CODEWITHSUNDEEP
phpurlexplode
Anonymouse
Anonymouse

Reputation: 25

URL not splitting into directories

I have a function that explodes a URL into directories. If I visit this script at example.com/test/directory, it should return test

function URLPart($part) {
    if ($part > 0) {
        $url = explode("/", $_SERVER['PHP_SELF']);
        return $url[$part - 1];
    } else {
        return $_SERVER['PHP_SELF'];
    }
}

URLPart(1);

Why is this not happening? I get nothing!

Upvotes: 2

Views: 31

Answers (2)

mickmackusa
mickmackusa

Reputation: 47854

There are two reasons you get nothing:

  1. Because a function call doesn't print to screen -- echo does. You need to echo your return value.

  2. @Syscall is correct that there is a leading / with $_SERVER['PHP_SELF'] so that is also throwing off your element targeting. 1 - 1 will have you accessing the 0 element explode()'s array. (I'm actually not sure why you were subtracting 1 from $part.)

While I am here I'll offer some additional advice.

  • $part is expected to be an integer, but let's force that to be true with intval().
  • !$part is used in the conditional to determine if the value is 0
  • I've extended your sample path string to illuminate my next point. I recommend adding $part+3 as the 3rd parameter of the explode() call so that the function doesn't have to do any unnecessary work. For this case, we are accessing element 2 of the array, so jamming the remainder of the path into element 3 is of no consequence. (see demo's output for how this works out)
  • Before attempting to access an array element by key, it is good practice to check that the key exists first -- to avoid a Notice. This is the reason for the !isset() condition.
  • ++$part means add one to $part.
  • I prefer to write the early returns (failures) first in my codes, and put the successful output as the lowest return. And as I said earlier, echo the return value.

Code: (Demo)

function URLPart($part){
    if(!$part=intval($part)){  // convert to integer and check if zero
        return '/example.com/test/directory/foo/bar/what'; // $_SERVER['PHP_SELF']
    }else{
        $url = explode("/",'/example.com/test/directory/foo/bar/what',$part+3); // $_SERVER['PHP_SELF']
        var_export($url);
        echo "\n\n";
        // when $part=1, offset 2 is desired
        if(!isset($url[++$part])){  // if incremented $part is an invalid / non-existent offset
            return '/example.com/test/directory/foo/bar/what';  // change this to your needs
        }else{
            return $url[$part];
        }
    }
}

echo URLPart(1);  // you must echo the return value if you want it to be printed to screen

Output:

array (
  0 => '',
  1 => 'example.com',
  2 => 'test',
  3 => 'directory/foo/bar/what',
)

test

Upvotes: 0

Syscall
Syscall

Reputation: 19780

The $_SERVER['PHP_SELF'] variable contains an initial /, e.g. "/test/directory".

You have to ltrim($_SERVER['PHP_SELF'],'/'), before.

$url = explode("/", ltrim($_SERVER['PHP_SELF'], '/'));

or just :

return $url[$part];

Upvotes: 1

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