preg_match to match url with multiple languages

Question

i was using standard preg_match for making url excluding

  http://domainlllll.com/

and it was working without any issue

  preg_match("/^[0-9a-z_\/.\|\-]+$/",$url)

but now i want to support multiple languages so i used this and it is also working without any problem

   preg_match("/\S+/",$url)

my url is

link/kn/some-word-গরম-এবং-সেক্সি-ইমেজ/611766

but i want to exclude some special characters which is hackers favorite like single quotes and other. i dont want to exclude all special character as few are part of few languages and it will break those languages in link

Any guide will be great

Do you want to match with the `^\S+$` pattern, but add some excluded chars to it? Like `'/^[^\s\'"]+$/'`? — Wiktor Stribiżew, Jun 30 '17 at 18:59
@WiktorStribiżew shows 157ms in matching https://regex101.com/r/xB0dZ6/1 this much time ????? — swetlana, Jun 30 '17 at 19:19

score 0 · Accepted Answer · answered Jun 30 '17 at 19:04

Look, the /^[0-9a-z_\/.\|\-]+$/ regex requires the whole string to match the patterns, 1+ chars from the defined ranges (digits, lowercase ASCII letters) and sets. The /\S+/ regex does not require a full string match since there are no anchors, ^ (start of string) and $ (end of string), and matches 1 or more non-whitespace chars anywhere inside a string.

If you plan to match strings that only contain non-whitespace symbols and NOT quotes, use

preg_match('~^[^\s\'"]+$~', $url)

The ^[^\s\'"]+$ matches

^ - start of string
[^\s\'"]+ - 1 or more chars other than whitespace (\s), ' and "
$ - end of string (better use \z instead if you need to validate a string).

preg_match to match url with multiple languages

1 Answers1