php call static function of class dynamically(function name is unknown)

Question

Here is my Class:

class ProcessUploadedExcel {
 public static function test1($a,$b)
    {
        dd('hi');
    }
 public static function test2($a,$b)
    {
        dd('hi');
    }

}

in another file I want to call one of the functions. So I included the file:

use App\Library\ProcessUploadedExcel;

the function is stored in a variable and when I use call_user_func_array I get an error:

ProcessUploadedExcel::test1(1,2);//works fine.

$func_name = 'test1';
call_user_func_array("ProcessUploadedExcel::" . $func_name, [1,2]);//gets error

error:

call_user_func_array() expects parameter 1 to be a valid callback, class 'ProcessUploadedExcel' not found

in another file Is the file containing the class ProcessUploadedExcel included/required in the calling class ? — Cid
– Cid, Commented Jan 14, 2020 at 14:00
The error tells you that the class isn't found which as @Cid has said the class isn't included, as that code works fine when both the class and query are included within the same file. — akaBase
– akaBase, Commented Jan 14, 2020 at 14:03
Please extract and provide a minimal reproducible example. There's too much guessing as to what you're doing exactly. — Ulrich Eckhardt
– Ulrich Eckhardt, Commented Jan 14, 2020 at 14:04

jspit · Accepted Answer · 2020-01-15 06:42:51Z

3

You can do this:

class ProcessUploadedExcel {
 public static function test1($a,$b)
    {
        //dd('hi');
        var_dump($a,$b);
    }
 public static function test2($a,$b)
    {
        dd('hi');
    }
}

$func_name = 'test1';
ProcessUploadedExcel::$func_name(1,2);

Output:

int(1) int(2)

or if you want to use call_user_func_array()

call_user_func_array([ProcessUploadedExcel::class,$func_name],[1,2]);

Both solutions work properly with use and namespaces.

edited Jan 15, 2020 at 6:42

answered Jan 14, 2020 at 14:08

jspit

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Comments

Amin Fazlali · Accepted Answer · 2020-01-14 14:14:03Z

2

call_user_func_array ignores the use and you have to provide the full path:

$func_name = 'test1';
call_user_func_array("App\Library\ProcessUploadedExcel::" . $func_name, [1,2]);

answered Jan 14, 2020 at 14:14

Amin Fazlali

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Comments

Dmitry Leiko · Accepted Answer · 2020-01-14 14:22:57Z

2

You can try this way:

class ProcessUploadedExcel {
    public static function test1($a,$b)
    {
        echo ('hi');
    }
    public static function test2($a,$b)
    {
        echo ('hi');
    }

}

$func_name = 'test1';
call_user_func_array([ ProcessUploadedExcel::class, $func_name ] , [1,2]);

answered Jan 14, 2020 at 14:22

Dmitry Leiko

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1 Comment

jspit Over a year ago

This is the best solution for working with namespaces. ProcessUploadedExcel::class is a constant which takes use into account and has here the content 'App\Library\ProcessUploadedExcel'.

Ulrich Eckhardt · Accepted Answer · 2020-01-14 14:13:18Z

First misconception is that you "included the file: use App\Library\ProcessUploadedExcel;". No, you just told PHP that whenever it encounters ProcessUploadedExcel, you actually mean App\Library\ProcessUploadedExcel. This doesn't include anything. Including something is a different mechanism using require or require_once. Actually including the file is done by the autoloader in most projects.

Now, you pass a string to call_user_func_array(). PHP doesn't look inside that string, so it can't tell that you mean something different than what you write there. Inside the called function, where the string is used, the above use ... is not effective, so the function fails since it can't find the according callback.

Remedies:

You could provide the fully qualified name "App\\Library\\ProcessUploadedExcel::test1" as parameter.
You could make PHP look up the fully qualified name for you using ProcessUploadedExcel::class . "::test1".

I'm not sure that either of these works though, but check the documentation which way to specify a callback exist. Maybe you need to specify class and function separately like [ProcessUploadedExcel::class, "test1"].

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