Dynamically Load A Class

How do you dynamically load a class using a variable?

[Sekka]

Hi,

I want to dynamically load a class using a variable, e.g.

$module = new Mod_{$tmp_Name} ();

Where $tmp_Name is the name of the class.

Is there anyway to achieve this kind of thing?

Thank you.

[eelixduppy]

I'm not sure if this can be done, although if it can, I assume that it follows the same concept as variable variables [php.net]. This works for function names too. I would give it a try and play around with it to see what you can achieve.

Best of luck!

[Sekka]

What about using eval ()?

[uk2.php.net...]

[Sekka]

Bingo! This works,

class Mod_Test {

function Mod_Test () {

echo "Running<br>";

}

function test2 () {

echo "Running2<br>";

}

}

$toPass = "Test";

$toRun = "\$module = new Mod_{$toPass} ();";

eval ($toRun);

// Echos "Running"

$module -> test2 ();

// Echos "Running 2"

[eelixduppy]

Nice work! Glad you got it ;)

[thing3b]

$toRun = "\$module = new Mod_{$toPass} ();";
eval ($toRun);

Eval() should not be used in situations like this. Unless a lot of pre-checking of variables is done, eval() leads to security flaws in PHP scripts and the possibility of your PHP script getting hacked.

Only in rare cases should eval be used, and I have avoid these cases as much as possible. Eval() is evil.

Instead, what I do is:

class Mod_Test {
function Mod_Test () {
echo "Running<br>";
}
function test2 () {
echo "Running2<br>";
}
}

$toPass = 'Test';
$classToLoad = 'Mod_' . $toPass;
$module = new $classToLoad();
$module -> test2 ();

[Sekka]

Hrm. I tried this initially but couldn't get it to work.

I did realise using eval() would cause a major security hole so I did make sure that was being passed to the eval() statement was properly checked and verified.

Thank you very much.