I need to extract some text that starts and ends with a double quote " but will not extract if it detects multiple double quotes.
This is my example
I tried using different look-arounds, positive/negative look-aheads and look-behinds, but it leads to an error.
In my example above, I would like to exclude the data
"XxXXXXX - "" """"XX""""""",
and
"XxXXXXX - ""XXXXX XXXXXXXX 1.4.90 """"X2""""""",
from being matched.
I saw some other answers here but I'm getting an error whenever I use a negative look-behind, no problems in positive look-ahead and negative look-ahead but it doesn't work.
Edit:
I've added some examples regex in the link provided, and also more example data.
However, I still don't want to match data above by the current regex.
What about using this:
"([^"]+?)"(,|$)
You can see it here
and also here
Thanks for this one. Strange, I think I've tried this one before. But didn't get the result I've expected. Maybe it's because I didn't wait for it to be matched again.
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$1FINISH would give you the following:
abcSTARTdefBEGINghiFINISHjkl
which would allow you to change your START/END tokens only when paired properly.
Each (x) is a group, but I have put (?:x) for each of the ones except the middle which marks it as a non-capturing group; the only one I left without a ?: was the middle; however, you could also conceivably capture the BEGIN/END tokens as well if you wanted to move them around or what-have-you.
See the Java regex documentation for full details on Java regexes.
I am trying to create a bunch of YAML files, mostly composed of strings of text. Now when using apostrophes in words, they must be escaped by typing a double apostrophe, because I’m using apostrophes to wrap the strings.
I want to create a regex that will check for apostrophes in the text that aren’t double. What I have is this:
^([^'\n]*?)'(([^'\n]*?)'(?!')([^'\n]+?))*?'$\n
https://regex101.com/r/v4nUTn/3
My issue is that as soon as my string has a double apostrophe, but also has an apostrophe which isn’t a double apostrophe, it doesn’t match because my negative lookahead doesn’t match as soon as it sees the double apostrophe. (for example the string t''e'st won’t match even though it is missing a double apostrophe after the e)
How can I make it so that my negative lookahead will not fail as soon as it sees one double apostrophe?
This regex should work:
\w'\w
Test here.
My guess is that maybe an expression similar to
('[^'\r\n]*'|[^\r\n\w']+)|([\w']*)
would be an option to look into.
If the second capturing group returns true, then the string is undesired.
If you wish to explore/simplify/modify the expression, it's been
explained on the top right panel of
regex101.com. If you'd like, you
can also watch in this
link, how it would match
against some sample inputs.
One suggestion would be to do this in two steps.
For example, if every 'candidate' value looks like this: - 'something here' (where you want to test the apostrophes in the something here content of the string, then first isolate out that content via:
/^\s*- '(.+)'$/im
And then make sure all apostrophe's appear as you want them to appear within match group 1 of the result.
Then, replace the original match with your 'sanitised' match.
Doing this means you don't have to be concerned with the bounding apostrophes causing complications to the check for apostrophes in the value.
Note: there may well be a perfect one-step regex to do this, but understanding that you can break tasks into several steps is useful if you spend a lot of time with regular expressions, and can help you sidestep 'perfect regex paralysis'.
If you want your string to match if there is at least one 'single quote' between your singlequote strings, then you should allow consumption of either a string which doesn't have any singlequote in it or consume if it contains two singlequotes and then you should modify your regex a bit to consume two singlequotes and add |'' in your regex, which will now consume either non-singlequote text or a portion which has at least two singlequotes.
Try this updated regex demo and see if this works like you wanted?
https://regex101.com/r/v4nUTn/4
How would I go about using Regex to extract the number from the following file:
abc_defg123_100aaa_abc_defg123
Where I want the 100 from the substring '_100aaa_'?
The closest have gotten is:
[0-9](?!(aaa_))*\w
but this matches up to the first underscore found!
Many thanks!
Try this:
(?<=_)\d+(?=aaa_)
See live demo.
This regex uses look arounds to assert, without capturing, the delimiting input either side of the target.
I have the following content with what I think are the possible cases of someone defining an link:
hello link what <a href=something.jpg>link</a>
I also have the following regular expression with a positive look behind:
(?<=href=["\'])something
The expression matches the word "something" in the first two links. In an attempt to capture the third instance of "something" in the link without any quotes, I thought making the ["\'] token optional (using ?) would capture it. The expression now looks like this:
(?<=href=["\']?)something
Unfortunately it now does not mach any of the instances of "something". What could I be doing incorrectly? I'm using http://gskinner.com/RegExr/ to test this out.
Many regex flavors only support fixed-length lookbehind assertions. If you have an optional token in your lookbehind, its length isn't fixed, rendering it invalid.
So the real question is: What regex flavor are you actually targeting with your regex?
Well, here I am back at regex and my poor understanding of it. Spent more time learning it and this is what I came up with:
/(.*)
I basically want the number in this string:
510973
My regex is almost good? my original was:
"/<a href=\"travis.php?theTaco(.*)\">(.*)<\/a>/";
But sometimes it returned me huge strings. So, I just want to get numbers only.
I searched through other posts but there is such a large amount of unrelated material, please give an example, resource, or a link directing to a very related question.
Thank you.
Try using a HTML parser provided by the language you are using.
Reason why your first regex fails:
[0-9999999] is not what you think. It is same as [0-9] which matches one digit. To match a number you need [0-9]+. Also .* is greedy and will try to match as much as it can. You can use .*? to make it non-greedy. Since you are trying to match a number again, use [0-9]+ again instead of .*. Also if the two number you are capturing will be the same, you can just match the first and use a back reference \1 for 2nd one.
And there are a few regex meta-characters which you need to escape like ., ?.
Try:
<a href=\"travis\.php\?theTaco=([0-9]+)\">\1<\/a>
To capture a number, you don't use a range like [0-99999], you capture by digit. Something like [0-9]+ is more like what you want for that section. Also, escaping is important like codaddict said.
Others have already mentioned some issues regarding your regex, so I won't bother repeating them.
There are also issues regarding how you specified what it is you want. You can simply match via
/theTaco=(\d+)/
and take the first capturing group. You have not given us enough information to know whether this suits your needs.