CODEWITHSUNDEEP
javascriptphpregex
Satish Kumar Singh
Satish Kumar Singh

Reputation: 116

regex for url some issue

i wrote regex to validate url, it works fine for most of url i for below url it does not works:

my regex :

/^(https?:\/\/)?([\da-z\.-]+)\.([a-z\.]{2,6})([?=\/\w \.-]*)\/?$/

https://www.facebook.com/permalink.php?story_fbid=802451379821615&id=298161773583914&pnref=story

how to make it work for all urls

Upvotes: 1

Views: 729

Answers (3)

Robert
Robert

Reputation: 20286

It does not make sense to write REGEX just use

filter_var($url, FILTER_VALIDATE_URL);

Validates value as URL (according to » http://www.faqs.org/rfcs/rfc2396), optionally with required components. Beware a valid URL may not specify the HTTP protocol http:// so further validation may be required to determine the URL uses an expected protocol, e.g. ssh:// or mailto:. Note that the function will only find ASCII URLs to be valid; internationalized domain names (containing non-ASCII characters) will fail.

For JavaScript check

https://code.google.com/p/google-caja/source/browse/trunk/src/com/google/caja/plugin/uri.js

Upvotes: 0

Adam
Adam

Reputation: 18807

Your regex must handle the following cases with ipv6, utf8 characters...

ipv4

http://192.168.1.1/test.htm

ipv6

http://[2a00:1450:4007:806::1007]/!voilà

international characters

http://bébé.fr/

Many complexes possibilities, in fact, so a better thing would be to test the protocol and the hostname:

if (preg_match("#^(https?)://([^/]+)/#", $url, $out)) {
   if (gethostbyname($out[2])) {
     return 1;
   }
}
return 0;

Or a far more simple solution is to not use regular expression and use PHP parse_url() functions which handle all cases.

Upvotes: 0

Michael Granowski
Michael Granowski

Reputation: 76

My understanding is that catering for every possible valid URL leads to major headaches. However, based on this resource, there is a pretty simple regex that should handle most edge cases.

Try this by @stephenhay, it works for your example anyway:

^(https?|ftp):\/\/[^\s/$.?#].[^\s]*$

Upvotes: 2

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