I found this related question : In perl, backreference in replacement text followed by numerical literal
but it seems entirely different.
I have a regex like this one
s/([^0-9])([xy])/\1 1\2/g
^
whitespace here
But that whitespace comes up in the substitution.
How do I not get the whitespace in the substituted string without having perl confuse the backreference to \11?
For eg.
15+x+y changes to 15+ 1x+ 1y.
I want to get 15+1x+1y.
\1 is a regex atom that matches what the first capture captured. It makes no sense to use it in a replacement expression. You want $1.
$ perl -we'$_="abc"; s/(a)/\1/'
\1 better written as $1 at -e line 1.
In a string literal (including the replacement expression of a substitution), you can delimit $var using curlies: ${var}. That means you want the following:
s/([^0-9])([xy])/${1}1$2/g
The following is more efficient (although gives a different answer for xxx):
s/[^0-9]\K(?=[xy])/1/g
Just put braces around the number:
s/([^0-9])([xy])/${1}1${2}/g
I am trying to find a pattern match as below
abc(xxxx):efg(xxxx):xyz(xxxx) where xxxx - [0-9] digits
I used
set string "my string is abc(xxxx):efg(xxxx):xyz(xxxx)"
regexp abc(....):efg(....):xyz(....) $string result_str
it returns 0. Can anyone help?
The problem you've got is that ( and ) have special meaning to regular expressions in Tcl (and many other RE engines besides) in that they denote a capturing sub-RE. To make the characters “normal”, they have to be escaped with a backslash, and that means that it's best to put the regular expression in braces (because backslashes are general Tcl metacharacters).
Thus:
% set string "my string is abc(xxxx):efg(xxxx):xyz(xxxx)"
% regexp {abc\(....\):efg\(....\):xyz\(....\)} $string
1
If you want to also capture the contents of those parentheses, you need a slightly more complex RE:
regexp {abc\((....)\):efg\((....)\):xyz\((....)\)} $string \
all abc_bit efg_bit xyz_bit
Note that those .... sequences always match exactly four characters, but it's better to be more specific. To match any number of digits in each case:
regexp {abc\((\d+)\):efg\((\d+)\):xyz\((\d+)\)} $string -> abc efg xyz
When using regexp to extract bits of a string, it's pretty common to use -> as a (rather strange) variable name for the whole string match; it looks mnemonically like it's saying “send the pieces extracted to these variables”.
Not worked with tcl but seems like you need to escape the ( and ). Also if you are sure that the x's would be digits, use \d{4} instead of ..... Based on this, the updated regex you could try is
abc\(\d{4}\):efg\(\d{4}\):xyz\(\d{4}\).
I need a regular expression that will find a number(s) that is not inside parenthesis.
Example abcd 1 (35) (df)
It would only see the 1.
Is this very complex? I've tried and had no luck.
Thanks for any help
An easy solution is to first remove the unwanted values:
my $string = "abcd 12 (35) (df) 2311,22";
$string =~ s/\(\d+\)//g; # remove numbers within parens
my #numbers = $string =~ /\d+/g; # extract the numbers
This is quite hard but something like this will probably do:
^(?:\()(\d+)(?:[^)])|(?:[^(0-9]|^)(\d+)(?:[^)0-9]|^)|(?:[^(])(\d+)(?:\))$
The problem is to match (123, 123) and also to not match the string 123 as the number 2 between the non-parentheses characters 1 and 3. Also there are probably some edge cases for start of and end of string.
My suggestion is to not use a regex for this. Maybe a regex that matches numbers and then use the capture info to check if the surrounding characters are not parentheses.
The regular expression would be:
^[a-z]+ ([0-9]+) \([0-9]+\) \([a-z]+\)$
The result is the first (and only) matching group of the regex.
Maybe you want to remove the ^ and $ if the regex should not match only if it’s the content of a whole single line. You can also use [a-zA-Z] or [[:alpha:]]. This depends on the regular expression engine you use and, of course, the content you want to match.
Example perl code:
if (m/^[a-z]+ ([0-9]+) \([0-9]+\) \([a-z]+\)$/) {
print("$1\n");
}
Please note that your question contains not enough information to make a good answer possible (you did not say anything about the general format of your expression, for example if you want to match integers or floating points)
How about
/(?:^|[^\d(])(\d+)(?:[^\d)]|$)/
? This matches a string of digits (\d+) that are
preceded by the beginning of the string, or a character that is not a digit or an open parenthesis ((?:^|[^\d(]))
succeeded by the end of the string, or by a character that is not a digit or a close parenthesis ((?:[^\d)]|$))
Given a string of identifiers separated by :, is it possible to construct a regular expression to extract the unique identifiers into another string, also separated by :?
How is it possible to achieve this using a regular expression? I have tried s/(:[^:])(.*)\1/$1$2/g with no luck, because the (.*) is greedy and skips to the last match of $1.
Example: a:b:c:d:c:c:x:c:c:e:e:f should give a:b:c:d:x:e:f
Note: I am coding in perl, but I would very much appreciate using a regex for this.
In .NET which supports infinite repetition inside lookbehind, you could search for
(?<=\b\1:.*)\b(\w+):?
and replace all matches with the empty string.
Perl (at least Perl 5) only supports fixed-length lookbehinds, so you can try the following (using lookahead, with a subtly different result):
\b(\w+):(?=.*\b\1:?)
If you replace that with the empty string, all previous repetitions of a duplicate entry will be removed; the last one will remain. So instead of
a:b:c:d:x:e:f
you would get
a:b:d:x:c:e:f
If that is OK, you can use
$subject =~ s/\b(\w+):(?=.*\b\1:?)//g;
Explanation:
First regex:
(?<=\b\1:.*): Check if you can match the contents of backreference no. 1, followed by a colon, somewhere before in the string.
\b(\w+):?: Match an identifier (from a word boundary to the next :), optionally followed by a colon.
Second regex:
\b(\w+):: Match an identifier and a colon.
(?=.*\b\1:?): Then check whether you can match the same identifier, optionally followed by a colon, somewhere ahead in the string.
Check out: http://www.regular-expressions.info/duplicatelines.html
Always a useful site when thinking about any regular expression.
$str = q!a:b:c:d:c:c:x:c:c:e:e:f!;
1 while($str =~ s/(:[^:]+)(.*?)\1/$1$2/g);
say $str
output :
a:b:c:d:x:e:f
here's an awk version, no need regex.
$ echo "a:b:c:d:c:c:x:c:c:e:e:f" | awk -F":" '{for(i=1;i<=NF;i++)if($i in a){continue}else{a[$i];printf $i}}'
abcdxef
split the fields on ":", go through the splitted fields, store the elements in an array. check for existence and if exists, skip. Else print them out. you can translate this easily into Perl code.
If the identifiers are sorted, you may be able to do it using lookahead/lookbehind. If they aren't, then this is beyond the computational power of a regex. Now, just because it's impossible with formal regex doesn't mean it's impossible if you use some perl specific regex feature, but if you want to keep your regexes portable you need to describe this string in a language that supports variables.
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
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
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.
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