Returning an array from inside a php function not working

Question

I'm trying to get the $keysList array out of the function, but seem be going wrong somewhere. I'm getting the error:

Passed Notice: Uninitialized string offset: 2 in C:\web\apache\htdocs\dev\case2.php on line 40 Notice: Uninitialized string offset: 1 in C:\web\apache\htdocs\dev\case2.php on line 40 Notice: Uninitialized string offset: 0 in C:\web\apache\htdocs\dev\case2.php on line 40 Output = a , a and a .

How to get this right?

<?php

$catHandle = "addCat";

function validCatKeys($catHandle,$keysList)
{

    switch($catHandle){

    case "addCat":

            $listCountryCode = 'US';
            $listUserName    = 'Norman';
            $listUserId      = '1';
            $keysList        = array($listCountryCode,$listUserName,$listUserId);
            return true;
        break;

    case "addSubCat":

        break;

    case "addElm":

        break;

    default:
       return false;
    }


}



if(validCatKeys($catHandle,$keysList = ''))
{

    echo 'Passed';
    list($a, $b, $c) = $keysList;
    echo "Output =  a $a, a $b and a $c.";

}else{echo 'Failed';
}



?>

Repeat question? http://stackoverflow.com/questions/5692568/php-function-return-array — Al., May 01 '13 at 08:20

score 1 · Answer 1 · answered May 01 '13 at 08:22

function validCatKeys($catHandle,$keysList)
{

    switch($catHandle){

    case "addCat":

            $listCountryCode = 'US';
            $listUserName    = 'Norman';
            $listUserId      = '1';
            $keysList        = array($listCountryCode,$listUserName,$listUserId);
            return $keysList;
        break;
    ....

Praveen kalal · Answer 2 · 2013-05-01T08:35:19.090

please try below code its working Demo http://codepad.viper-7.com/2Eyym1 the array $keyList is local variable and has local scope so declare it global it will work.

<?php
    $catHandle = "addCat";

    function validCatKeys($catHandle,$keysList)
    {
        global $keysList;

        switch($catHandle){

        case "addCat":

                $listCountryCode = 'US';
                $listUserName    = 'Norman';
                $listUserId      = '1';
                $keysList        = array($listCountryCode,$listUserName,$listUserId);
                return true;
            break;

        case "addSubCat":

            break;

        case "addElm":

            break;

        default:
           return false;
        }


    }



    if(validCatKeys($catHandle,$keysList = array()))
    {

        echo 'Passed';
        list($a, $b, $c) = $keysList;
        echo "Output =  a $a, a $b and a $c.";

    }else{echo 'Failed';
    }

score 1 · Accepted Answer · answered May 01 '13 at 08:25

1

Define the $keysList variable as passed by reference in your function definition:

function validCatKeys($catHandle,&$keysList)

(notice the &)

This will make any internal changes to the $keysList variable available outside the function.

answered May 01 '13 at 08:25

Gareth Cornish

4,357
1
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Returning an array from inside a php function not working

3 Answers3