parse_url() returns an error when example.com is passed

Question

According to following code if $host_name is something like example.com PHP returns a notice: Message: Undefined index: host but on full URLs like http://example.com PHP returns example.com. I tried if statements with FALSE and NULL but didn't work.

$host_name = $this->input->post('host_name');
$parse = parse_url($host_name);
$parse_url = $parse['host'];

How can I modify the script to accept example.com and return it?

Well, it's not a valid URL. The protocol prefix is not optional, if the hostname is to be detected as such. — mario, Dec 23 '12 at 11:32

score 5 · Accepted Answer · answered Dec 23 '12 at 11:10

5

Upgrade your php. 5.4.7 Fixed host recognition when scheme is ommitted and a leading component separator is present.
Add scheme manually: if(mb_substr($host_name, 0, 4) !== 'http') $host_name = 'http://' . $host_name;

answered Dec 23 '12 at 11:10

Alex

11,479
6
28
50

2

In a second thought, you don't have a component separator, so only the second option left. – Alex Dec 23 '12 at 11:11

score 5 · Answer 2 · answered Dec 23 '12 at 11:21

5

You could just check the scheme is present using filter_var and prepend one if not present

$host_name = 'example.com';
if (!filter_var($host_name, FILTER_VALIDATE_URL, FILTER_FLAG_SCHEME_REQUIRED)) {
    $host_name = 'http://' . $host_name;
}
$parse = parse_url($host_name);

var_dump($parse);

array(2) {
  ["scheme"]=>
  string(4) "http"
  ["host"]=>
  string(11) "example.com"
}

answered Dec 23 '12 at 11:21

Crisp

11,417
3
38
41

In some cases wont work. In your example `filter_var` checks not only mising schema – Sergey Chizhik Sep 30 '15 at 13:32

score 4 · Answer 3 · answered Dec 23 '12 at 11:10

4

Just add a default scheme in that case:

if (strpos($host_name, '://') === false) {
    $host_name = 'http://' . $host_name;
}
$parse = parse_url($host_name);

answered Dec 23 '12 at 11:10

Niko

26,516
9
93
110

This is not a particularly robust piece of code though, is it? – Lightness Races in Orbit Dec 23 '12 at 11:12

score 0 · Answer 4 · answered Sep 17 '15 at 09:54

this is a sample function that returns the real host regardless of the scheme..

function gettheRealHost($Address) { 
   $parseUrl = parse_url(trim($Address)); 
   return trim($parseUrl[host] ? $parseUrl[host] : array_shift(explode('/', $parseUrl[path], 2))); 
} 

gettheRealHost("example.com"); // Gives example.com 
gettheRealHost("http://example.com"); // Gives example.com 
gettheRealHost("www.example.com"); // Gives www.example.com 
gettheRealHost("http://example.com/xyz"); // Gives example.com

parse_url() returns an error when example.com is passed

4 Answers4

Linked