How to convert json to string in php

Question

I want to convert json data to String in php. This is my json data

    {
        "": [{
            "description": "hello, this is description speaking"
        }, {
            "fieldName": "myfile",
            "originalFilename": "image.png",
            "path": "/var/folders/rq/q_m4_21j3lqf1lw48fqttx_80000gn/T/ttzVoNPfBxMMirec1tJsnrd2.png",
            "headers": "[Object]",
            "size": "82745"
        }]
    }

I want to print "hello, this is description speaking" in my php file and also i want to upload my file. My php code is given below please correct it.

if ($_SERVER['REQUEST_METHOD'] == 'POST'){
 $json = file_get_contents("php://input");
    $json_obj = json_decode($json);

    echo $json_obj;

    echo $_POST['description'];

    $name = $_FILES['myfile']['name'];
$tmp_name = $_FILES['myfile']['tmp_name'];
$error = $_FILES['myfile']['error'];

if (!empty($name)) {
    $location = 'uploads/';

    if  (move_uploaded_file($tmp_name, $location.$name)){
        echo 'Uploaded';
    }

} else {
    echo 'please choose a file';
}
}  else {
    echo "error 2";
}

In this code errors are Undefined index: file in D:\xampp\htdocs\work. And without the fileupload code it shows nothing.

It looks like your JSON is not valid. Try validating it jsonlint.com first and debugging it once it is valid. — Kevin
– Kevin, Commented Oct 21, 2016 at 14:12
Instead of echo $json_obj to learn it's contents, do the following: echo ( "<pre>" ); print_r($json_obj); echo ( "</pre>" ); using that you will be able to easily see what items you are attempting to get. - for example: $json_obj->myfile (or it might be $json_obj['myfile'], i forget right now) — Mark Phillips
– Mark Phillips, Commented Oct 21, 2016 at 14:13
I have corrected my json and now tell me the php code for getting the values. — fazil tm
– fazil tm, Commented Oct 22, 2016 at 6:07
You have to check out PHP manual again about json_decode. You don't have to use third party tool to check the validity of the JSON encoded string, because, json_decode() will return null if it is not encoded properly. The second parameter of json_decode(), that you have not used it, will determine if the returned value of json_decode is associative array or an object — SaidbakR
– SaidbakR, Commented Oct 22, 2016 at 17:01

Vinicius Dias · Accepted Answer · 2016-10-22 16:23:10Z

3

If you just want to access the description property of your json, you can do the following:

$json_obj = json_decode($json, true);
$description = $json_obj[''][0]['description'];

edited Oct 22, 2016 at 16:23

answered Oct 21, 2016 at 14:17

Vinicius Dias

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5 Comments

fazil tm Over a year ago

It shows error -- Notice: Trying to get property of non-object in D:\xampp\htdocs\work\filesharing\2.php on line 12

Vinicius Dias Over a year ago

As I said, if that is your json, it's not a valid one. You need to format it properly before using with PHP.

fazil tm Over a year ago

I have corrected my json and now tell me the php code for getting the values.

Vinicius Dias Over a year ago

Your json can be better, but anyway... I edited the answer

Mamdouh Saeed Over a year ago

Yes, that i have tested

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