array_push() not affecting array defined outside of function scope [duplicate]

Question

$files = array();
    
function listFolderFiles($dir) {
    $ffs = scandir($dir);
    echo '<ol>';
        foreach ($ffs as $ff) {
            if ($ff != '.' && $ff != '..') {
                if (is_dir($dir . '/' . $ff))
                    listFolderFiles($dir . '/' . $ff);
                else
                    echo '<li>' . $ff;
                
                array_push(
                    $files,
                    array(
                        'file' => $dir . '/' . $ff,
                        'hash' => hash_file('md5', $dir . '/' . $ff)
                    )
                );
                echo '</li>';
            } 
        }
    echo '</ol>';
}
    
listFolderFiles('/var/www');
var_dump($files);

Why am I getting empty output here?

array(0) { };

Aren't you getting warnings about accessing an uninitialized variable? If not, make sure you have error reporting turned up fully. — Barmar
– Barmar, Commented Apr 17, 2014 at 8:17

Community · Accepted Answer · 2020-06-20 09:12:55Z

6

This is a scope issue. Your $files array variable is outside of your function.

You need to pass that as a parameter as a reference (Tipped by @Barmar ;) ) to it..

The right way..

function listFolderFiles($dir,&$files){

and call it like.. listFolderFiles('/var/www',$files);

You are passing it as a reference because you are not returning the array , so whatever changes done inside the function to the array will remain there itself. So you need to add reference as shown in the method signature. This will modify the original array itself.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Apr 17, 2014 at 8:15

Shankar Narayana Damodaran

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5 Comments

Barmar Over a year ago

It needs to be passed by reference.

underscore Over a year ago

This answer worked after adding the reference.Why is that & mean ? @Barmar

Shankar Narayana Damodaran Over a year ago

@samitha, I have explained in the answer. Please see.

Ivan Yonkov Over a year ago

@samitha the reference (&) means that it changes the variable after processing the function

Ivan Yonkov Over a year ago

@samitha I'd suggest to read much more the PHP manual: php.net/manual/en/language.references.pass.php

Barmar · Accepted Answer · 2014-04-17 08:25:26Z

3

You can pass a variable by reference, as in Shankar's answer. Or you can have the function return a result:

$files = listFolderFiles($dir);    

function listFolderFiles($dir){
    $files = array();
    $ffs = scandir($dir);
    echo '<ol>';

    foreach($ffs as $ff){
        if($ff != '.' && $ff != '..'){
            if(is_dir($dir.'/'.$ff)) listFolderFiles($dir.'/'.$ff);
            else echo '<li>'.$ff;

            array_push($files, array('file' => $dir.'/'.$ff, 'hash' => hash_file('md5', $dir.'/'.$ff)));
            echo '</li>';
        } 
    }
    echo '</ol>';
    return $files;
}

answered Apr 17, 2014 at 8:25

Barmar

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Comments

jackfrankland · Accepted Answer · 2014-04-17 08:24:49Z

0

For you purposes here, it may be better to declare the files variable inside the function itself, and return it at the end. This way, the function will be re-usable:

function listFolderFiles($dir){
    $files = array();
    $ffs = scandir($dir);
    echo '<ol>';

    foreach($ffs as $ff){
        if($ff != '.' && $ff != '..'){
            if(is_dir($dir.'/'.$ff)) listFolderFiles($dir.'/'.$ff);
            else echo '<li>'.$ff;

            array_push($files, array('file' => $dir.'/'.$ff, 'hash' => hash_file('md5', $dir.'/'.$ff)));
            echo '</li>';
        } 
    }
    echo '</ol>';
    return $files;
}

$files = listFolderFiles('/var/www');
var_dump($files);

answered Apr 17, 2014 at 8:24

jackfrankland

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Comments

Abdullah Aden · Accepted Answer · 2014-04-17 08:17:50Z

-1

you can use globals

function listFolderFiles($dir){

global $files; // global var

    $ffs = scandir($dir);

    echo '<ol>';
    foreach($ffs as $ff){

        if($ff != '.' && $ff != '..'){

            if(is_dir($dir.'/'.$ff)) listFolderFiles($dir.'/'.$ff);
            else echo '<li>'.$ff;
            array_push($files, array('file' => $dir.'/'.$ff, 'hash' => hash_file('md5', $dir.'/'.$ff)));
            echo '</li>';

        }

answered Apr 17, 2014 at 8:17

Abdullah Aden

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8 Comments

Ivan Yonkov Over a year ago

bad idea, do not follow it

Abdullah Aden Over a year ago

@royal why, i used it jst yestiday to solve a scope issue i had. is there a better way?

Barmar Over a year ago

@longJOURNEY See Shankar's answer.

Bowdzone Over a year ago

Although it would work, it is nearly always bad practice to use globals. Why define something to be available everywhere when you just need it at a much smaller scope?

Ivan Yonkov Over a year ago

it resolves the scope issue for the current need, but leaves the code leaning on globals. A well structured app will not lean on globals. Better way is Shankar's answer. And best is to use objects and instance variables which will change i.e. private $_files = array(); public function listFolderFiles($dir) { array_push($this->_files, array(....)); }

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