instantiate a class from a variable in PHP?


Question

I know this question sounds rather vague so I will make it more clear with an example:

$var = 'bar';
$bar = new {$var}Class('var for __construct()'); //$bar = new barClass('var for __construct()');

This is what I want to do. How would you do it? I could off course use eval() like this:

$var = 'bar';
eval('$bar = new '.$var.'Class(\'var for __construct()\');');

But I'd rather stay away from eval(). Is there any way to do this without eval()?

1
131
2/10/2009 8:52:31 PM

Accepted Answer

Put the classname into a variable first:

$classname=$var.'Class';

$bar=new $classname("xyz");

This is often the sort of thing you'll see wrapped up in a Factory pattern.

See Namespaces and dynamic language features for further details.

190
2/12/2015 11:15:07 PM

How to pass dynamic constructor parameters too

If you want to pass dynamic constructor parameters to the class, you can use this code:

$reflectionClass = new ReflectionClass($className);

$module = $reflectionClass->newInstanceArgs($arrayOfConstructorParameters);

More information on dynamic classes and parameters

PHP >= 5.6

As of PHP 5.6 you can simplify this even more by using Argument Unpacking:

// The "..." is part of the language and indicates an argument array to unpack.
$module = new $className(...$arrayOfConstructorParameters);

Thanks to DisgruntledGoat for pointing that out.


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