Pass Javascript Object to PHP JSON decode

Question

First, I realize there are a lot of other similar posts. I have read through many, and I have still not been able to this it to work. That being said.....I have a 2 dimensional javascript object that is created dynamically. I am trying to pass it to PHP so I can save insert it into a MySQL table. It looks like the easiest way to do this is with an Ajax post. Here is my javascript:

   var jsonString = JSON.stringify(TableData);

               $.ajax({
                    type: 'POST',
                    url: 'submit.php',
                    data: jsonString, 

                    success: function(){
                        alert("OK");
                    } 
                });

I always get the success alert so I don't think the problem is there. Here is my PHP file.

<?php
$servername = "localhost";
$username = "SME";
$password = "mypass";
$db = "p3";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $db);
if (!$conn) {

         die("Could not connect to database");
        } else echo "hey hey";


$data = json_decode("jsonString");


     print $data;

 ?>

I am not really sure what I am doing wrong. Both of the files are in the same folder so I don't think it's a URL problem. Any help would be appreciated.

you cant echo / print an array ( at least in a meaningful way) try print_r(), var_dump() or var_export() next time.. — ArtisticPhoenix
– ArtisticPhoenix, Commented Aug 1, 2017 at 0:53

Difster · Accepted Answer · 2017-08-01 00:04:59Z

2

I see the problem. In order to access the what you're passing on, you need to get it from the $_POST variable which is an array. You can iterate over that array to get what you need and insert it in to your database after doing '$data = json_decode($_POST);`

Notice that your type in AJAX is post this is the same as the form method. It might also help in your ajax if you added `dataType: "json" to your parameters.

$.ajax({
                type: 'POST',
                url: 'submit.php',
                data: jsonString,
                dataType: 'json'

                success: function(){
                    alert("OK");
                }

answered Aug 1, 2017 at 0:04

Difster

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14 Comments

Evan Lalo Over a year ago

I have made your suggested changes and it appears that my data is null. I used var_dump($data) and there is nothing in there. The odd thing is that when I look at my array in the debugger, it appears to be filled. Any ideas?

Difster Over a year ago

Did you do var_dump($_POST)? What does that show?

Evan Lalo Over a year ago

var_dump($_POST) shows array(0) { }. I checked the value of my JSON.stringify variable and it has the appropriate values. One odd thing is that when I have my dataType: 'JSON' I don't get my success alert.

Difster Over a year ago

Ok, have you checked the console log when you try to post? It doesn't seem that the data is ever making it to your PHP page.

Evan Lalo Over a year ago

I used console.log(jsonString); and it shows my array values and I get my success alert. I am not sure of a way to check if it did indeed go to my submit page other than checking the console there as well or using var_dump().

|

giolliano sulit · Accepted Answer · 2017-08-01 02:01:57Z

0

So it looks like in your PHP you're not actually grabbing the posted data.

This is how I would recommend getting the data:

<?php
    // Check if the posted value is not empty
    if (!empty($_POST('jsonString')) {
        $jsonData = json_decode($_POST('jsonString'));
    }
?>

Then you can do what you need to do with the $jsonData.

answered Aug 1, 2017 at 2:01

giolliano sulit

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Evan Lalo · Accepted Answer · 2017-08-06 22:51:55Z

0

The main problem was in my PHP code. Here's the answer.

$data = json_decode($_POST['denied'], true);

I was then able to insert into my MySQL table like this:

foreach($data as $user) {

        $sql = "INSERT INTO deniedusers (firstname, lastname, username, email, password) values ('".$user['fname']."', '".$user['lname']."', '".$user['uname']."', '".$user['email']."', '".$user['password']."')";

I appreciate everyone's answers. Thanks!

answered Aug 6, 2017 at 22:51

Evan Lalo

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