This question already has answers here:
How can I match on, but exclude a regex pattern?
(6 answers)
Closed 2 years ago.
How do i extract this from a regex:
mymatch[ someother char ]
What i want is mymatch when followed by a [ but i don't want the square bracket in the match.
I'm stuck on this but it gets also the square bracket:
\b.*?\[
More in general, how can i exclude from the match some portion of the pattern?
For example here (abc2)mymatch i want a regex returning my match only when it is preeceds by (abc2).
in this you need to use lookahead and lookbehind , in first one use positive lookahead :
\w+(?=\[)
?=[ : mean followed by "[" and not catch in matches
in second example use positive lookbehind :
(?<=\(abc2\))\w+
(?<=(abc2)) : mean leaded by "(abc2)" and not catch in matches
for more information about lookahead and lookbehind
Related
This question already has answers here:
Regular expression to match a line that doesn't contain a word
(34 answers)
Closed 2 years ago.
I would like to come out with a regex expression that negate the matched results of regex expression: .google.*search. And, is it possible to achieve it with regex from the regex expression I am trying to negate?
Test data
[1] https://www.google.com/search?newwindow=1&sxsrf=ALeKk02MzEfbUp3jO4Np
[2] https://github.com/redis/redis-rb
[3] https://web.whatsapp.com/
Expected result
Row 2, 3 match the regex pattern and are part of the results.
the following regex does the trick
^(?!.+google.*search)
basically matching the beginning of the line then negating (?!) (negative lookahead) your regex.
You may use a negative lookahead here:
https?:\/\/(?!.*\.google\..*search).*
Demo
The "secret sauce" here is (?!.*\.google\..*search), which asserts that .google. followed by search does not occur anywhere within the URL to the right of the https:// portion.
This question already has answers here:
What is a non-capturing group in regular expressions?
(18 answers)
Closed 2 years ago.
I have this regexp:
^(?<FOOTER_TYPE>[ a-zA-Z0-9-]+)?(?<SEPARATOR>:)?(?<FOOTER>(?<=:)(.|[\r\n](?![\r\n]))*)?
Which I'm using to match text like:
BREAKING CHANGE: test
my multiline
string.
This is not matched
You can see the result here https://regex101.com/r/gGroPK/1
However, why is there the last Group 4 ?
You will need to make last group non-capturing:
^(?<FOOTER_TYPE>[ a-zA-Z0-9-]+)?(?<SEPARATOR>:)?(?<FOOTER>(?<=:)(?:.|[\r\n](?![\r\n]))*)?
Make note of:
(?:.|[\r\n](?![\r\n]))*)?
(?: at the start makes this optional group non-capturing.
Updated Demo
it is group 4 because the fourth parentheses you defined is:
(.|[\r\n](?![\r\n]))*)
it translate to
"either dot, or the following regex"
and in the example you have, it ends on a dot.
string.
so as regex is usually greedy, it captures dot as the forth group
This question already has answers here:
regular expression for anything but an empty string
(9 answers)
Closed 4 years ago.
I have the following code
https://stackblitz.com/edit/angular-uvxifq-qrjtpg
pattern="^\s+$"
I need to negate this, because as of now I got error when I put a letter, but it should only show error IF there is only white space.
I tried using ?! but it shows an error
You can use the following regex:
^(?: *[^\s] *)+$
demo: https://regex101.com/r/th0A3A/3/
If you want to use a negative lookahead (?!, you could check if from the start till the end of the string there are no whitespace characters:
pattern="^(?!\s*$)[\s\S]+$"
Regex demo
That will match
^ Start of string
(?!\s*$) Negative lookahead, assert that what follows is not 0+ whitespace characters and end of string
[\s\S]+ Match 1+ times any character including new lines
$ End of string
This question already has answers here:
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
Closed 5 years ago.
https://regex101.com/r/rnr6SC/2
I am writing a regex to find any instances of words that start with & and end with ;. There could be multiple copies on the same line, on different lines, or whatever. I just want to find anything that looks like
&
&dog;
&cat; and &dog;
or similar. I think my check for the ; is too greedy in my example. How do I fix my regex to find the specific words I need without selecting anything else?
This regex should work for you:
(?<=&)[^;&]+(?=;)
RegEx Demo
(?<=&): Lookbehind to assert previous position has &
[^;&]+: Match 1+ character that is not ; and not &
(?=;): Lookahead to assert next position has ;
Lookarounds just make sure that a matching text is surrounded by & and ;, if you want these markers also to be part of match then remove lookarounds and use:
&[^;&]+;
You might remove the lookaheads (?= and make the .* non greedy .*?
&.*?;
That would match:
ampersand &,
.*? any character zero or more times non greedy
a semicolon ;
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 8 years ago.
What is the use of
?=
in perl regex
please tell the exact meaning and give some regex example.
(?=...)
is a positive lookahead, a type of zero-width assertion. What it's saying is that the match must be followed by whatever is within the parentheses but that part isn't captured.
Example:
.*(?=bar)
This pattern matches all the characters upto the string bar. When bar is detected then it stops matching. If a line contains more than one bar means it matches upto the last bar because .* does a greedy match.
DEMO