php url validator shows valid address as not valid

Question

if (!filter_var($site, FILTER_VALIDATE_URL, FILTER_FLAG_HOST_REQUIRED))...

This validator shows:
https://www.g - as valid address !
www.google.com - as invalid address !

If you type some address into address bar, and you see that it works - it should be a valid address. And vice versa. I need such one validator. Or - I'm missing something in the above code ?

Musa · Accepted Answer · 2013-01-15 22:05:50Z

4

www.google.com is not a valid url. It needs a protocol for it to be valid i.e. http://www.google.com

answered Jan 15, 2013 at 22:05

Musa

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6 Comments

Alegro Over a year ago

Musa, I can't expect from user (during registration) that in the You site form input - type http...

Alegro Over a year ago

@ithcy, yes, I can. But I expect that the validator do that.

ithcy Over a year ago

@Alegro validators are for validating, not modifying. They only answer the question "is this correct?"

ithcy Over a year ago

Also please note that http://www.g is a valid URL. It's not a valid internet URL, as g is not a valid TLD, but it is valid.

Musa Over a year ago

@Alegro you're going to have to add a default protocol if not is specified then validate the url.

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Community · Accepted Answer · 2023-11-17 19:32:19Z

3

A user comment on php.net states:

"Please note that FILTER_FLAG_HOST_REQUIRED and FILTER_FLAG_SCHEME_REQUIRED have disappeared. They were previously mentioned in the constants page but it seems that host and scheme are now required by default for the validate_url filter."

Seems like the protocol is a must with recent PHP versions and cannot be disabled.

edited Nov 17, 2023 at 19:32

CommunityBot

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answered Jan 15, 2013 at 22:11

Sven

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