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
I am trying to open a pdf file that is located on the network :
I call the file like this in c# (Sorry for sending my code as a picture because of breakpoint i have to)
But it can't find the path .Another thing that i should add is when i call the file outside the c# like this \\127.0.0.1\dccfile\test\dcc1\1.pdf it works .
The value you're looking at in the debugger tooltip is a C# literal, not a string. C# literals delimit strings with straight quotes " and escapes metacharacters with backslashes \. See the quotes at the start and end of the literal in the tooltip? They're not part of the string. Backslashes are C# metacharacters, to include one in a string you have to precede it with another backslash. The C# literal "\\" encodes a string containing a single backslash character. The first \ you see in "\\127.0.0... is a metacharacter telling C# that the next character is a literal backslash, not a metacharacter. The code "\\127.0.0.1\\DCCFile\\test\\dcc1\\1.pdf" you see in the tooltip encodes the C# string \127.0.0.1\DCCFile\test\dcc1\1.pdf with no quotes and single backslashes.
Your problem is the value of Configuration.AccountDetail.DCCFileAddress needs to start with two backslashes and it does not.
Your code pathString.Replace(#"\\", #"\") will have no effect because there are no double backslashes in your string; the debugger is displaying the backslashes doubled so you know they are literal backslashes and not metacharacters.
Hey I have an issue with Regex.Escape I'm trying to feed it an Email from TextBox Controll. The function recieves "test#test.test". What I expect to get is this "test#test\.test" Regex.Escape escapes the dot character. Hovever what I get instead is "test#test\\.test" which is very confusing. I plan on handing that string down to an SQL query and I'm worried abut users misbehaving.
holder.address = Regex.Escape(EmailAddressInput.Text);
This is how I assign resulting string to field in holder class.
I have been researching this problem on my own but most sources (including MSDN) suggest to prefix the dot ("the special character") with one backslash.
As it is right now backslash escapes backslash and result is a badly formatted email address.
var s = "test#test\\.test"; means the s holds the test#test\.test string. Your issue does not exist. There is a single backslash. Click the magnifier button on the right - you will see that in the Text Visualizer.
Regex has to have \\ because its escaping the \
the string itself actually only has one \ in it.
string DownloadDirectoryPath = #"C:\\Program Files\\companyname\\productname\\username\\0\\2012081617085746"
(this is the path i get from the sql server)
but my application uses single slashes so I try to use
DownloadDirectoryPath= DownloadDirectoryPath.Replace(#"\\", #"\");
but this doesn't work and I get the same string.
any advice?
PLEASE NOTICE:
the value above is what i see in watch window
You've not got a # in front of your value for DownloadDirectoryPath, so it's not actually got any \\'s in it, only \'s
Do a console.WriteLine(DownloadDirectoryPath) to check what it really has in it.
Edit (OP updated question):
If you hover over the variable containing it (or use the watch window) while debugging then VS will show a single \ as \\ to disambiguate it from an escape character. Write it to the console, a file or some other output to check what it really is.
The # tells the compiler to read the string Verbatim. The person that wrote the SQL intended for the string to be interpreted. In order to work with what you have remove the #
string DownloadDirectoryPath = "C:\\Program Files\\companyname\\productname\\username\\0\\2012081617085746"
Also note that when you see a verbatim string in the debugger watch window, you will see the escape characters that were added by the compiler, not the verbatim version from your source code.
Today I found out that putting strings in a resource file will cause them to be treated as literals, i.e putting "Text for first line \n Text for second line" will cause the escape character itself to become escaped, and so what's stored is "Text for first line \n Text for second line" - and then these come out in the display, instead of my carriage returns and tabs
So what I'd like to do is use string.replace to turn \\ into \ - this doesn't seem to work.
s.Replace("\\\\", "\\");
doesn't change the string at all because the string thinks there's only 1 backslash
s.Replace("\\", "");
replaces all the double quotes and leaves me with just n instead of \n
also, using # and half as many \ chars or the Regex.Replace method give the same result
anyone know of a good way to do this without looping through character by character?
Since \n is actually a single character, you cannot acheive this by simply replacing the backslashes in the string. You will need to replace each pair of \ and the following character with the escaped character, like:
s.Replace("\\n", "\n");
s.Replace("\\t", "\t");
etc
You'd be better served adjusting the resx files themselves. Line breaks can be entered via two mechanisms: You can edit the resx file as XML (right-click in Solution Explorer, choose "Open As," and choose XML), or you can do it in the designer.
If you do it in the XML, simply hit Enter, backspace to the beginning of the newline you've created, and you're done. You could do this with Search and Replace, as well, though it will be tricky.
If you use the GUI resx editor, holding down SHIFT while pressing ENTER will give you a line break.
You could do the run-time replacement thing, but as you are discovering, it's tricky to get going -- and in my mind constitutes a code smell. (You can also make a performance argument, but that would depend on how often string resources are called and the scale of your app in general.)
I'm actually going with John's solution and editing the XML directly as that's the better solution for the project, but codelogic answered the question that was driving me insane.