Create instance of class based on string for name and array for parameters

Question

The title is really explicit, but I am going to give an example:

class Foo
{
    public function __construct($param1, $param2) {
        echo 'Class name: ' . __CLASS__ . "\n" .
             'param1 = ' . $param1 . "\n" .
             'param2 = ' . $param2;
    }
}

$class_name = 'Foo';
$params =
    [
    'content1',
    'content2'
    ];

Having this code, how can I create an instance of Foo, using the parameters in $params.

I know I can do

$class = new $class_name($params);

But this would pass the parameters as one array, I would like it to expand the parameters. The following would be the behavior I would like:

$class = new $class_name($param[0], $param[1]);

Thank you!

Other than the answer posted, anything else would be a hack. You either want 2 args or you want an array of args. — AbraCadaver
– AbraCadaver, Commented Mar 18, 2015 at 21:11

Dracony · Accepted Answer · 2015-03-18 20:38:01Z

1

If you have 5.6:

$instance = new $class(...$args);

If not use reflection:

$reflect  = new ReflectionClass($class);
$instance = $reflect->newInstanceArgs($args);

answered Mar 18, 2015 at 20:38

Dracony

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Comments

AbraCadaver · Accepted Answer · 2015-03-18 21:25:35Z

0

I hesitate to post this because really you should design to either accept an array of parameters or individual parameters. However, here are to hackish ways:

Using another function:

public function init($params) {
    echo 'Class name: ' . __CLASS__ . "\n" .
         'param1 = ' . $params[0] . "\n" .
         'param2 = ' . $params[1];
}

public function __construct($param1, $param2=null) {
    if(!is_array($param1)) {
        $param1 = func_get_args();
    }
    call_user_func_array(array($this, 'init'), $param1);
}

Doing it all in the __construct:

public function __construct($param1, $param2=null) {
    if(!is_array($param1)) {
        $param1 = func_get_args();
    }
    echo 'Class name: ' . __CLASS__ . "\n" .
         'param1 = ' . $param1[0] . "\n" .
         'param2 = ' . $param1[1];
}

answered Mar 18, 2015 at 21:25

AbraCadaver

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1 Comment

tleb Over a year ago

Thank you for you answer. My example was only ... an example, I am actually creating a class which executes the specified method in the specified class, and which passes the arguments to this class. This is why I can't modify my receiving class.

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