CODEWITHSUNDEEP
phparraysjsonobject
roz333
roz333

Reputation: 715

json_decode() returns null instead an array

I am going to INSERT an array of objects into mysql via AJAX but on server side json_decode() returns null

How can i solve this?

This is the ajax codes:

let mainObj = [

    { username: 'david', password: 33456, email: '[email protected]' },
    { username: 'rose', password: 3333, email: '[email protected]' },
    { username: 'adam', password: 95112, email: '[email protected]' },
    { username: 'lisa', password: 'sarlak', email: '[email protected]' },

]



let sendMe = JSON.stringify(mainObj);



let xhr = new XMLHttpRequest();
xhr.onreadystatechange = function () {

    if (this.readyState == 4 && this.status == 200) {


        document.getElementById('result').innerHTML = xhr.responseText;
    }


}

xhr.open("GET", "check.php?x=" + sendMe, true);
xhr.send();

And the php codes (check.php):

$obj= json_decode($_GET['x'], true);

$b= $obj[1]->username;


var_dump($b);

It returns null but i need it returns an array of objects which be usable in database.

Upvotes: 0

Views: 64

Answers (3)

JureW
JureW

Reputation: 683

Something like this (others already wrote it..):

$json = '[{"username":"david","password":33456,"email":"[email protected]"},{"username":"rose","password":3333,"email":"[email protected]"},{"username":"adam","password":95112,"email":"[email protected]"},{"username":"lisa","password":"sarlak","email":"[email protected]"}]';

$decodedJson = json_decode($json, true);


// $b= $decodedJson[1]->username; // Wrong, you have an array, not object
// Correct
foreach($decodedJson as $single) {
    print_r($single["username"]."\n");
}

// Prints out:

david rose adam lisa

Upvotes: 1

AH.Pooladvand
AH.Pooladvand

Reputation: 2059

You are trying to treat an array as an object, json_decode returns an array not an object or stdclass... instead of

$b= $obj[1]->username;

should be

$b= $obj[1]['username'];

I'm assuming that you are not using any framework since the thing that you did should throw and exception so it's better to enable error reporting 

ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);

Upvotes: 1

user8034901
user8034901

Reputation:

Since you're using the second parameter true in $obj= json_decode($_GET['x'], true); your returned $obj will be an array. You either use:

$obj = json_decode($_GET['x']);
$b = $obj[1]->username;

or

$obj = json_decode($_GET['x'], true);
$b = $obj[1]['username'];

to get "rose".

https://www.php.net/manual/en/function.json-decode.php

Upvotes: 1

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