In PHP, how do I check if a function exists?

Question

How can I check if the function my_function already exists in PHP?

Answer 1

Using function_exists:

if(function_exists('my_function')){
    // my_function is defined
}

Answer 2

http://php.net/manual/en/function.function-exists.php

<?php
  if (!function_exists('myfunction')) {

     function myfunction()
     {
       //write function statements
     }

}
 ?>

Answer 3

var_dump( get_defined_functions() );

DISPLAYS all existing functions

Answer 4

I want to point out what kitchin has pointed out on php.net:

<?php
// This will print "foo defined"
if (function_exists('foo')) {
  print "foo defined";
} else {
  print "foo not defined";
}
//note even though the function is defined here, it previously was told to have already existed
function foo() {}

If you want to prevent a fatal error and define a function only if it has not been defined, you need to do the following:

<?php
// This will print "defining bar" and will define the function bar
if (function_exists('bar')) {
  print "bar defined";
} else {
  print "defining bar";
  function bar() {}
} 

Answer 5

And if my_function is in namespace:

namespace MyProject;

function my_function() {
    return 123;
}

You can check if it exists

function_exists( __NAMESPACE__ . '\my_function' );

in the same namespace or

function_exists( '\MyProject\my_function' );

outside of the namespace.

P.S. I know this is a very old question and PHP documentation improved a lot since then, but I believe people still taking a peek here, and this might be helpful.

Answer 6

Checking Multiple function_exists

$arrFun = array('fun1','fun2','fun3');
if(is_array($arrFun)){
  $arrMsg = array();
  foreach ($arrFun as $key => $value) {
    if(!function_exists($value)){
      $arrMsg[] = $value;
    }
  }
  foreach ($arrMsg as $key => $value) {
    echo "{$value} function does not exist <br/>";
  }
}
function fun1(){
}

Output

fun2 function does not exist 
fun3 function does not exist