Call variable class with PHP without call_user_func

Question

Is there a way to call a function in a class without using 'call_user_func' and without eval?

here is why I ask:

When I use 'call_user_func' I get the '$this not in context' error:

$this->id = $selectId['id'];
$file = 'score_' . $this->sid . '.php'; //the file with the class

if (@include_once($file)) { //including the file

    $scoreClass = 'Score_' . $this->id;
    call_user_func($scoreClass .'::selectData', $this->user_id);
}

I can overcome this 'not in context' error calling the function like this: but now my name is not variable.

$this->id = $selectId['id'];
$file = 'score_' . $this->sid . '.php'; //the file with the class

if (@include_once($file)) { //including the file

    $test = new Score_98673();
    $test->selectData($this->user_id);

}

Static methods are not in an object-context "Because static methods are callable without an instance of the object created, the pseudo-variable $this is not available inside the method declared as static." — Touki
– Touki, Commented Apr 16, 2013 at 13:22

deceze · Accepted Answer · 2013-04-16 13:39:03Z

3

call_user_func(array($score, 'selectData'), $this->user_id);

This is the correct syntax for the callable pseudotype.
http://php.net/manual/en/language.types.callable.php

You can also do this, BTW:

$method = 'selectData';
$score->$method($this->user_id);

edited Apr 16, 2013 at 13:39

answered Apr 16, 2013 at 13:25

deceze♦

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3 Comments

bwoebi Over a year ago

Don't forget to indicate that he can instantiate the class with $class = new $scoreClass;

BonifatiusK Over a year ago

When I use the first example I still get: "Using $this when not in object context" When I use the second example I will get this error: "Call to a member function selectPDFData() on a non-object"

deceze Over a year ago

@Boni $score will of course have to be an instance of the class, an object. Without instantiating an object you cannot call instance methods on it. In your example that's $test.

bwoebi · Accepted Answer · 2013-04-16 13:22:37Z

1

if (@include_once($file)) { //including the file
    $scoreClass = 'Score_' . $this->id;
    $test = new $scoreClass();
    $test->selectData($this->user_id);
}

This is all you have to do. As you can use variable class names when using new.

The above fails as you call a non-static method (should throw some error) which uses $this.

answered Apr 16, 2013 at 13:22

bwoebi

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