I am trying to capture a string that can contain any character but must always be followed by ';'
I want to capture it and trim the white space around it. I've tried using positive lookahead but that does not seem to exclude the whitespace.
Example:
this is a match ;
this is not a match
regex:
.+(?=\s*;)
result:
"this is a match " gets captured with trailing white space behind.
expected result:
"this is a match" (without whitespace)
You have to make sure the first and the last characters of your match are not spaces. Thus we use the non-whitespace character match (\S) before and after the all character match (.*). As spaces might be optional, the any character match (.) must be optional, thus we use * instead of +.
\S.*\S(?=\s*;)
If the string can start with space use .*\S(?=\s*;).
Demonstration
Thanks to #CarySwoveland for improving the answer.
You can match
.*(?<!\s)(?=\s*;)
provided the regex engine supports negative lookbehinds.
Demo
Note that this returns an empty string if the string is " ;".
You can make the dot non greedy and start the match with a non whitespace character:
\S.*?(?=\s*;)
Regex demo
If the non whitespace character itself should also not be a semicolon:
[^\s;].*?(?=\s*;)
Related
I cannot make a regex that only captures a trailing space or N of spaces, followed by a single letter s.
((\s)+(s){1,1})
Works but breaks when you start to stress test it, for example it greedily captures words beginning with s.
word s word s
word s
word suffering
word spaces
word s some ss spaces
there's something wrong
words S s
If you want a single letter s to be captured, as opposed to an s at the beginning of a longer word, you need to specify a word break \b after s:
\s+s\b
Demo on regex101
If you for example do not want to match in s# you can also assert a whitespace boundary to the right.
Note that for a match only, you can omit all the capture groups, and using (s){1,1} is the same as (s){1} which by itself can be omitted and would leave just s
\s+s(?!\S)
Regex demo
As \s can also match a newline, if you want to match spaces without newlines:
[^\S\n]+s(?!\S)
Regex demo
I want to regex match the last word in a string where the string ends in ... The match should be the word preceding the ...
Example: "Do not match this. This sentence ends in the last word..."
The match would be word. This gets close: \b\s+([^.]*). However, I don't know how to make it work with only matching ... at the end.
This should NOT match: "Do not match this. This sentence ends in the last word."
If you use \s+ it means there must be at least a single whitespace char preceding so in that case it will not match word... only.
If you want to use the negated character class, you could also use
([^\s.]+)\.{3}$
( Capture group 1
[^\s.]+ Match 1+ times any char except a whitespace char or dot
) Close group
\.{3} Match 3 dots
$ End of string
Regex demo
You can anchor your regex to the end with $. To match a literal period you will need to escape it as it otherwise is a meta-character:
(\S+)\.\.\.$
\S matches everything everything but space-like characters, it depends on your regex flavor what it exactly matches, but usually it excludes spaces, tabs, newlines and a set of unicode spaces.
You can play around with it here:
https://regex101.com/r/xKOYa4/1
I need to match only those words which doesn't have special characters like # and :.
For example:
git#github.com shouldn't match
list should return a valid match
show should also return a valid match
I tried it using a negative lookahead \w+(?![#:])
But it matches gi out of git#github.com but it shouldn't match that too.
You may add \w to the lookahead:
\w+(?![\w#:])
The equivalent is using a word boundary:
\w+\b(?![#:])
Besides, you may consider adding a left-hand boundary to avoid matching words inside non-word non-whitespace chunks of text:
^\w+(?![\w#:])
Or
(?<!\S)\w+(?![\w#:])
The ^ will match the word at the start of the string and (?<!S) will match only if the word is preceded with whitespace or start of string.
See the regex demo.
Why not (?<!\S)\w+(?!\S), the whitespace boundaries? Because since you are building a lexer, you most probably have to deal with natural language sentences where words are likely to be followed with punctuation, and the (?!\S) negative lookahead would make the \w+ match only when it is followed with whitespace or at the end of the string.
You can use negative lookbehind and negative lookahead patterns around a word pattern to make sure that the word is not preceded or followed by a non-space character, or in other words, to make sure that it is surrounded by either a space or a string boundary:
(?<!\S)\w+(?!\S)
Demo: https://regex101.com/r/cjhUUM/2
There are a thousand regular expression questions on SO, so I apologize if this is already covered. I did look first.
I have string:
Name Subname 11X22 88X620 AB33(20) YA5619 77,66
I need to capture this string: YA5619
What I am doing is just finding AB33(20) and after this I am capturing until first white space. But AB33(20) can be AB-33(20) or AB33(-20) or AB33(-1).
My preg_match regex is: (?<=\bAB\d{2}\(\d{2}\)\s).+?(?=\s)
Why I am getting error when I change from \d{2} to \d+?
For final result I was thinking this regix will work but no:
(?<=\bAB-?\d+\(-?\d+\)\s).+?(?=\s)
Any ideas what I am doing wrong?
With most regex flavors, lookbehind needs to evaluate to a fixed-length sequence, so you can't use variable quantifiers like * or + or even {1,2}.
Instead of using lookaround, you can simply match your marker pattern and then forget it with \K.
AB-?\d+(?:\(-?\d+\))? \K[^ ]+
demo: https://regex101.com/r/8XXngH/1
It depends on the language. If it is in .NET for example, it matches due to the various length in the lookbehind.
Another solution might be to use a character class and add the character you would allow to match. Then match a whitespace character and capture in a group matching \S+ which matches 1+ times not a whitespace character.
\bAB[()\d-]+\s\K\S+
Explanation
\bAB Match literally prepended with word boundary to prevent AB being part of a larger match.
[()\d-]+ Match 1+ times any of the listed character in the character class
\s Match a whitespace char (or \s+ to match 1 or more)
\K Reset the starting point of the reported match( Forget what was matched)
\S+ Match in a group 1+ times not a whitespace character
Regex demo | Php demo
I am trying to write a regex for finding slashes only that are not followed by special characters.
For example, if the string is,
/PErs/#loc/g/2, then I regex should find slashes (/) that are before P, g and 2. It should not return slash before # as # is a special character.
I could write \/\w but it is returning me /P, /g and /2.
Simplest one by using word boundary \b.
\/\b
\b matches between a word character and a non-word character.
DEMO
You want to use the lookahead operator.
Positive lookahead or detect if something is present after (ahead)
Try this regex instead:
\/(?=\w)
DEMO
We use here the positive lookahead operator (?=). It will "detect" the position of a given expression but won't match the expression.
Negative lookahead or detect if something is NOT present after (ahead)
Alternatively, you can also use the negative look ahead operator (?!).
\/(?![#])
DEMO
Negative lookahead with multiple special characters
This will match any / NOT followed by #. If you have more special characters, simply add them to the character class.
For example, if # and % were special characters, the regular expression above would become:
\/(?![##%])
DEMO
Matching slashes NOT followed by NON word character is not the same than followed by word character.
Have a try with:
/(?!\W)
This matches slashes NOT followed by NON word character
It matches the final slash in string: PErs/