Using parse_url in php, how do we check whether input is a url or just a string

Question

Here's my code

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="text" name="text"/>
<input type="submit" name="submit" value="submit"/>
</form>
<?php
if(isset($_POST['submit'])){
$url = $_POST['text'];
$parse = parse_url($url);
echo $parse['host'];
}
?>

I am using parse_url to get the host name to be outputted when somebody inputs the url, now if I try to enter some garbage input, it gives an error: "Undefined index: host" , I want to know how do we check whether its a valid url or just a string before echoing the output of parse_url function

use preg_match() instead or see this example

xkeshav
– xkeshav

2011-11-08 06:08:51 +00:00
Commented Nov 8, 2011 at 6:08 — xkeshav
– xkeshav, Commented Nov 8, 2011 at 6:08

Community · Accepted Answer · 2017-05-23 12:16:19Z

6

You can check to see if host is defined:

echo isset($parse['host']) ? $parse['host'] : 'Invalid input';

Or, you can create a regex to check to see if the string you're passing to parse_url is correctly formatted. This is a bit harder, but can be done. See: Parse URL with Regex using PHP or PHP url validation + detection

edited May 23, 2017 at 12:16

CommunityBot

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answered Nov 8, 2011 at 6:10

nickb

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Comments

Hardik Raval · Accepted Answer · 2011-11-08 06:28:42Z

1

Use Regular Expression

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="text" name="text"/>
<input type="submit" name="submit" value="submit"/>
</form>
<?php
if(isset($_POST['submit'])){
$url = $_POST['text'];

define('REGEX_URL','/http\:\/\/[a-zA-Z0-9\-\.]+\.[a-zA-Z]{2,3}(\/\S*)?/');

if(preg_match(REGEX_URL, $url, $url)) {
   echo "<a href=\"{$url[0]}\">{$url[0]}</a>";
} else {
   echo "<p>No Match Found</p>";
}
}
?>

answered Nov 8, 2011 at 6:28

Hardik Raval

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Comments

matthiasmullie · Accepted Answer · 2011-11-08 08:47:08Z

1

parse_url() will return false if the variable could not be parsed as url. Checking the parse_url() result for !== false would let you know that the string has been successfully parsed by the function and thus is a valid url.

$parse = parse_url($url);
if($parse !== false)
{
    echo 'valid url';
    if(isset($parse['host'])) echo $parse['host'];
}

After that, it's up to you to figure out what exactly you mean with "valid url". PHP's parse_url function will be able to parse a string that's nothing more than e.g. just a path. Depending upon your given problem/context, you may feel that a "valid url" consists out of at least scheme+host, or in a different context, only a relative path (without host) might also be valid. Adding those checks in yourself - just checking if the key is present in the parsed array - could look like:

$parse = parse_url($url);
if($parse !== false)
{
    if(isset($parse['scheme']) && isset($parse['host']))
    {
        echo 'valid url';
        echo $parse['host'];
    }
}

answered Nov 8, 2011 at 8:47

matthiasmullie

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1 Comment

user895378 Over a year ago

It should be noted that prior to PHP5.3.3 parse_url will also trigger an E_WARNING when it returns false so you may need to suppress this with @parse_url in < PHP 5.3.3

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