Php: Array returning empty

Question

I am stumped on my array is not being saved after having values added to it in the functions. I believe it is a problem with the variables not being in the scope of the function. I have the following variables:

$oldInterestIdArray = array();
$newInterestsIdArray = array();

and I have some functions to populate the arrays:

    function oldInterestIds($value, $key)
    {
        $data = fetchInterestDetail($value);
        $id = $data['id'];
        $oldInterestIdArray[] = $id;
    }


    function getNewInterestIds($value,$key)
    {
        if(interestExists($value))
        {
            $data = fetchInterestDetail($value);
            $id = $data['id'];
            $newInterestsIdArray[] = $id;
        }
        else
        {
            addInterest($value);
            $data = fetchInterestDetail($value);
            $id = $data['id'];
            $newInterestsIdArray[] = $id;
        }
    }

        if(count($errors) == 0)
        {
            $newInterests = array_diff($interestsArray, $interests);
            $common = array_intersect($interestsArray, $interests);
            $toChangeId = array_diff($interests, $common);
            array_walk($toChangeId, "oldInterestIds");
            array_walk($newInterests, "getNewInterestIds");
            echo json_encode($oldInterestIdArray);
        }

    }

But the $oldInterestIdArray is returning blank. But if I were to echo json inside the oldInterestIds function it works, which leaves me to believe this is a problem with the variables scope. I have tried changing the variable to:

global $oldInterestIdArray;
global $newInterestsIdArray;

But that is returning null. Any suggestions?

What if you return $oldInterestIdArray? Have you tried that? — Basit
– Basit, Commented Aug 11, 2014 at 5:08

Satish Sharma · Accepted Answer · 2014-08-11 05:11:32Z

1

declare global $oldInterestIdArray; inside the function like

  function oldInterestIds($value, $key)
    {
        global $oldInterestIdArray; // set here global;
        $data = fetchInterestDetail($value);
        $id = $data['id'];
        $oldInterestIdArray[] = $id;
    }

See Example

answered Aug 11, 2014 at 5:11

Satish Sharma

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Comments

SeanCannon · Accepted Answer · 2014-08-11 05:15:37Z

0

Pass your array in as a reference :

function oldInterestIds(&$arr, $value, $key)
{
    $data  = fetchInterestDetail($value);
    $id    = $data['id'];
    $arr[] = $id;
}

answered Aug 11, 2014 at 5:15

SeanCannon

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2 Comments

user1502224 Over a year ago

I am getting Warning: Missing argument 3 for oldInterestIds() and Fatal error: [] operator not supported for strings

SeanCannon Over a year ago

Right, you need to pass it in like I said :)

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