douyun1546 2013-12-14 20:22
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PHP错误:调用未定义的函数

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?

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  • duanbo1659 2013-12-14 20:23
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    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.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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