CODEWITHSUNDEEP
phparraysfunction
user1122940
user1122940

Reputation:

php function in array broken

I am trying to setup an array that pulls the filename and function name to run, but it not fully working.

The code is

$actionArray = array(
    'register' => array('Register.php', 'Register'),
);

if (!isset($_REQUEST['action']) || !isset($actionArray[$_REQUEST['action']])) {
    echo '<br><br>index<br><br>';
    echo '<a href="?action=register">test</a>';
    exit;
}

require_once($actionArray[$_REQUEST['action']][0]);
return $actionArray[$_REQUEST['action']][1];

Register.php has

function Register()
{
    echo 'register';

}


echo '<br>sdfdfsd<br>';

But it does not echo register and just sdfdfsd.

If I change the first lot of code from

return $actionArray[$_REQUEST['action']][1];

to

return Register();

It works, any ideas?

Thanks

Upvotes: 0

Views: 159

Answers (2)

apelsinapa
apelsinapa

Reputation: 585

You'll find something usefull here: How to call PHP function from string stored in a Variable

Call a function name stored in a string is what you want...

Upvotes: 0

Brandon - Free Palestine
Brandon - Free Palestine

Reputation: 16656

Change the last line to:

return call_user_func($actionArray[$_REQUEST['action']][1]);

This uses the call_user_func function for more readable code and better portability. The following also should work (Only tested on PHP 5.4+)

return $actionArray[$_REQUEST['action']][1]();

It's almost the same as your code, but I'm actually invoking the function instead of returning the value of the array. Without the function invocation syntax () you're just asking PHP get to get the value of the variable (in this case, an array) and return it.

Upvotes: 3

Related Questions