json_decode() returns null instead an array

Question

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.

print_r($obj) to see what is it.

u_mulder
– u_mulder

2020-04-03 07:26:56 +00:00
Commented Apr 3, 2020 at 7:26 — u_mulder
– u_mulder, Commented Apr 3, 2020 at 7:26

JureW · Accepted Answer · 2020-04-03 07:46:10Z

1

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

answered Apr 3, 2020 at 7:46

JureW

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1 Comment

roz333 Over a year ago

Thanks, Issue has been solved and i can see username and other paramers in html page but in check.php it still return null do you have any idea why doesn't it print in php page as well?

user8034901 · Accepted Answer · 2020-04-03 07:36:38Z

1

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

answered Apr 3, 2020 at 7:36

user8034901

1 Comment

roz333 Over a year ago

Thanks, Issue has been solved and i can see username and other paramers in html page but in check.php it still return null do you have any idea why doesn't it print in php page as well?

AH.Pooladvand · Accepted Answer · 2020-04-03 07:42:21Z

1

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);

edited Apr 3, 2020 at 7:42

answered Apr 3, 2020 at 7:36

AH.Pooladvand

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