instantiate a class from a variable in PHP?

php class variables eval
Pim Jager picture Pim Jager · Feb 10, 2009 · Viewed 124.9k times · Source

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()?

Answer

Paul Dixon picture Paul Dixon · Feb 10, 2009

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.

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