PHP Error: Call to Undefined Function

Question

I have two methods in a PHP class:

    public static function sID($urlID) {
            return base_convert($urlID, 10, 62) - 239000;
    }

    public static function load($urlID) {
            if ($urlID = "" || !is_numeric($urlID)) return 0;
            $id = sID($urlID); // ERROR ON THIS LINE
    }

I am getting the error:

Fatal error: Call to undefined function sID()

I have looked around online but amazingly enough, haven't found any solution to this seemingly simple problem. Is there a syntax error or otherwise?

Michael Hampton · Accepted Answer · 2013-12-14 20:23:57Z

4

It's a static class method, so you have to reference it with self::.

$id = self::sID($urlID);

You can also use the class name: (here, assuming class Foo):

$id = Foo::sID($urlID);

Since you're doing this within the class itself, using self:: is probably cleaner and easier to understand.

answered Dec 14, 2013 at 20:23

Michael Hampton

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Comments

Madara's Ghost · Accepted Answer · 2013-12-14 20:23:36Z

1

That's because you haven't declared a sID() function, you declared a static method. Which can be called with className::sID() (or self::sID() from inside the class itself).

answered Dec 14, 2013 at 20:23

Madara's Ghost

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