CODEWITHSUNDEEP
php
theHack
theHack

Reputation: 2014

Call function within a function PHP

1    function foo($i){
2       return bar($i)*4;
3       function bar($i){
4           return $i*4;
5           }
6       }
7    echo foo(4);

return 

Fatal error: Call to undefined function bar() in /var/www/index.php on line 2

why doesn't it work? it works well in javascript, while it works when i do this:

function foo($i){
   return bar($i)*4;
   }

function bar($i){
   return $i*4;
   }

Upvotes: 7

Views: 51740

Answers (6)

jpgerb
jpgerb

Reputation: 1120

For the sake of having a "recent" answer, here's what I did:

function printRx($conn, $patient)
{
    function rows($rows, $tod)
    {
        for ($x = 0; $x < count($tod); $x++) {
            if ($tod[$x] == 1) {
                $rows++;
                break;
            }
        }
        return $rows;
    }
    $tod['prn'] = (explode('|', $row['prn'])) ?? null;
    $rows = rows($rows, $tod['prn']);
    return;
}

Works beautifully

Upvotes: 0

Rajasekar Gunasekaran
Rajasekar Gunasekaran

Reputation: 1839

Execution not execute after return statement

change your code like this

 function foo($i){
 function bar($i){
  return $i*4;
 }
 return bar($i)*4;     
}
echo foo(4);

Upvotes: 0

Percy
Percy

Reputation: 1057

Define the function above your return value, otherwise it never gets executed.

<?php
function foo($i){
    function bar($i){
        return $i*4;
    }
    return bar($i)*4;
}
echo foo(4);
?>

Upvotes: 21

Tarek
Tarek

Reputation: 3798

If you define function within another function, it can be accessed directly, but after calling parent function. For example:

function a () {
    function b() {
    echo "I am b.";
   }
echo "I am a.<br/>";
}
//b(); Fatal error: Call to undefined function b() in E:\..\func.php on line 8
a(); // Print I am a.
b(); // Print I am b.

Upvotes: 1

vascowhite
vascowhite

Reputation: 18430

It doesn't work as you are calling bar() before it has been created. See example 2 here:- http://www.php.net/manual/en/functions.user-defined.php

Upvotes: 3

maaudet
maaudet

Reputation: 2358

Your code never reach function bar(...), therefore it's never defined.

You need to put your bar() function before your foo() or before the return bar. (Answer based on your first example).

With your second example you probably define bar() before you use foo() therefore bar() is defined and it works fine.

So as long as you have your function defined when you hit a certain spot in your code it works fine, no matter if it's called from within another function.

Upvotes: 1

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