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
nextLink1.replace(""",()), so basically I want to replace " with a blank. Any help would be greatly appreciated.
Thanks
You need to escape the " sign. Like this:
nextLink1.replace("\"","");
The compiler will recognize the first two quote marks, but the third one will produce a syntax error.
Using an escape sequence will place a double quote as such:
nextLink1.replace("\"","");
You can find more escape sequences here http://docs.oracle.com/javase/tutorial/java/data/characters.html
" is Java's metacharacter used to start or end Strings literals. If you want to use it inside String literal you need to escape it first with \ like \" (which is another Java's metacharacter used for example to create new lines mark "\n").
Also blank String is not () but "". So try this way
nextLink1.replace("\"","");
BTW Strings are immutable which means this method will not affect original String, but create new one with replaced character. If you want nextLink1 to contain String with replaced characters you will need to use
nextLink1 = nextLink1.replace("\"","");
I am looking to write a regex that can remove any characters upto the first &emsp and if there is a (new section) following &emsp then remove that as well. But the following regex doesn't seem to work. Why? How do I correct this?
String removeEmsp =" “[<centd>[</centd>]§ 431:10A–126 (new section)[<centd>]Chemotherapy services.</centd>] <centa>Cancer treatment.</centa>test snl.";
Pattern removeEmspPattern1 = Pattern.compile("(.*( (\\(new section\\)))?)(.*)", Pattern.MULTILINE);
System.out.println(removeEmspPattern1.matcher(removeEmsp).replaceAll("$2"));
Have you tried String Split? This creates an array of strings from a string, based on a deliminator.
Once you have the string split, just select the elements of the array that you need for print statement.
Read more here
Your regex is very long and I do not want to debug it. However the tip is that some characters have special meaning in regular expressions. For example & means "and". Squire brackets allow defining characters groups etc. Such characters must be escaped if you want them to be interpreted as just characters and not regex commands. To escape special character you have to write \ in front of it. But \ is escape character for java too, so it should be duplicate.
For example to replace ampersand by letter A you should write str.replaceAll("\\&", "A")
Now you have all information you need. Try to start from simpler regex and then expand it to what you need. Good luck.
EDIT
BTW parsing XML and/or HTML using regular expressions is possible but is highly not recommended. Use special parser for such formats.
Try this:
String removeEmsp =" “[<centd>[</centd>]§ 431:10A–126 (new section)[<centd>]Chemotherapy services.</centd>] <centa>Cancer treatment.</centa>test snl.";
System.out.println(removeEmsp.replaceFirst("^.*?\\ (\\(new\\ssection\\))?", ""));
System.out.println(removeEmsp.replaceAll("^.*?\\ (\\(new\\ssection\\))?", ""));
Output:
[<centd>]Chemotherapy services.</centd>] <centa>Cancer treatment.</centa>test snl.
[<centd>]Chemotherapy services.</centd>] <centa>Cancer treatment.</centa>test snl.
It will remove everything up to " " and optionally, the following "(new section)" text if any.
I need to remove comments from a String. Comments are specified using quotes. For example:
removeComments("1+4 'sum'").equals("1+4 ");
removeComments("this 'is not a comment").equals("this 'is not a comment");
removeComments("1+4 'sum' 'unclosed comment").equals("1+4 'unclosed comment");
I could iterate through the characters of the String, keeping track of the indexes of the quotes, but I would like to know if there is a simpler solution (maybe a regex?)
You can use replaceAll:
str = str.replaceAll("\\'.*?\\'", "");
This will replace the first and the second ' and everything between them with "" (Thus, will remove them).
Edit: As stated on the comments, there is no need to backslash the single quote.
If you don't need to be able to have quotes inside the comment, this will do it:
input.replaceAll("'[^']+'", "");
It matches a quote, at least one of anything that isn't a quote, then a quote.
Working example
I want to convert the directory path from:
C:\Users\Host\Desktop\picture.jpg
to
C:\\Users\\Host\\Desktop\\picture.jpg
I am using replaceAll() function and other replace functions but they do not work.
How can I do this?
I have printed the statement , it gives me the one which i wanted ie
C:\Users\Host\Desktop\picture.jpg
but now when i pass this variable to open the file, i get this exception why?
java.io.FileNotFoundException: C:\Users\Host\Desktop\picture.jpg
EDIT: Changed from replaceAll to replace - you don't need a regex here, so don't use one. (It was a really poor design decision on the part of the Java API team, IMO.)
My guess (as you haven't provided enough information) is that you're doing something like:
text.replace("\\", "\\\\");
Strings are immutable in Java, so you need to use the return value, e.g.
String newText = oldText.replace("\\", "\\\\");
If that doesn't answer your question, please provide more information.
(I'd also suggest that usually you shouldn't be doing this yourself anyway - if this is to include the information in something like a JSON response, I'd expect the wider library to perform escaping for you.)
Note that the doubling is required as \ is an escape character for Java string (and character) literals. Note that as replace doesn't treat the inputs as regular expression patterns, there's no need to perform further doubling, unlike replaceAll.
EDIT: You're now getting a FileNotFoundException because there isn't a filename with double backslashes in - what made you think there was? If you want it as a valid filename, why are you doubling the backslashes?
You have to use :
String t2 = t1.replaceAll("\\\\", "\\\\\\\\");
or (without pattern) :
String t2 = t1.replace("\\", "\\\\");
Each "\" has to be preceeded by an other "\". But it's also true for the preceeding "\" so you have to write four backslashes each time you want one in regex.
In strings \ is bydefault used as escape character therefore in order to select "\" in a string you have to use "\" and for "\" (i.e blackslack two times) use "\\". This will solve your problem and thos will also apply to other symbols also like "
Two explanations:
1. Replace double backslashes to one (not what you asked)
You have to escape the backslash by backslashes. Like this:
String newPath = oldPath.replaceAll("\\\\\\\\", "\\");
The first parameter needs to be escaped twice. Once for the Java Compiler and once because you use regular expressions. So you want to replace two backslashes by one. So, since we have to escape a backslash add one backslash. Now you have \\. This will be compiled to \. BUT!! you have to escape the backslash once again because the first parameter of the replaceAll method uses regular expressions. So to escape it, add a backslash, but that backslash needs to be escaped, so we get \\\\. These for backslashes represents one backslash in the regex. But you want to replace the double backslash to one. So use 8 backslashes.
The second parameter of the replaceAll method isn't using regular expressions, but it has to be escaped as well. So, you need to escape it once for the Java Compiler and once for the replace method: \\\\. This is compiled to two backslashes, which are being interpreted as 1 backslash in the replaceAll method.
2. Replace single backslash to a pair of backslashes (what you asked)
String newPath = oldPath.replaceAll("\\\\", "\\\\\\\\");
Same logic as above.
3. Use replace() instead of replaceAll().
String newPath = oldPath.replace("\\", "\\\\");
The difference is that the replace() method doesn't use regular expressions, so you don't have to escape every backslash twice for the first parameter.
Hopefully, I explained well...
-- Edit: Fixed error, as pointed out by xehpuk --