PHP Array to Json Object

Question

I need to convert a PHP array to JSON but I don't get what I expect. I want it to be an object that I can navigate easily with a numeric index. Here's an example code:

$json = array();
$ip = "192.168.0.1";
$port = "2016";
array_push($json, ["ip" => $ip, "port" => $port]);
$json = json_encode($json, JSON_PRETTY_PRINT);
// ----- json_decode($json)["ip"] should be "192.168.0.1" ----
echo $json;

This is what I get

[  
   [  
      "ip" => "192.168.0.1",
      "port" => "2016"
   ]
]

But I want to get an object instead of array:

{  
   "0": {  
      "ip": "192.168.0.1",
      "port": "2016"
   }
}

Why do you want the outer array to become a JSON object? If it's only got numeric keys, an array delivers the same information in slightly less space and is easier to work with on the JS side of things (you can use Array::forEach and Array::map for example). — 00dani
– 00dani, Commented Jan 30, 2016 at 12:14
@00Davo Thanks, because I thought it was easier to handle it when it's an object hhh, and I use it mostly on server-side (PHP) only :) P.S. Upvoted — Blanket431
– Blanket431, Commented Jan 30, 2016 at 13:09

jbafford · Accepted Answer · 2016-01-30 08:46:26Z

41

You want to json_encode($json, JSON_FORCE_OBJECT).

The JSON_FORCE_OBJECT flag, as the name implies, forces the json output to be an object, even when it otherwise would normally be represented as an array.

You can also eliminate the use of array_push for some slightly cleaner code:

$json[] = ['ip' => $ip, 'port' => $port];

edited Jan 30, 2016 at 8:46

answered Jan 30, 2016 at 8:38

jbafford

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2 Comments

Blanket431 Over a year ago

Thanks but this is what I get: { "0": { "0": "ip => 192.168.0.1", "1": "port => 2016" } }

jbafford Over a year ago

Your strings are not quite on the right place in your array_push. Try array_push($json, ["ip" => $ip, "port" => $port]);

Veshraj Joshi · Accepted Answer · 2016-01-30 08:44:50Z

10

just use only

$response=array();
$response["0"]=array("ip"     => "192.168.0.1",
                     "port"   => "2016");
$json=json_encode($response,JSON_FORCE_OBJECT);

answered Jan 30, 2016 at 8:44

Veshraj Joshi

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Comments

MR_AMDEV · Accepted Answer · 2019-04-06 18:40:52Z

6

Just in case if you want to access your objectivitized json whole data OR a specific key value:

PHP SIDE

 $json = json_encode($yourdata, JSON_FORCE_OBJECT);

JS SIDE

 var json = <?=$json?>;
 console.log(json);            // {ip:"192.168.0.1", port:"2016"}
 console.log(json['ip']);      // 192.168.0.1
 console.log(json['port']);    // 2016

answered Apr 6, 2019 at 18:40

MR_AMDEV

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Comments

Cihan Uygun · Accepted Answer · 2016-01-30 08:55:58Z

5

To get array with objects you can create stdClass() instead of array for inner items like below;

<?PHP

    $json = array();
    $itemObject = new stdClass();
    $itemObject->ip = "192.168.0.1";
    $itemObject->port = 2016;

    array_push($json, $itemObject);
    $json = json_encode($json, JSON_PRETTY_PRINT);
    echo $json;

?>

A working example http://ideone.com/1QUOm6

answered Jan 30, 2016 at 8:55

Cihan Uygun

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1 Comment

Stefan Teunissen Over a year ago

imo this is the only proper answer. Especially if you want an object inside of an array.

Antonis Charalambous · Accepted Answer · 2018-01-09 13:07:03Z

5

In order to get an object and not just a json string try:

$json = json_decode(json_encode($yourArray));

If you want to jsonise the nested arrays as well do:

$json =json_decode(json_encode($yourArray, JSON_FORCE_OBJECT));

answered Jan 9, 2018 at 13:07

Antonis Charalambous

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Comments

Sanju Kaniyamattam · Accepted Answer · 2017-11-03 05:16:18Z

-1

$ip   = "192.168.0.1";
$port = "2016";
$json = array("response" => array("ip" => $ip, "port" => $port)); 
//IF U NEED AS JSON OBJECT
$json = [array("ip" => $ip, "port" => $port)]; //IF U NEED AS JSON ARRAY
$json = json_encode($json);
echo $json;

answered Nov 3, 2017 at 5:16

Sanju Kaniyamattam

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