hashtages between quotation marks

Question

I am working on this little code. I managed to make it work with the double quotation mark on one side but not on the other:

/(?<!\\S)#([0-9\\p{L}]+)+(?=[\\s,!?.\\n][^"]|$)/

Here's what I mean: https://regex101.com/r/yN4tJ6/307

The last " #action should not be converting into a hashtag. How do I add this function to the code above?

Answer 1

This expression seems to work:

(?<!\S)(?<!".)#([0-9\p{L}]+)+(?=[\s,!?.\n][^"]|$)

DEMO

Answer 2

The problem with your current pattern, which almost works, is that you would need a variable width lookbehind to correctly check for the presence/absence of a double quote before each hashtag. The approach I took was to use preg_match_all with a pattern which just consumes enough information to make a decision as to whether a hashtag is a match or not. Consider the following script:

preg_match_all('/(?:^|[^"]\s+)(#[0-9\p{L}]+)[.;,!?]?(?=$|\s+[^"])/', $input, $matches);
print_r($matches[1]);

 Array
(
    [0] => #action
    [1] => #Action
    [2] => #cool
    [3] => #000000
    [4] => #ffffff
)

Here is an explanation of the pattern:

(?:^|[^"]\s+)   match the start of the input, OR
                a single non quote character, followed by one or more whitespaces
(#[0-9\p{L}]+)  then match and capture a hashtag
[.;,!?]?        followed by an optional punctuation character
(?=$|\s+[^"])   finally lookahead and assert either the end of the input, OR
                one or more whitespaces followed by a single non quote character

Note that while we do match some content which we don't really want, that doesn't matter, because the first capture group only contains the hashtag.

Answer 3

My guess is that you might wanted to design an expression similar to:

(?<!"\s)#([0-9\p{L}]+)(?=[\s,!?.\n][^"]|$)

The expression is explained on the top right panel of regex101.com , if you wish to explore/simplify/modify it, and in this link , you can watch how it would match against some sample inputs, if you like.

Test

$re = '/(?<!"\s)#([0-9\p{L}]+)(?=[\s,!?.\n][^"]|$)/m';
$str = 'I enjoy #action movies! #Action
movies are #cool.

Color #000000;  #ffffff; work fine

<div style=" #something "> - works

#action " - works

" #action  - doesn\'t work


';

preg_match_all($re, $str, $matches, PREG_SET_ORDER, 0);

var_dump($matches);