I'm using the following regex in c# to match some input cases:
^
(?<entry>[#])?
(?(entry)(?<id>\w+))
(?<value>.*)
$
The options are ignoring pattern whitespaces.
My input looks as follows:
hello
#world
[xxx]
This all can be tested here: DEMO
My problem is that this regex will not match the last line. Why?
What I'm trying to do is to check for an entry character. If it's there I force an identifier by \w+. The rest of the input should be captured in the last group.
This is a simplyfied regex and simplyfied input.
The problem can be fixed if I change the id regex to something like (?(entry)(?<id>\w+)|), (?(entry)(?<id>\w+))? or (?(entry)(?<id>\w+)?).
I try to understand why the conditional group doesn't match as stated in original regex.
I'm firm in regex and know that the regex can be simplyfied to ^(\#(?<id>\w+))?(?<value>.*)$ to match my needs. But the real regex contains two more optional groups:
^
(?<entry>[#])?
(\?\:)?
(\(\?(?:\w+(?:-\w+)?|-\w+)\))?
(?(entry)(?<id>\w+))
(?<value>.*)
$
That's the reason why I'm trying to use a conditional match.
UPDATE 10/12/2018
I tested a little arround it. I found the following regex that should match on every input, even an empty one - but it doesn't:
(?(a)a).*
DEMO
I'm of the opinion that this is a bug in .net regex and reported it to microsoft: See here for more information
There is no error in the regex parser, but in one's usage of the . wildcard specifier. The . specifier will consume all characters, wait for it, except the linefeed character \n. (See Character Classes in Regular Expressions "the any character" .])
If you want your regex to work you need to consume all characters including the linefeed and that can be done by specify the option SingleLine. Which to paraphrase what is said
Singline tells the parser to handle the . to match all characters including the \n.
Why does it still fail when not in singleline mode for the other lines are consumed? That is because the final match actually places the current position at the \n and the only option (as specified is use) is the [.*]; which as we mentioned cannot consume it, hence stops the parser. Also the $ will lock in the operations at this point.
Let me demonstrate what is happening by a tool I have created which illustrates the issue. In the tool the upper left corner is what we see of the example text. Below that is what the parser sees with \r\n characters represented by ↵¶ respectively. Included in that pane is what happens to be matched at the time in yellow boxes enclosing the match. The middle box is the actual pattern and the final right side box shows the match results in detail by listening out the return structures and also showing the white space as mentioned.
Notice the second match (as index 1) has world in group capture id and value as ↵.
I surmise your token processor isn't getting what you want in the proper groups and because one doesn't actually see the successful match of value as the \r, it is overlooked.
Let us turn on Singline and see what happens.
Now everything is consumed, but there is a different problem. :-)
I have a regex to match the following:
somedomain.com/services/something
Basically I need to ensure that /services is present.
The regex I am using and which is working is:
\/services*
But I need to match /services OR /servicos. I tried the following:
(\/services|\/servicos)*
But this shows 24 matches?! https://regex101.com/r/jvB1lr/1
How to create this regex?
The (\/services|\/servicos)* matches 0+ occurrences of /services or /servicos, and that means it can match an empty string anywhere inside the input string.
You can group the alternatives like /(services|servicos) and remove the * quantifier, but for this case, it is much better to use a character class [oe] as the strings only differ in 1 char.
You want to use the following pattern:
/servic[eo]s
See the regex demo
To make sure you match a whole subpart, you may append (?:/|$) at the pattern end, /servic[eo]s(?:/|$).
In C#, you may use Regex.IsMatch with the pattern to see if there is a match in a string:
var isFound = Regex.IsMatch(s, #"/servic[eo]s(?:/|$)");
Note that you do not need to escape / in a .NET regex as it is not a special regex metacharacter.
Pattern details
/ - a /
servic[eo]s - services or servicos
(?:/|$) - / or end of string.
Well the * quantifier means zero or more, so that is the problem. Remove that and it should work fine:
(\/services|\/servicos)
Keep in mind that in your example, you have a typo in the URL so it will correctly not match anything as it stands.
Here is an example with the typo in the URL fixed, so it shows 1 match as expected.
First off you specify C# (really .Net is the library which holds regex not the language) in this post but regex101 in your example is set to PHP. That is providing you with invalid information such as needed to escape a forward slash / with \/ which is unnecessary in .Net regular expressions. The regex language is the same but there are different tools which behave differently and php is not like .Net regex.
Secondly the star * on the ( ) is saying that there may be nothing in the parenthesis and your match is getting null nothing matches on every word.
Thirdly one does not need to split the whole word. I would just extract the commonality in the words into a set [ ]. That will allow the "or-ness" you need to match on either services or servicos. Such as
(/servic[oe]s)
Will inform you if services are found or not. Nothing else is needed.
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 currently using the following regular expression:
(?<!&)[^&;]*(?!;)
To match text like this:
match1<match2>
And extract:
match1
match2
However, this seems to match an extra five empty strings. See Regex Storm.
How can I only match the two listed above?
Note the existing pattern ((?<=^|;)[^&]+) by #xanatos will only match matches 1 to 3 in the following string and not match4:
match1<e;match2<match;3+match&4
Try changing the * to a +:
(?<!&)[^&;]+(?!;)
Test here
More correct regex:
(?<=^|;)[^&]+
Test here
The basic idea here is that a "good" substring starts at the beginning of the string (^) or right after the ;, and ends when you encounter a & ([^&]+).
Third version... But here we are showing how if you have a problem, and you decide to use regexes, now you have two problems:
(?<=^|;)([^&]|&(?=[^&;]*(?:&|$)))+
Test here
I have managed it with:
(?<Text>.+?)(?:&[^&;]*?;|$)
This seems to match all of the corner cases but it might not work with a case I can't think of at the moment.
This won't work if the string starts with a &...; pattern or is only that.
See Regex Storm.
I wrote a very simple regular expression that need to match the next pattern:
word.otherWord
- Word must have at least 2 characters and must not start with digit.
I wrote the next expression:
[a-zA-Z][a-zA-Z](.[a-zA-Z0-9])+
I tested it using Regex tester and it seems to be working at most of the cases but when I try some inputs that ends with 'e' it's not working.
for example:
Hardware.Make does not work but Hardware.Makee is works fine, why? How can I fix it?
That's because your regex looks for inputs which length is even.
You have two characters matched by [a-zA-Z][a-zA-Z] and then another two characters matched by (.[a-zA-Z0-9]) as a group which is repeated one or more times (because of +).
You can see it here: http://regex101.com/r/fW2bC1
I think you need that:
[a-zA-Z]+(\.[a-zA-Z0-9]+)+
Actually, the dot is a regex metacharacter, which stands for "any character". You'll need to escape the dot.
For your situation, I'd do this:
[a-zA-Z]{2,}\.[a-zA-Z0-9]+
The {2,} means, at least 2 characters from the previous range.
In regex, the dot period is one of the most commonly used metacharacters and unfortunately also commonly misused metacharacter. The dot matches a single character without caring what that character is...
So u would also re-write it like
[a-zA-Z]+(\.[a-zA-Z0-9]+)+