We have a regular expression containing (?!\s) but the regex engine used does not allow to use lookahead assertions. Complete regex is
^(?!\s)(.*)(\S)$
Can anyone please suggest any alternative ways of achieving same functionality with out using a lookahead.
You may use
^\S(.*\S)?$
It will match
^ - start of string
\S - a non-whitespace char
(.*\S)? - 1 or 0 occurrences of
.* - any 0+ chars other than line break chars, as many as possible
\S - a non-whitespace char
$ - end of string.
See a regex demo using the Go regex engine (RE2) that does not allow lookaheads.
Related
I'm trying to find a solution to a regex that can match anything after a string or nothing, but if there's something it can't be a dot .
is it possible to do without negative lookahead?
here's an example regex:
.*\.(cpl)[^.].*
now the string:
C:\Windows\SysWOW64\control.exe mlcfg32.cpl sounds
this one is matched, but if there's only:
C:\Windows\SysWOW64\control.exe mlcfg32.cpl
it's not matched because due to the dot blacklist it's searching for any character after cpl,if i use ? after the [^.] however it won't blacklist the . in case there's something else after, so it will capture this even if it shouldn't:
C:\Windows\SysWOW64\control.exe mlcfg32.cpl. sounds
can it be done without using negative lookaheads? - ?!
You may use this regex:
.*\.cpl(?:[^.].*|$)
RegEx Demo
RegEx Breakdown:
.*: Match 0 or more of any character
\.cpl: Match .cpl
(?:[^.].*|$): Match end of string or a non-dot followed by any text
You can use
.*\.(cpl)(?:[^.].*)?$
See the regex demo. Details:
.* - zero or more chars other than line break chars as many as possible
\. - a dot
(cpl) - Group 1: cpl
(?:[^.].*)? - an optional non-capturing group that matches a char other than . char and then zero or more chars other than line break chars as many as possible
$ - end of string.
I have this kind of text:
other text opt1 opt2 opt3 I_want_only_this_text because_of_this
And am using this regex:
(?<=opt1|opt2|opt3).*?(?=because_of_this)
Which returns me:
opt2 opt3 I_want_only_this_text
However, I want to match only "I_want_only_this_text".
What is the best way to achieve this?
I don't know in what order the opt's will appear and they are only examples. Actual words will be different and there will be more of them.
Test screenshot
Actual data:
regex
(?<=※|を|備考|町|品は|。).*(?=のお届けとなります|でお届けします|にてお届け致します|にてお届けいたします)
text
こだわり豚には通常の豚よりビタミンB1が2倍以上あります。私たちの育てた愛情たっぷりのこだわり豚をぜひ召し上がってください。商品説明名称えびの産こだわり豚切落し産地宮崎県えびの市内容量500g×8パック合計4kg賞味期限90日保存方法-15℃以下で保存すること提供者株式会社さつま屋産業備考・本お礼品は冷凍でのお届けとなります
what I want to get:
冷凍で
You can use
(?<=※|を|備考|町|品は|。)(?:(?!※|を|備考|町|品は|。).)*?(?=のお届けとなります|でお届けします|にてお届け致します|にてお届けいたします)
See the regex demo. The scheme is the same as in (?<=opt1|opt2|opt3)(?:(?!opt1|opt2|opt3).)*?(?=because_of_this) (see demo).
The tempered greedy token solution allows you to match multiple occurrences of the same pattern in a longer string.
Details
(?<=※|を|備考|町|品は|。) - a positive lookbehind that matches a location that is immediately preceded with one of the alternatives listed in the lookbehind
(?:(?!※|を|備考|町|品は|。).)*? - any char other than a line break char, zero or more but as few as possible occurrences, that is not a starting point of any of the alternative patterns in the negative lookahead
(?=のお届けとなります|でお届けします|にてお届け致します|にてお届けいたします) - a positive lookahead that requires one of the alternative patterns to appear immediately to the right of the current location.
You could add a negative lookahead (?!\s*opt\d) to assert that there is no opt and a digit to the right. You can use a character class to list the digits 1, 2 and 3 instead of using the alternation with |.
(?<=\bopt[123]\s(?!\s*opt\d)).*?(?=\s*\bbecause_of_this\b)
Regex demo
It might be a bit more efficient to use a match with a capture group:
\bopt[123]\s(?!\s*opt\d)(.*?)\s*\bbecause_of_this\b
Regex demo
What about:
.*\bopt[123]\b\s*(.*?)\s*because_of_this\b
See the online demo.
.* - A greedy match of any character other than newline upto the last occurence of:
\bopt[123]\b - A word boundary followed by literally "opt" with a trailing number 1, 2 or 3 and another word boundary.
\s* - 0+ whitespace characters.
(.*?) - A 1st capture group with a lazy match of 0+ characters upto:
\s* - 0+ whitespace characters.
because_of_this\b - Literally "because_of_this" followed by a word-boundary.
If you need to have this written out in alternations:
.*\b(?:opt1|opt2|opt3)\b\s*(.*?)\s*because_of_this\b
See that demo.
I am trying to match a string the 2nd word after "Vores ref.:" using positive lookbehind. It works in online testers like https://regexr.com/, but my tool Alteryx dont allow quantifiers like + in a lookbehind.
"ABC This is an example Vores ref.: 23244-2234 LW782837673 Test 2324324"
(?<=Vores\sref.:\s\d+-\d+\s+)\w+ is correctly matching the LW78283767, on regexr.com but not in Alteryx.
How can I rewrite the lookahead expression by using quantifiers but still get what I want?
You can use a replacement approach here using
.*?\bVores\s+ref\.:\s+\d+-\d+\s+(\w+).*
Replace with $1.
See the regex demo.
Details:
.*? - any 0+ chars other than line break chars, as few as possible
\bVores - whole word Vores
\s+ - one or more whitespaces
ref\.: - ref.: substring
\s+ - one or more whitespaces
\d+-\d+ - one or more digits, - and one or more digits
\s+ - one or more whitespaces
(\w+) - Capturing group 1: one or more word chars.
.* - any 0+ chars other than line break chars, as many as possible.
You can use a capture group instead.
Note to escape the dot \. to match it literally.
\bVores\sref\.:\s\d+-\d+\s+(\w+)
The pattern matches:
\bVores\sref\.:\s\d+-\d+\s+ Your pattern turned into a match
(\w+) Capture group 1, match 1+ word characters
Regex demo
I'm trying to match all the non whitespace characters after a string in Regex. In this example, I want to match "b" without the whitespaces and the slashes around it:
a: /b/
I tried using (?<=a:)([^\s\/]+) but it doesn't work.
You still need to account for / before b, not just for whitespace.
You may use a \K based regex (if your regex flavor is PCRE/Onigmo/Boost):
a:\s*\/\K[^\s\/]+
See the regex demo.
Also, if you are using a regex engine that supports unknown width lookbehind patterns, you may use
(?<=a:\s*\/)[^\s\/]+
See this regex demo.
Else, you need to capture your substring with parentheses:
a:\s*\/([^\s\/]+)
See this regex demo.
Details
a: - a a: string
\s* - 0+ whitespaces
\/ - a / char
\K - a match reset operator
[^\s\/]+ - 1+ chars other than whitespace and /.
I cannot figure out how to add two regex together, I have these requirements:
Letters and space ^[\p{L} ]+$
Cannot be whitespace ^[^\s]+$
I cannot figure out how to write one regex that will combine both? There is perhaps some other solution?
You may use
^(?! +$)[\p{L} ]+$
^(?!\s+$)[\p{L}\s]+$
^\s*\p{L}[\p{L}\s]*$
Details
^ - start of string
(?!\s+$) - no 1 or more whitespaces are allowed till the end of the string
[\p{L}\s]+ - 1+ letters or whitespaces
$ - end of string.
See the regex demo.
The ^\s*\p{L}[\p{L}\s]*$ is a regex that matches any 0+ whitespaces at the start of the string, then requires a letter that it consumes, and then any 0+ letters/whitespaces may follow.
See the regex demo.