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.

• Death-is-the-real-truth: $data = json_decode($_POST);
• ArtisticPhoenix: you cant echo / print an array ( at least in a meaningful way) try print_r(), var_dump() or var_export() next time..

Answer 1

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");
 } 

• Evan Lalo: I have made your suggested changes and it appears that my data is null. I used var_dump($data) and there is nothing in there. The odd thing is that when I look at my array in the debugger, it appears to be filled. Any ideas?
• Difster: Did you do var_dump($_POST)? What does that show?
• Evan Lalo: 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: 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: 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().

Answer 2

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.

Answer 3

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!