How to create an object from String?

Question

I tried the code below:

$dyn = "new ". $className . "(" .$param1 . ", ". $param2 . ");";
$obj = eval($dyn);

It compiles, but is null.

How can you instantiate an object in PHP dynamically?

+3

Patrick desjardins Jan 17 '10 at 4:07

3 answers

If you really want to use eval- what chances you should not, if you are new to PHP;) - you would do something more ...

$dyn = "new ". $className . "(" .$param1 . ", ". $param2 . ");";
eval("\$obj = $dyn");

+4

Richard JP Le Guen Jan 17 '10 at 4:33

? eval , , , .

, factory , , , ,

+1

AndrewMurphy 17 . '10 7:16

prodigitalson · Accepted Answer · 2010-01-17T04:09:06+0000

$class = 'ClassName';
$obj = new $class($arg1, $arg2);