So I have the following:
^[a-zA-Z]+\b(myword+-)\b*
which I thought would match
^ start of string
[a-zA-Z] any alpha character
+ of one or more characters
\b followed by a word break
(myword+-) followed by myword which could include one or more special characters
\b followed by a word break
\* followed by anything at all
One: it does not work - it does not match anything
Two: any special characters included in {myword+-) throws an error
I could escape the special characters, but I don't know in advance what they might be, so I would have to escape all the possibilites, or perhaps I could just escape every character in {\m\y\w\o\r\d\\+\\-)
Edited to add:
Sorry, I knew I should have given more information
I have a series of strings to seach through in the form:
extra android-sdk and more that is of no interest
extra android-ndk and more that is of no interest
extra anjuta-extra and more that is of no interest
community c++-gtk-utils and more that is of no interest
and I have a list of items to search for in the strings:
android-sdk
android-ndk
extra
c++-gtk-utils
The item should only match if the second word in the string is an exact match to the item, so:
android-sdk will match the first string
android-ndk will match the second string
extra wuill NOT match the third string
c++-gtk-utils will match the fourth string
So (myword+-) is the item I am searching for "which could include one or more special characters"
Thanks for the help
Andrew
OK, with the help from above I worked it out.
This regex does exactly what I wanted, bear in mind that I am working in tcl (note the spaces to delimit the search word):
^[a-zA-Z]+\y extra \y *
where the search word is "extra".
It is necessary to escape any characters in the search string which may be interpreted by regex as qualifiers etc e.g +
So this will also work:
^[a-zA-Z]+\y dbus-c\+\+ \y *
Andrew
Strong recommendation: if you want to match literal strings, don't use regular expressions.
If we have this sample data:
set strings {
{extra android-sdk and more that is of no interest}
{extra android-ndk and more that is of no interest}
{extra anjuta-extra and more that is of no interest}
{community c++-gtk-utils and more that is of no interest}
}
set search_strings {
android-sdk
android-ndk
extra
c++-gtk-utils
}
Then, to find matches in the 2nd word of each string, we'll just use the eq string equality operator
foreach string $strings {
foreach search $search_strings {
if {[lindex [split $string] 1] eq $search} {
puts "$search matches $string"
}
}
}
outputs
android-sdk matches extra android-sdk and more that is of no interest
android-ndk matches extra android-ndk and more that is of no interest
c++-gtk-utils matches community c++-gtk-utils and more that is of no interest
If you insist on regular expression matching, you can escape any special characters to take away their usual regex meaning. Here, we'll take the brute force approach: any non-word chars will get escaped, so that the pattern may look like ^\S+\s+c\+\+\-gtk\-utils
foreach string $strings {
foreach search $search_strings {
set pattern "^\\S+\\s+[regsub -all {\W} $search {\\&}]"
if {[regexp $pattern $string]} {
puts "$search matches $string"
}
}
}
I was hoping to be able to make a portion of a regular expression to be a literal string, like
set pattern "^\\S+\\s+(***=$string)"
set pattern "^\\S+\\s+((?q)$string)"
but both failed.
Tcl regular expressions are documented at
https://www.tcl.tk/man/tcl8.6/TclCmd/re_syntax.htm
Also note your pattern ^[a-zA-Z]+\b(myword+-)\b* does not provide for any whitespace between the first and second words.
Disclaimer: Since your question lacks information what input and output is expected, I will give it a try to tell you why your Regex isn't working at all. Since it's not a full answer you might not want to mark it as accepted and possibly wait for someone to give you an example of working solution, as soon as you provide necessary information.
Notes:
quantifier characters (*, +, ? etc.) are applied to literal character or character class (a.k.a character group, namely characters/ranges inside [ ]) - when in your regex you write (myword+-) the only thing the + sign is applied to is letter 'd', nothing else.
what is myword in your regex? If you want a set of characters use [ ] combined with character ranges and/or character tokens such as \w (all word characters, such as letters and some special characters) or \d (all digit characters)
you also seem to misunderstand and misuse groups ("( )"), character classes ("[ ]") and quantifier notation ("{ }")
I am using Notepad++ to find (".*)"(.*) and replace it with \1\"\2 but it doesn't seem to work. I don't know why.
Example:
Someone said "My name is "sean""
I want it to be:
Someone said "My name is \"sean\""
Edit: In my case the closing quote is always on the end of line so will (".*)"(.*"$) work?
Edit2: Also the first quote is preceded with a comma so I will use (,".*)"(.*"$) though it may not work in some cases but I think it will work with my file.
Now there is the problem with the replace it doesn't add \" it just add some space.
It should work... you just need to do a little fixing...
The Find what regex should be ("[^"]*)("\w*)(")([^"]*")
The Replace with expression should be \1\\\2\\\3\4
Make sure you select the Search Mode to be "Regular expression"
Explanation...
This is quite tricky - I've assumed that the quoted text WITHIN quotes is just a single word. If you assume something else it becomes very hard to pin down.
You need to find a
" followed by
[^"]* - any number of characters that are NOT a " and then
("\w*)(") - a quoted word, and then finally
([^"]*") - any additional number of non-quote characters + a final quote
This is important because regular expression matching is greedy by default, and a .* would continue to match all characters, including " until the end of the string (see link )
In the replacement string you need to have \\ to represent a single \
I am using Notepad++ to remove some unwanted strings from the end of a pattern and this for the life of me has got me.
I have the following sets of strings:
myApp.ComboPlaceHolderLabel,
myApp.GridTitleLabel);
myApp.SummaryLabel + '</b></div>');
myApp.NoneLabel + ')') + '</label></div>';
I would like to leave just myApp.[variable] and get rid of, e.g. ,, );, + '...', etc.
Using Notepad++, I can match the strings themselves using ^myApp.[a-zA-Z0-9].*?\b (it's a bit messy, but it works for what I need).
But in reality, I need negate that regex, to match everything at the end, so I can replace it with a blank.
You don't need to go for negation. Just put your regex within capturing groups and add an extra .*$ at the last. $ matches the end of a line. All the matched characters(whole line) are replaced by the characters which are present inside the first captured group. .
matches any character, so you need to escape the dot to match a literal dot.
^(myApp\.[a-zA-Z0-9].*?\b).*$
Replacement string:
\1
DEMO
OR
Match only the following characters and then replace it with an empty string.
\b[,); +]+.*$
DEMO
I think this works equally as well:
^(myApp.\w+).*$
Replacement string:
\1
From difference between \w and \b regular expression meta characters:
\w stands for "word character", usually [A-Za-z0-9_]. Notice the inclusion of the underscore and digits.
(^.*?\.[a-zA-Z]+)(.*)$
Use this.Replace by
$1
See demo.
http://regex101.com/r/lU7jH1/5
I'm trying to write a regular expression which will match a string. For simplicity, I'm only concerned with double quote (") strings for the moment.
So far I have this: "\"[^\"]*\""
This works for most strings but fails when there is an escaped double quote such as this:
"a string \" with an escaped quote"
In this case, it only matches up to the escaped quote.
I've tried several things to allow an escaped quote but so far I've been unsuccessful, can anyone give me a hand?
I've managed to solve it myself:
"\"(\\.|[^\"\\])*\""
Try this:
"[^"\\\r\n]*(?:\\.[^"\\\r\n]*)*"
If you want a multi-line escaped string you can use:
"[^"\\]*(?:\\.[^"\\]*)*"
You need a negative lookbehind. Check if this works?
"\"[^\"]*(?<!\\)"
(?<!\\)" is supposed to match " that's not followed by \.
Try:
"((\\")|[^"(\\")])+"
From Regular Expression Library.
Usually you want to accept escaped anything.
" [^"\\]* (?: \\. [^"\\]* )* " would be the fastest.
"[^"\\]*(?:\\.[^"\\]*)*" compressed.
POSIX does not, AFAIK, support lookaround - without it, there is really no way to do this with just regular expressions. However, according to a POSIX emulator I have (no access to a native environment or library), This might get you close, in certain cases:
"[^\"]*"|"[^\]*\\|\\[^\"]*[\"]
it will capture the part before and the part after the escaped quote... with this source string (ignore the line breaks, an imagine it's all in one string):
I want to match "this text" and "This text, where there is an escaped
slash (\\), and an \"escaped quote\" (\")", but I also want to handle\\ escaped
back-slashes, as in "this text, with a \\ backslash: \\" -- with a little
text behind it!
it will capture these groups:
"this text" -- simple, quoted string
"This text, where there is an escaped slash (\ -- part 1 of quoted string
\), and an \ -- part 2
"escaped quote\ -- part 3
" (\ -- part 4
")" -- part 5, and ends with a quote
\\ -- not part of a quoted string
"this text, with a \ -- part 1 of quoted string
\ backslash: \ -- part 2
\" -- part 3, and ends with a quote
With further analysis you can combine them, as appropriate:
If the group starts and ends with a ", then it is fine on its own
If the group starts with a ", and ends with a \, then it needs to be IMMEDIATELY followed by another match group that either ends with a quote character itself, or recursively continues to be IMMEDIATELY followed by another match group
If the group does not immediately follow another match, it is not part of a quoted string
I think that's all the analysis that you need - but make sure to test it!!!
Let me know if this idea helps!
EDIT:
Additional note: just to be clear, for this to work all quotes in the entire source string must be escaped if they are not to be used as delimiters, and backslashes must be escaped everywhere as well
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