Using a variable as class name

Question

I have the following:

User::model()->exists($someParamsHere);

Is there a way to make the class name 'User' dynamic? So something like this:

$className::model()->exists($someParamsHere);

But that doesn't seem to work.

I also read something about the ReflectionClass, but i'm not really sure how to use it.

I tried this, but off course the model() method is never called this way:

$reflectionMethod  = new ReflectionMethod($className, 'exists');
$reflectionMethod->invoke(null, $someParamsHere);

DaSourcerer · Accepted Answer · 2013-11-20 14:04:42Z

2

$className::model() works with PHP 5.3 and above, if I'm not mistaken. A workaround is to use CActiveRecord::model($className). See the documentation for CActiveRecord.model().

answered Nov 20, 2013 at 14:04

DaSourcerer

6,6465 gold badges35 silver badges58 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

davey Over a year ago

My mistake! I used $this->className as the variable, but it gives an error on the '->'. Putting this in another variable works. $className = $this->className; and then $className::model()->exists();. Great answer though.

Nikolay Krasnov · Accepted Answer · 2013-11-20 14:28:53Z

1

In PHP >= 5.3 works fine:

<?php
class Foo {

    static function Bar() {
        return "Bar";
    }

    static function getFoo() {
        return new static();
    }

    function getBar() {
        return static::Bar();
    }
}

$class = "Foo";

print $class::Bar() . "\n";

print $class::getFoo()->Bar() . "\n";

Result:

Bar
Bar

edited Nov 20, 2013 at 14:28

answered Nov 20, 2013 at 14:12

Nikolay Krasnov

3731 silver badge14 bronze badges

Collectives™ on Stack Overflow

Using a variable as class name

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related