In my LaTeX work I need to do Regex search with \|(.*?)\| to capture |whatever| and replace it with \somecommand{$1}. But I do not want to capture || (That is, there is nothing between them.) How should I refine my regex search?
(By the way, what should my title be, so that it is useful for others?)
Change your regex to,
\|[^|]+\|
OR
\|.+\|
If you want to also capture pipes in between searched content
You have to change the asterix (which matches 0+ times) to a plus sign make the quantifier match at least 1 character.
\|(.+?)\|
^
Related
I want to use regex in notepad to find this pattern: "[0-9]+[\.][0-9]+[,][0-9]+" e.g. 1.010,80260
However from these kind of numbers I just want to remove the '.' , so the new value should be 1010,80260 .
So far I can only replace the whole pattern. Is there a way to do it?
Thank you in advance!
You can make use of the \K meta escape since PCRE doesn't support variable width lookbehinds:
regex:
[0-9]+\K[\.](?=[0-9]+[,][0-9]+)
[0-9]+ - capture digits
\K - forget what we've captured
[\.] - capture a period; just \. can be used, no need for the char class brackets
(?=[0-9]+[,][0-9]+) - ahead of me should be digits followed by a comma and digits
replace:
Nothing
\K is bugged in Notepad++ so you could use this regex instead since you only care that at least one digit is behind the period:
(?<=\d)\.(?=[0-9]+[,][0-9]+)
You can use \K, which basically says throw away whatever was matched up until that point, then add a lookahead. Like so
[0-9]+\K\.(?=[0-9]+[,][0-9]+)
Change the regular expression to: ([0-9]+)[\.]([0-9]+[,][0-9]+)
The () pieces are groups which you can refer to in the replace with \1 for the first group, and \2 for the second group.
The docs also explain this here: https://npp-user-manual.org/docs/searching/#substitution-grouping (even better, and in more detail, than my usage in this answer...)
EDIT: I just wanted to share the animated gif showing that 'Replace' in Notepad++ 7.9.5. does not seem to work.
I'm using the below regex string to match the word "kohls" which is located in a group of other words.
\W*((?i)kohls(?-i))\W*
It works great when the word is alone, but if the word is in a url, the match includes a period on both sides.
See the below examples:
Thank you for shopping at Kohls - returns a match for kohls.
https://www.kohls.com - returns a match for .kohls.
Edit. https://www.KohlsAndMichaels.com - doesn't return any match for kohls.
I want it to only extract the exact match for kohls without periods or any other symbols/text in front or behind it. Can you tell me what I'm doing wrong?
In cases like that you can always use a site like regex101.com, which explains the regular expression and shows the matches with colors. So this is how your regular expression currently works:
As you can see in blue color, the problem with the dots is in the \W*, which matches any non-word character. In order to fix this, you can use the following regular expression:
\b((?i)kohls(?-i))\b
The \b (before and after the word you want to match) is used to assert the position at a word boundary. See how this work on that website now:
If you still have questions, look at the explanation of the regular expression provided by that website. It is worth looking.
The \W metacharacter is used to find non-word characters. So adding a star operator will match 0 or more of these non-word characters (like periods). Did you meant to add a word boundary instead?
\b(?i)kohls(?-i)\b
Replace both \W* with [\W,\.\-]* etc.
Should be enough.
My organization has an in-house language, with syntax like:
cmo/create/mo1///tri
createpts/brick/xyz/2,2,2/0.,0.,0./1.,1.,1./1,1,1
I am writing a Vim syntax file, and would like to capture the first instance of a word enclosed by two characters (in this case, /), without capturing the characters themselves.
I.e., the regex would capture, from the lines above,
create
brick
My solution so far is to use this pattern:
[,/=" "].\{-}[,/=" "]
But from /this/and/this/and/this, it will capture /this/and/this/and/this/.
As you can see, the issue is two-fold: (i) my current solution is greedy, and (ii) captures the / characters as well, when I just want the words enclosed by these.
Thanks!
One possible solution:
^[^\/]\+\/\zs[^\/]\+\ze\/
^ anchor the search to the BOL,
[^\/]\+ one or more non-slash characters, as many as possible,
\/ a slash,
\zs start the match here,
[^\/]\+ one or more non-slash characters, as many as possible.
I'm trying to match first occurrence of window.location.replace("http://stackoverflow.com") in some HTML string.
Especially I want to capture the URL of the first window.location.replace entry in whole HTML string.
So for capturing URL I formulated this 2 rules:
it should be after this string: window.location.redirect("
it should be before this string ")
To achieve it I think I need to use lookbehind (for 1st rule) and lookahead (for 2nd rule).
I end up with this Regex:
.+(?<=window\.location\.redirect\(\"?=\"\))
It doesn't work. I'm not even sure that it legal to mix both rules like I did.
Can you please help me with translating my rules to Regex? Other ways of doing this (without lookahead(behind)) also appreciated.
The pattern you wrote is really not the one you need as it matches something very different from what you expect: text window.location.redirect("=") in text window.location.redirect("=") something. And it will only work in PCRE/Python if you remove the ? from before \" (as lookbehinds should be fixed-width in PCRE). It will work with ? in .NET regex.
If it is JS, you just cannot use a lookbehind as its regex engine does not support them.
Instead, use a capturing group around the unknown part you want to get:
/window\.location\.redirect\("([^"]*)"\)/
or
/window\.location\.redirect\("(.*?)"\)/
See the regex demo
No /g modifier will allow matching just one, first occurrence. Access the value you need inside Group 1.
The ([^"]*) captures 0+ characters other than a double quote (URLs you need should not have it). If these URLs you have contain a ", you should use the second approach as (.*?) will match any 0+ characters other than a newline up to the first ").
I'm searching the pattern (.*)\\1 on the text blabl with regexec(). I get successful but empty matches in regmatch_t structures. What exactly has been matched?
The regex .* can match successfully a string of zero characters, or the nothing that occurs between adjacent characters.
So your pattern is matching zero characters in the parens, and then matching zero characters immediately following that.
So if your regex was /f(.*)\1/ it would match the string "foo" between the 'f' and the first 'o'.
You might try using .+ instead of .*, as that matches one or more instead of zero or more. (Using .+ you should match the 'oo' in 'foo')
\1 is the backreference typically used for replacement later or when trying to further refine your regex by getting a match within a match. You should just use (.*), this will give you the results you want and will automatically be given the backreference number 1. I'm no regex expert but these are my thoughts based on my limited knowledge.
As an aside, I always revert back to RegexBuddy when trying to see what's really happening.
\1 is the "re-match" instruction. The question is, do you want to re-match immediately (e.g., BLABLA)
/(.+)\1/
or later (e.g., BLAahemBLA)
/(.+).*\1/