I have a feed in Yahoo Pipes and want to match everything after a question mark.
So far I've figured out how to match the question mark using..
\?
Now just to match everything that is after/follows the question mark.
\?(.*)
You want the content of the first capture group.
Try this:
\?(.*)
The parentheses are a capturing group that you can use to extract the part of the string you are interested in.
If the string can contain new lines you may have to use the "dot all" modifier to allow the dot to match the new line character. Whether or not you have to do this, and how to do this, depends on the language you are using. It appears that you forgot to mention the programming language you are using in your question.
Another alternative that you can use if your language supports fixed width lookbehind assertions is:
(?<=\?).*
With the positive lookbehind technique:
(?<=\?).*
(We're searching for a text preceded by a question mark here)
Input: derpderp?mystring blahbeh
Output: mystring blahbeh
Example
Basically the ?<= is a group construct, that requires the escaped question-mark, before any match can be made.
They perform really well, but not all implementations support them.
\?(.*)$
If you want to match all chars after "?" you can use a group to match any char, and you'd better use the "$" sign to indicate the end of line.
?(.*\n)+
With this you can get everything Even a new line
Check out this site: http://rubular.com/ Basically the site allows you to enter some example text (what you would be looking for on your site) and then as you build the regular expression it will highlight what is being matched in real time.
str.replace(/^.+?\"|^.|\".+/, '');
This is sometimes bad to use when you wanna select what else to remove between "" and you cannot use it more than twice in one string. All it does is select whatever is not in between "" and replace it with nothing.
Even for me it is a bit confusing, but ill try to explain it. ^.+? (not anything OPTIONAL) till first " then | Or/stop (still researching what it really means) till/at ^. has selected nothing until before the 2nd " using (| stop/at). And select all that comes after with .+.
Related
I've been looking around and could not make this happen. I am not totally noob.
I need to get text delimited by (including) START and END that doesn't contain START. Basically I can't find a way to negate a whole word without using advanced stuff.
Example string:
abcSTARTabcSTARTabcENDabc
The expected result:
STARTabcEND
Not good:
STARTabcSTARTabcEND
I can't use backward search stuff. I am testing my regex here: www.regextester.com
Thanks for any advice.
Try this
START(?!.*START).*?END
See it here online on Regexr
(?!.*START) is a negative lookahead. It ensures that the word "START" is not following
.*? is a non greedy match of all characters till the next "END". Its needed, because the negative lookahead is just looking ahead and not capturing anything (zero length assertion)
Update:
I thought a bit more, the solution above is matching till the first "END". If this is not wanted (because you are excluding START from the content) then use the greedy version
START(?!.*START).*END
this will match till the last "END".
START(?:(?!START).)*END
will work with any number of START...END pairs. To demonstrate in Python:
>>> import re
>>> a = "abcSTARTdefENDghiSTARTjlkENDopqSTARTrstSTARTuvwENDxyz"
>>> re.findall(r"START(?:(?!START).)*END", a)
['STARTdefEND', 'STARTjlkEND', 'STARTuvwEND']
If you only care for the content between START and END, use this:
(?<=START)(?:(?!START).)*(?=END)
See it here:
>>> re.findall(r"(?<=START)(?:(?!START).)*(?=END)", a)
['def', 'jlk', 'uvw']
The really pedestrian solution would be START(([^S]|S*S[^ST]|ST[^A]|STA[^R]|STAR[^T])*(S(T(AR?)?)?)?)END. Modern regex flavors have negative assertions which do this more elegantly, but I interpret your comment about "backwards search" to perhaps mean you cannot or don't want to use this feature.
Update: Just for completeness, note that the above is greedy with respect to the end delimiter. To only capture the shortest possible string, extend the negation to also cover the end delimiter -- START(([^ES]|E*E[^ENS]|EN[^DS]|S*S[^STE]|ST[^AE]|STA[^RE]|STAR[^TE])*(S(T(AR?)?)?|EN?)?)END. This risks to exceed the torture threshold in most cultures, though.
Bug fix: A previous version of this answer had a bug, in that SSTART could be part of the match (the second S would match [^T], etc). I fixed this but by the addition of S in [^ST] and adding S* before the non-optional S to allow for arbitrary repetitions of S otherwise.
May I suggest a possible improvement on the solution of Tim Pietzcker?
It seems to me that START(?:(?!START).)*?END is better in order to only catch a START immediately followed by an END without any START or END in between. I am using .NET and Tim's solution would match also something like START END END. At least in my personal case this is not wanted.
[EDIT: I have left this post for the information on capture groups but the main solution I gave was not correct.
(?:START)((?:[^S]|S[^T]|ST[^A]|STA[^R]|STAR[^T])*)(?:END)
as pointed out in the comments would not work; I was forgetting that the ignored characters could not be dropped and thus you would need something such as ...|STA(?![^R])| to still allow that character to be part of END, thus failing on something such as STARTSTAEND; so it's clearly a better choice; the following should show the proper way to use the capture groups...]
The answer given using the 'zero-width negative lookahead' operator "?!", with capture groups, is: (?:START)((?!.*START).*)(?:END) which captures the inner text using $1 for the replace. If you want to have the START and END tags captured you could do (START)((?!.*START).*)(END) which gives $1=START $2=text and $3=END or various other permutations by adding/removing ()s or ?:s.
That way if you are using it to do search and replace, you can do, something like BEGIN$1FINISH. So, if you started with:
abcSTARTdefSTARTghiENDjkl
you would get ghi as capture group 1, and replacing with BEGIN$1FINISH would give you the following:
abcSTARTdefBEGINghiFINISHjkl
which would allow you to change your START/END tokens only when paired properly.
Each (x) is a group, but I have put (?:x) for each of the ones except the middle which marks it as a non-capturing group; the only one I left without a ?: was the middle; however, you could also conceivably capture the BEGIN/END tokens as well if you wanted to move them around or what-have-you.
See the Java regex documentation for full details on Java regexes.
I have this regex which scans a text for the word very: (?i)(?:^|\W)(very)[\W$] which works. My goal is to upgrade it and avoid doing a match if very is within quotes, standalone or as part of a longer block.
Now, I have this other regex which is matching anything NOT inside curly quotes: (?<![\S"])([^"]+)(?![\S"]) which also works.
My problem is that I cannot seem to combine them. For example the string:
Fred Smith very loudly said yesterday at a press conference that fresh peas will "very, very defintely not" be served at the upcoming county fair. In this bit we have 3 instances of very but I'm only interested in matching the first one and ignore the whole Smith quotation.
What you describe is kind of tricky to handle with a regular expression. It's difficult to determine whether you are inside a quote. Your second regex is not effective as it only ignores the first very that is directly to the right of the quote and still matches the second one.
Drawing inspiration from this answer, that in turn references another answer that describes how to regex match a pattern unless ... I can capture the matches you want.
The basic idea is to use alternation | and match all the things you don't want and then finally match (and capture) what you do want in the final clause. Something like this:
"[^"]*"|(very)
We match quoted strings in the first clause but we don't capture them in a group and then we match (and capture) the word very in the second clause. You can find this match in the captured group. How you reference a captured group depends on your regex environment.
See this regex101 fiddle for a test case.
This regex
(?i)(?<!(((?<DELIMITER>[ \t\r\n\v\f]+)(")(?<FILLER>((?!").)*))))\bvery\b(?!(((?<FILLER2>((?!").)*)(")(?<DELIMITER2>[ \t\r\n\v\f]+))))
could work under two conditions:
your regex engine allows unlimited lookbehind
quotes are delimited by spaces
Try it on http://regexstorm.net/tester
I'm creating some reports in Google Analytics.
I am trying to write a RegEx that will match
www.website.com/
www.website.com/?_string_begins_with_question_mark
But will not match
www.website.com/string_doesnt_begin_with_question_mark
Using Reggy (with POSIX Extended), I tried to create an optional group that would match a string beginning with a question mark, followed by any number of characters. I thought
(\?.+)?
would do the trick, but it ignores the question mark requirement, and matches any string.
I tried some variations:
www.website.com/(\?(.+))?
www.website.com/(\?.+)?
www.website.com/(?.+)?
Et cetera.
Any help is appreciated - Sorry if this has already been asked! I'm new to RegEx.
Thank you!
Your regexp
www.website.com/(\?.+)?
will still match anything that contains www.website.com/, no matter what comes after the slash. Have you tried appending a $ (end-of-input marker)?
www\.website\.com/(?:\?.+)?$
(Escape the dots for more precision; the ?: is just a way of indicating that the group is of no special meaning and does not have to be remembered -- if you omit this, you can access the contents of the group by \1, e.g., in a replace operation.)
Try this one
www.website.com/\?(.+)
I would suggest
www.website.com/($|\?.+$)
Either empty or question mark plus some string
I've been looking around and could not make this happen. I am not totally noob.
I need to get text delimited by (including) START and END that doesn't contain START. Basically I can't find a way to negate a whole word without using advanced stuff.
Example string:
abcSTARTabcSTARTabcENDabc
The expected result:
STARTabcEND
Not good:
STARTabcSTARTabcEND
I can't use backward search stuff. I am testing my regex here: www.regextester.com
Thanks for any advice.
Try this
START(?!.*START).*?END
See it here online on Regexr
(?!.*START) is a negative lookahead. It ensures that the word "START" is not following
.*? is a non greedy match of all characters till the next "END". Its needed, because the negative lookahead is just looking ahead and not capturing anything (zero length assertion)
Update:
I thought a bit more, the solution above is matching till the first "END". If this is not wanted (because you are excluding START from the content) then use the greedy version
START(?!.*START).*END
this will match till the last "END".
START(?:(?!START).)*END
will work with any number of START...END pairs. To demonstrate in Python:
>>> import re
>>> a = "abcSTARTdefENDghiSTARTjlkENDopqSTARTrstSTARTuvwENDxyz"
>>> re.findall(r"START(?:(?!START).)*END", a)
['STARTdefEND', 'STARTjlkEND', 'STARTuvwEND']
If you only care for the content between START and END, use this:
(?<=START)(?:(?!START).)*(?=END)
See it here:
>>> re.findall(r"(?<=START)(?:(?!START).)*(?=END)", a)
['def', 'jlk', 'uvw']
The really pedestrian solution would be START(([^S]|S*S[^ST]|ST[^A]|STA[^R]|STAR[^T])*(S(T(AR?)?)?)?)END. Modern regex flavors have negative assertions which do this more elegantly, but I interpret your comment about "backwards search" to perhaps mean you cannot or don't want to use this feature.
Update: Just for completeness, note that the above is greedy with respect to the end delimiter. To only capture the shortest possible string, extend the negation to also cover the end delimiter -- START(([^ES]|E*E[^ENS]|EN[^DS]|S*S[^STE]|ST[^AE]|STA[^RE]|STAR[^TE])*(S(T(AR?)?)?|EN?)?)END. This risks to exceed the torture threshold in most cultures, though.
Bug fix: A previous version of this answer had a bug, in that SSTART could be part of the match (the second S would match [^T], etc). I fixed this but by the addition of S in [^ST] and adding S* before the non-optional S to allow for arbitrary repetitions of S otherwise.
May I suggest a possible improvement on the solution of Tim Pietzcker?
It seems to me that START(?:(?!START).)*?END is better in order to only catch a START immediately followed by an END without any START or END in between. I am using .NET and Tim's solution would match also something like START END END. At least in my personal case this is not wanted.
[EDIT: I have left this post for the information on capture groups but the main solution I gave was not correct.
(?:START)((?:[^S]|S[^T]|ST[^A]|STA[^R]|STAR[^T])*)(?:END)
as pointed out in the comments would not work; I was forgetting that the ignored characters could not be dropped and thus you would need something such as ...|STA(?![^R])| to still allow that character to be part of END, thus failing on something such as STARTSTAEND; so it's clearly a better choice; the following should show the proper way to use the capture groups...]
The answer given using the 'zero-width negative lookahead' operator "?!", with capture groups, is: (?:START)((?!.*START).*)(?:END) which captures the inner text using $1 for the replace. If you want to have the START and END tags captured you could do (START)((?!.*START).*)(END) which gives $1=START $2=text and $3=END or various other permutations by adding/removing ()s or ?:s.
That way if you are using it to do search and replace, you can do, something like BEGIN$1FINISH. So, if you started with:
abcSTARTdefSTARTghiENDjkl
you would get ghi as capture group 1, and replacing with BEGIN$1FINISH would give you the following:
abcSTARTdefBEGINghiFINISHjkl
which would allow you to change your START/END tokens only when paired properly.
Each (x) is a group, but I have put (?:x) for each of the ones except the middle which marks it as a non-capturing group; the only one I left without a ?: was the middle; however, you could also conceivably capture the BEGIN/END tokens as well if you wanted to move them around or what-have-you.
See the Java regex documentation for full details on Java regexes.
Here's the input string:
loadMedia('mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', 'http://www.something.com/videos/JohnsAwesomeCaption.xml', '/videos/video-splash-image.gif)
With this RegExp: \'.+.xml\'
... we get this:
'mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', 'http://www.something.com/videos/JohnsAwesomeCaption.xml'
... but I want to extract only this:
http://www.something.com/videos/JohnsAwesomeCaption.xml
Any suggestions? I'm sure this problem has been asked before, but it's difficult to search for. I'll be happy to Accept a solution.
Thanks!
If you want to get everything within quotes that starts with http:
(?<=')http:[^']+(?=')
If you only want those ending with .xml
(?<=')http:[^']+\.xml(?=')
It doesn't select the quotation marks (as you asked)
It's fast!
Fair warning: it only works if the regex engine you're using can handle lookbehind
Knowing the language would be helpful. Basically, you are having a problem because the + quantifier is greedy, meaning it will match the largest part of the string that it can. you need to use a non-greedy quantifier, which will match as little as possible.
We will need to know the language you're in to know what the syntax for the non-greedy quantifier should be.
Here is a perl recipe. Just as a sidenote, instead of .+, you probably want to match [^.]+.xml.
\'.+?.xml\'
should work if your language supports perl-like regexes.
This should work (tested in javascript, but pretty sure it would work in most cases)
'[^']+?\.xml'
it looks for these rules
starts with '
is followed by anything but '
ends in .xml'
you can demo it at http://RegExr.com?2tp6q
in .net this regex works for me:
\'[\w:/.]+\.xml\'
breaking it down:
a ' character
followed by a word character or ':' or '/' or '.' any number of times (which matches the url bit)
followed by '.xml' (which differentiates the sought string from the other urls which it will match without this)
followed by another ' character
I tested it here
Edit
I missed that you don't want the quotes in the result, in which case as has been pointed out you need to use look behind and look ahead to include the quotes in the search, but not in the answer. again in .net:
(?<=')[\w:/.]+\.xml(?=')
but I think the best solution is a combination of those offered already:
(?<=')[^']+\.xml(?=')
which seems the simplest to read, at least to me.