Function call inside a function does not work - PHP

Question

This is simple but I am trying to echo a value from function-1 inside function-2

function testing(){
   $files1 = "dir";  
}

function tested() {
    testing();
    echo "testing".$files1;
}

Any help on why its not working would be appreciated.

Answer: https://stackoverflow.com/a/45387323/257705

Possible duplicate of [Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?](https://stackoverflow.com/questions/16959576/reference-what-is-variable-scope-which-variables-are-accessible-from-where-and) — user3942918, Jul 29 '17 at 08:44
@PaulCrovella That possible is accessing one .php file inside another. — X10nD, Jul 29 '17 at 08:48
@ArunpandianM there is no indication that he is using an object. — Mohammed Joraid, Jul 29 '17 at 08:49
The answer is, you cannot reference a variable that is being declared in another function the way you did. — Mohammed Joraid, Jul 29 '17 at 08:50
@MJoraid ok, I cannot access a variable in another function inside a second function then how do I get it done because that is what I want? — X10nD, Jul 29 '17 at 08:52

B. Desai · Accepted Answer · 2017-07-29T08:51:48.620

3

For this you have to return value from function

function testing(){
   $files1 = "dir";  
   return $files1;
}

function tested() {
    $files1 = testing();
    echo "testing".$files1;
}
tested();

edited Jul 29 '17 at 08:51

answered Jul 29 '17 at 08:47

B. Desai

16,414
5
26
47

score 2 · Answer 2 · answered Jul 29 '17 at 08:52

they function run but the second function doesn't understand the local variable from function one.

I would do the following

<?php
function testing(){
   $files1 = "dir"; 
    return $files1;
}

function tested() {
   $files1 =  testing();
    echo "testing".$files1;
}

tested();
?>

or you could use the global variable.

<?php
function testing(){
   $GLOBALS['files1'] = "dir"; 
   // return $files1;
}

function tested() {
    testing();
    echo "testing ". $GLOBALS['files1'];
}

tested();
?>

Function call inside a function does not work - PHP

2 Answers2