PHP pass function name as param then call the function?

I need to pass a function as a parameter to another function and then call the passed function from within the function...This is probably easier for me to explain in code..I basically want to do something like this:

function ($functionToBeCalled)
{
   call($functionToBeCalled,additional_params);
}

Is there a way to do that.. I am using PHP 4.3.9

Thanks!

Solution 1:

I think you are looking for call_user_func.

An example from the PHP Manual:

<?php
function barber($type) {
    echo "You wanted a $type haircut, no problem";
}
call_user_func('barber', "mushroom");
call_user_func('barber', "shave");
?>

Solution 2:

function foo($function) {
  $function(" World");
}
function bar($params) {
  echo "Hello".$params;
}

$variable = 'bar';
foo($variable);

Additionally, you can do it this way. See variable functions.

Solution 3:

In php this is very simple.

<?php

function here() {
  print 'here';
}


function dynamo($name) {
 $name();
}

//Will work
dynamo('here');
//Will fail
dynamo('not_here');

Solution 4:

You could also use call_user_func_array(). It allows you to pass an array of parameters as the second parameter so you don't have to know exactly how many variables you're passing.