PHP preg_match function not working as expected

Question

I want to find the index of the pattern from the string and find all Unicode characters like "\u2026","\u2021","\u2011" and so on.

Below is the code snippet I am currently using. But it is resulting in else situation.

if(preg_match('/[\\^]u[0-9]{4}/gi',$data['title'],$matches,PREG_OFFSET_CAPTURE)){
     print_r($matches);
}
else{
     echo "Not Found";
}

Thanks.

Show the text input. Note that `g` flag is not supported, you need to use `preg _match_all` — Wiktor Stribiżew, Dec 21 '19 at 10:14
Thank you, @WiktorStribiżew for suggesting preg_match_all, using that and modified pattern it works!! — Monil Bhavsar, Dec 21 '19 at 17:32

bobble bubble · Accepted Answer · 2019-12-21T11:06:58.000

There are several issues with your code.

If you're using single quotes for the pattern and want to match a literal backslash, you need to use at least \\\ or even \\\\ to produce an escaped backslash \\. Just echo your pattern if unsure.
Instead of using the global flag g which is not available in PHP use preg_match_all. If it matches, it returns the number of matches. You can check match condition by preg_match_all(...) > 0
Unsure about ^ in [\\^]. if you don't need it, drop it. Further [0-9] can be reduced to \d. Also I would add a word boundary \b after \d{4} if something like \u12345 should not be matched.

See this PHP demo at tio.run

$pattern = '/\\\u\d{4}\b/i';
# echo $pattern;

if(preg_match_all($pattern, $data['title'], $matches, PREG_OFFSET_CAPTURE) > 0){
  print_r($matches[0]);
} else{
  echo "Not Found";
}

PHP preg_match function not working as expected

1 Answers1