How do you decode a returned JSON array in php function

Question

I have two php functions, one returns a JSON array and the other attempts to decode it and access it's contents in order to strip a URL out of it. My problem is that I get the following errors..

1) json_decode() expects parameter 1 to be string

2) Invalid argument supplied for foreach() on line 77

Here is my code

php file 1

<?php

function getAnimation($userid, $db) {

    include('connect.php');

    $db = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
    // Check connection
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error() . "please contact [email protected] for technical assistance";
        echo "<br>";
    }

    $box_num = 1;

    $select = "SELECT card_id, order_num FROM decks WHERE box_num=$box_num AND id=$userid ORDER BY order_num";
    $result = mysqli_query($db, $select) or die("SQL Error 1: " . mysqli_error($db));

    while ($row = mysqli_fetch_array($result, MYSQL_ASSOC)) {
        $users[] = array(
            'card_id' => $row['card_id'],
            'order_num' => $row['order_num'],
        );
    }

    json_encode($users);

    return $users;

    mysqli_close($db);
}

?>

snippet of php file 2 that is attempting to decode the JSON array and access its contents.

include('getAnimation.php');
$animation = getAnimation($result_array[0], $db); //<< atempting to access returned array here
$obj = json_decode($animation);  //FIRST ERROR HERE
foreach ($obj->card_id as $item) { // SECOND ERROR HERE
$url = ($item->card_id);
echo $url; //show me the money
}

thanks for any help you can give!

jterry · Accepted Answer · 2013-06-12 22:21:13Z

2

Modify the return to be return json_encode($users). You're not actually setting the variable $users... you could also set it with $users = json_encode($users), then return $users.

To answer your second question, modify your code as such:

foreach($obj as $item) {
    $url = $item->card_id;
}

edited Jun 12, 2013 at 22:21

answered Jun 12, 2013 at 22:14

jterry

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4 Comments

David Folksman Over a year ago

Thanks :) I don;t quite understand whats going on here. COuld you elaborate a bit for me.

jterry Over a year ago

Sure- after you create your array with your function and json_decode it, you'll have an array of objects that you then just want to walk through (foreach) and access their properties (card_id, in this case)!

David Folksman Over a year ago

Just as an added thought, if i wanted to store the contents of the JSON array in an array, would I have to do something like..

foreach ($obj as $item) {                 $url[$i] = $item->card_id;                 $i++;                 echo $url[$i];             }

jterry Over a year ago

Doesn't look like $i would point to anything in that case... you'd have to do foreach ($obj as $i => $item) { } if you wanted to access the keys.

Mr. B. · Accepted Answer · 2013-06-12 22:17:33Z

1

Change

json_encode($users);
return $users;

to:

$users = json_encode($users);
return $users;

That's it! :-)

answered Jun 12, 2013 at 22:17

Mr. B.

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Comments

HorusKol · Accepted Answer · 2013-06-12 22:50:41Z

1

json_decode() expects parameter 1 to be string

json_encode() returns a value, it doesn't act on the parameter by reference, so when you have:

json_encode($users);

return $users;

you're returning the original array, and not the JSON representation.

Fix it by doing:

return json_encode($users);

Invalid argument supplied for foreach() on line 77

You encode an array, so you should be decoding an array - so you should be looping over the array, not an object:

foreach ($obj->card_id as $item)

should just be:

foreach ($obj as $item) {
    $url = $item["card_id"];
    echo $url;
}

Side note

Avoid using generic variables names like $obj and $item - they are not descriptive, and PHP can have some unusual scoping quirks, especially if you are using buffering.

edited Jun 12, 2013 at 22:50

answered Jun 12, 2013 at 22:15

HorusKol

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5 Comments

David Folksman Over a year ago

THanks, that fixed the first error, but I still get the second error. Any ideas?

HorusKol Over a year ago

sry - didn't read past the first error, as I thought the second was caused by the first. have addressed that now

David Folksman Over a year ago

Thanks for that answer it was very good, but I had already given best answer. Just as an added thought, if i wanted to store the contents of the JSON array in an array, would I have to do something like..

foreach ($obj as $item) {                 $url[$i] = $item->card_id;                 $i++;                 echo $url[$i];             }

HorusKol Over a year ago

well, once you've decoded the JSON string into a variable, it will be an array...

David Folksman Over a year ago

Oh right :) I thought you had to assign it to a php array to access its members. I want to pass each url to a javascript function in the same page. Thanks for your help. :)

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