Match.Groups[{1-N}].Value is the match captures in the order of specification in the pattern for anything in a ( ) parenthesis set(s). If there is only one ( ) there will be two indexed groups; 0 as mentioned above, and 1 of the items specified to be captured to N.
You only have one ( ) set so the data you want is found in match capture group 1. Group 0 has the non match capture items along with the match capture data.
If one names the match capture such as (?<MyNameHere> ) one can also access the match via Match.Groups["MyNameHere"].Value.
Suggestion on your pattern away from the answer
Usage of * (zero or more) in patterns can be problematic in that it can significantly increase the time of the parser takes due to backtracking false scenarios.
If one knows there is text to be found, don't tell the parser zero items may happen when that is impossible, change it to + one or more. That slight change can greatly affect the parsing operations, both in time and operations.
Change ^.*\[(.*)\]$ to ^.+\[(.+)\]$.
But to even increase the efficiency of the pattern, focus on the knowns of the characters [ and ] as anchors.
Pattern Restructure To Use Anchors
^[^[]+\[([^\]]+)[\s\]]+$
Why is this pattern better? Because we will look for "[" and "]" as anchors.
Let us break it down
^ - Beginning of the pattern (a hard anchor)
[^ ]+ This is a set notation where the ^ says NOT.
[^\[]+ So we want to capture all text + (one or more) that is NOT a [. This tells the pattern to match up to our anchor [ in the text. Note that we don't have to escape it for regex parser treats all characters in a set [ ] as a literal so [^[] is valid. (To be clear this is a match but don't capture text anchor so we will not find this text in an index above the 0 index; only in 0).
\[ Our literal anchor the "[" character.
([^\]]+) This is our match capture which says match this set where any character is valid but not an "]". Here we have to escape the ] because otherwise it would signify the end of our set.
[\s\]]+ we know the end of our text there will be spaces and the "]" character, so let us match (but not to capture) any combination of spaces and a ] before the end.
$ our final anchor, the end of the file/buffer indicator (or line if the right parser rule is set).
I have kind of a weird problem that I am attempting to resolve with some elegant regular expressions.
The system I am working on was initially designed to accept an incoming string and through a pattern matching method, alter the string which it then returns. A very simplistic example is:
Incoming string:
The dog & I went to the park and had a great time...
Outgoing string:
The dog {&} I went to the park and had a great time {...}
The punctuation mapper wraps key characters or phrases and wraps them in curly braces. The original implementation was a one way street and was never meant for how it is currently being applied and as a result, if it is called incorrectly, it is very easy for the system to "double" wrap a string as it is just doing a simple string replace.
I spun up Regex Hero this morning and started working on some pattern matches and having not written a regular expression in nearly a year, quickly hit a wall.
My first idea was to match a character (i.e. &) but only if it wasn't wrapped in braces and came up with [^\{]&[^\}], which is great but of course catches any instance of the ampersand so long as it is not preceded by a curly brace, including white spaces and would not work in a situation where there were two ampersands back to back (i.e. && would need to be {&}{&} in the outgoing string. To make matters more complicated, it is not always a single character as ellipsis (...) is also one of the mapped values.
Every solution I noodle over either hits a barrier because there is an unknown number of occurrences of a particular value in the string or that the capture groups will either be too greedy or finally, cannot compensate for multiple values back to back (i.e. a single period . vs ellipsis ...) which the original dev handled by processing ellipsis first which covered the period in the string replace implementation.
Are there any regex gurus out there that have any ideas on how I can detect the undecorated (unwrapped) values in a string and then perform their replacements in an ungreedy fashion that can also handle multiple repeated characters?
My datasource that I am working against is a simple key value pair that contains the value to be searched for and the value to replace it with.
Updated with example strings:
Undecorated:
Show Details...
Default Server:
"Smart" 2-Way
Show Lender's Information
Black & White
Decorated:
Show Details{...}
Default Server{:}
{"}Smart{"} 2-Way
Show Lender{'}s Information
Black {&} White
Updated With More Concrete Examples and Datasource
Datasource (SQL table, can grow at any time):
TaggedValue UntaggedValue
{:} :
{&} &
{<} <
{$} $
{'} '
{} \
{>} >
{"} "
{%} %
{...} ...
{...} …
{:} ：
{"} “
{"} ”
{'} `
{'} ’
Broken String: This is a string that already has stuff {&} other stuff{!} and {...} with {_} and {#} as well{.} and here are the same characters without it & follow by ! and ... _ & . &&&
String that needs decoration: Show Details... Default Server: "Smart" 2-Way Show Lender's Information Black & White
String that would pass through the method untouched (because it was already decorated): The dog {&} I went to the park and had a great time {...}
The other "gotcha" in moving to regex is the need to handle escaping, especially of backslashes elegantly due to their function in regular expressions.
Updated with output from #Ethan Brown
#Ethan Brown,
I am starting think that regex, while elegant might not be the way to go here. The updated code you provided, while closer still does not yield correct results and the number of variables involved may exceed the regex logics capability.
Using my example above:
'This is a string that already has stuff {&} other stuff{!} and {...} with {_} and {#} as well{.} and here are the same characters without it & follow by ! and ... _ & . &&&'
yields
This is a string that already has stuff {&} other stuff{!} and {...} with {_} and {#} as well{.} and here are the same characters without it {&} follow by {!} and {...} {_} {&} . {&&}&
Where the last group of ampersands which should come out as {&}{&}{&} actually comes out as {&&}&.
There is so much variability here (i.e. need to handle ellipsis and wide ellipsis from far east languages) and the need to utilize a database as the datasource is paramount.
I think I am just going to write a custom evaluator which I can easily enough write to perform this type of validation and shelve the regex route for now. I will grant you credit for your answer and work as soon as I get in front of a desktop browser.
This kind of problem can be really tough, but let me give you some ideas that might help out. One thing that's really going to give you headaches is handling the case where the punctuation appears at the beginning or end of the string. Certainly that's possible to handle in a regex with a construct like (^|[^{])&($|[^}]), but in addition to that being painfully hard to read, it also has efficiency issues. However, there's a simple way to "cheat" and get around this problem: just pad your input string with a space on either end:
var input = " " + originalInput + " ";
When you're done you can just trim. Of course if you care about preserving input at the beginning or end, you'll have to be more clever, but I'm going to assume for argument's sake that you don't.
So now on to the meat of the problem. Certainly, we can come up with some elaborate regular expressions to do what we're looking for, but often the answer is much much simpler if you use more than one regular expression.
Since you've updated your answer with more characters, and more problem inputs, I've updated this answer to be more flexible: hopefully it will meet your needs better as more characters get added.
Looking over your input space, and the expressions you need quoted, there are really three cases:
Single-character replacements (! becomes {!}, for example).
Multi-character replacements (... becomes {...}).
Slash replacement (\ becomes {})
Since the period is included in the single-character replacements, order matters: if you replace all the periods first, then you will miss ellipses.
Because I find the C# regex library a little clunky, I use the following extension method to make this more "fluent":
public static class StringExtensions {
public static string RegexReplace( this string s, string regex, string replacement ) {
return Regex.Replace( s, regex, replacement );
}
}
Now I can cover all of the cases:
// putting this into a const will make it easier to add new
// characters in the future
const string normalQuotedChars = #"\!_\\:&<\$'>""%:`";
var output = s
.RegexReplace( "(?<=[^{])\\.\\.\\.(?=[^}])", "{$&}" )
.RegexReplace( "(?<=[^{])[" + normalQuotedChars + "](?=[^}])", "{$&}" )
.RegexReplace( "\\\\", "{}" );
So let's break this solution down:
First we handle the ellipses (which will keep us from getting in trouble with periods later). Note that we use a zero-width assertions at the beginning and end of the expression to exclude expressions that are already quoted. The zero-width assertions are necessary, because without them, we'd get into trouble with quoted characters right next to each other. For example, if you have the regex ([^{])!([^}]), and your input string is foo !! bar, the match would include the space before the first exclamation point and the second exclamation point. A naive replacement of $1!$2 would therefore yield foo {!}! bar because the second exclamation point would have been consumed as part of the match. You'd have to end up doing an exhaustive match, and it's much easier to just use zero-width assertions, which are not consumed.
Then we handle all of the normal quoted characters. Note that we use zero-width assertions here for the same reasons as above.
Finally, we can find lone slashes (note we have to escape it twice: once for C# strings and again for regex metacharacters) and replace that with empty curly brackets.
I ran all of your test cases (and a few of my own invention) through this series of matches, and it all worked as expected.
I'm no regex god, so one simple way:
Get / construct the final replacement string(s) - ex. "{...}", "{&}"
Replace all occurrences of these in the input with a reserved char (unicode to the rescue)
Run your matching regex(es) and put "{" or whatever desired marker(s).
Replace reserved char(s) with the original string.
Ignoring the case where your original input string has a { or } character, a common way to avoid re-applying a regex to an already-escaped string is to look for the escape sequence and remove it from the string before applying your regex to the remainders. Here's an example regex to find things that are already escaped:
Regex escapedPattern = new Regex(#"\{[^{}]*\}"); // consider adding RegexOptions.Compiled
The basic idea of this negative-character class pattern comes from regular-expressions.info, a very helpful site for all thing regex. The pattern works because for any inner-most pair of braces, there must be a { followed by non {}'s followed by a }
Run the escapedPattern on the input string, find for each Match get the start and end indices in the original string and substring them out, then with the final cleaned string run your original pattern match again or use something like the following:
Regex punctPattern = new Regex(#"[^\w\d\s]+"); // this assumes all non-word,
// digit or space chars are punctuation, which may not be a correct
//assumption
And replace Match.Groups[1].Value for each match (groups are a 0 based array where 0 is the whole match, 1 is the first set of parentheses, 2 is the next etc.) with "{" + Match.Groups[1].Value + "}"
I have a string:
<graphic id="8374932">Translating Cowl (Inner/Outer Bondments</graphic>
And my pattern:
"<graphic id=\"(.*?)\">(.*?)</graphic>"
But it fails for second group, saying: "Not enough )'s." How should I prevent it?
EDIT: First off, if you goal is to parse HTML or XML I strongly advise against it. If your goal is to learn or to surgically grab an element node then regex may, and I say may be a tool to use. I am answering this with the thought that you are using the html pattern to learn from....
I believe you have confused your data with your pattern and the regex pattern is failing.
I recommend these things
Don't use .*? to get text. It is too nebulous for the regex parser. Be more succinct in your pattern.
Since you know that the text is enclosed in quotes or by >xxx< use those as anchors.
Once anchors are determined extract the text
Place captured text into named capture groups.
How to get the text? Tell the regex parser to get everthing that is not an anchor character by using the set operation with the ^ (which means not when in a set [ ]) such as ([^\"]+) which says match everything that is not a quote.
Change your pattern to this which demonstrates the above suggestions:
string data = #"<graphic id=""8374932"">Translating Cowl (Inner/Outer Bondments</graphic>";
// \x22 is the hex escape for the quote, makes it easier to read.
string pattern = #"
(?:graphic\s+id=\x22) # Match but don't capture (MBDC) the beginning of the element
(?<ID>[^\x22]+) # Get all that is not a quote
(?:\x22>) # MBDC the quote
(?<Content>[^<+]+) # Place into the Content match capture group all text that is not + or <
(?:\</graphic) # MBDC The graphic";
// Ignore Pattern whitespace only allows us to comment, does not influence regex processing.
var mt = Regex.Match(data, pattern, RegexOptions.IgnorePatternWhitespace);
Console.WriteLine ("ID: {0} Content: {1}", mt.Groups["ID"], mt.Groups["Content"]);
// Outputs:
// ID: 8374932 Content: Translating Cowl (Inner/Outer Bondments
How to remove ,(comma) which is between "(double inverted comma) and "(double inverted comma). Like there is "a","b","c","d,d","e","f" and then from this, between " and " there is one comma which should be removed and after removing that comma it should be "a","b","c","dd","e","f" with the help of the regex in C# ?
EDIT: I forgot to specify that there may be double comma between quotes like "a","b","c","d,d,d","e","f" for it that regex does not work. and there can be any number of comma between quotes.
And there can be string like a,b,c,"d,d",e,f then there should be result like a,b,c,dd,e,f and if string like a,b,c,"d,d,d",e,f then result should be like a,b,c,ddd,e,f.
Assuming the input is as simple as your examples (i.e., not full-fledged CSV data), this should do it:
string input = #"a,b,c,""d,d,d"",e,f,""g,g"",h";
Console.WriteLine(input);
string result = Regex.Replace(input,
#",(?=[^""]*""(?:[^""]*""[^""]*"")*[^""]*$)",
String.Empty);
Console.WriteLine(result);
output: a,b,c,"d,d,d",e,f,"g,g",h
a,b,c,"ddd",e,f,"gg",h
The regex matches any comma that is followed by an odd number of quotation marks.
EDIT: If fields are quoted with apostrophes (') instead of quotation marks ("), the technique is exactly the same--except you don't have to escape the quotes:
string input = #"a,b,c,'d,d,d',e,f,'g,g',h";
Console.WriteLine(input);
string result = Regex.Replace(input,
#",(?=[^']*'(?:[^']*'[^']*')*[^']*$)",
String.Empty);
Console.WriteLine(result);
If some fields were quoted with apostrophes while others were quoted with quotation marks, a different approach would be needed.
EDIT: Probably should have mentioned this in the previous edit, but you can combine those two regexes into one regex that will handle either apostrophes or quotation marks (but not both):
#",(?=[^']*'(?:[^']*'[^']*')*[^']*$|[^""]*""(?:[^""]*""[^""]*"")*[^""]*$)"
Actually, it will handle simple strings like 'a,a',"b,b". The problem is that there would be nothing to stop you from using one of the quote characters in a quoted field of the other type, like '9" Nails' (sic) or "Kelly's Heroes". That's taking us into full-fledged CSV territory (if not beyond), and we've already established that we're not going there. :D
They're called regular expressions for a reason — they are used to process strings that meet a very specific and academic definition for what is "regular". It looks like you have some fairly typical csv data here, and it happens that csv strings are outside of that specific definition: csv data is not formally "regular".
In spite of this, it can be possible to use regular expressions to handle csv data. However, to do so you must either use certain extensions to normal regular expressions to make them Turing complete, know certain constraints about your specific csv data that is not promised in the general case, or both. Either way, the expressions required to do this are unwieldly and difficult to manage. It's often just not a good idea, even when it's possible.
A much better (and usually faster) solution is to use a dedicated CSV parser. There are two good ones hosted at code project (FastCSV and Linq-to-CSV), there is one (actually several) built into the .Net Framework (Microsoft.VisualBasic.TextFieldParser), and I have one here on Stack Overflow. Any of these will perform better and just plain work better than a solution based on regular expressions.
Note here that I'm not arguing it can't be done. Most regular expression engines today have the necessary extensions to make this possible, and most people parsing csv data know enough about the data they're handling to constrain it appropriately. I am arguing that it's slower to execute, harder to implement, harder to maintain, and more error-prone compared to a dedicated parser alternative, which is likely built into whichever platform you're using, and is therefore not in your best interests.
var input = "\"a\",\"b\",\"c\",\"d,d\",\"e\",\"f\"";
var regex = new Regex("(\"\\w+),(\\w+\")");
var output = regex.Replace(input,"$1$2");
Console.WriteLine(output);
You'd need to evaluate whether or not \w is what you want to use.
You can use this:
var result = Regex.Replace(yourString, "([a-z]),", "$1");
Sorry, after seeing your edits, regular expressions are not appropriate for this.
This should be very simple using Regex.Replace and a callback:
string pattern = #"
"" # open quotes
[^""]* # some not quotes
"" # closing quotes
";
data = Regex.Replace(data, pattern, m => m.Value.Replace(",", ""),
RegexOptions.IgnorePatternWhitespace);
You can even make a slight modification to allow escaped quotes (here I have \", and the comments explain how to use "":
string pattern = #"
\\. # escaped character (alternative is be """")
|
(?<Quotes>
"" # open quotes
(?:\\.|[^""])* # some not quotes or escaped characters
# the alternative is (?:""""|[^""])*
"" # closing quotes
)
";
data = Regex.Replace(data, pattern,
m => m.Groups["Quotes"].Success ? m.Value.Replace(",", "") : m.Value,
RegexOptions.IgnorePatternWhitespace);
If you need a single quote replace all "" in the pattern with a single '.
Something like the following, perhaps?
"(,)"