PHP Array to Json Object
Solution 1
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];
Solution 2
just use only
$response=array();
$response["0"]=array("ip" => "192.168.0.1",
"port" => "2016");
$json=json_encode($response,JSON_FORCE_OBJECT);
Solution 3
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
Solution 4
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
Solution 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));
Hassan Ila
Updated on October 20, 2022Comments
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Hassan Ila over 1 year
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 $json;
This is what I get
[ [ "ip" => "192.168.0.1", "port" => "2016" ] ]
But I want to get an object instead of array:
{ "0": { "ip": "192.168.0.1", "port": "2016" } }
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Hassan Ila over 8 yearsThanks but this is what I get: { "0": { "0": "ip => 192.168.0.1", "1": "port => 2016" } }
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jbafford over 8 yearsYour strings are not quite on the right place in your
array_push
. Tryarray_push($json, ["ip" => $ip, "port" => $port]);
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Tayyab Hayat almost 3 years
JSON_FORCE_OBJECT
solve my issue, thanks