jquery ajax sending json to php error - php

Question

Okay so here is my ajax request:
$("#scoreForm").submit(function(e){
e.preventDefault();
var nickName = $('#nickName').val();
var gameScore = parseInt($('#gameScore').text());
var result = { 'result' : '{ "nick":"'+nickName+'", "score":'+gameScore+' }' };
//var result = { "nick":nickName, "score":gameScore };
$.ajax({
url: "http://localhost:8888/snake/addScore.php",
type: "POST",
data: result,
//dataType: "jsonp",
success: function(data){
alert("siker");
},
error: function(jqXHR, textStatus, errorThrown) {
alert("bukta " + textStatus);
//console.log(data);
}
});
return false;
});
and my php process code:
$json_decoded = json_decode($_POST['result'],true);
//$nick = $_GET['nick'];
//$score = $_GET['score'];
$mysqli = new mysqli("localhost", "root", "root", "snake",8889);
//$mysqli->query("INSERT INTO scores(nickName,score) VALUES('".$nick."', ".$score.")");
$mysqli->query("INSERT INTO scores(nickName,score) VALUES('".$json_decoded['nick']."', ".$json_decoded['score'].")");
echo "true";
Now i got the data inserted into the database, but ajax still firing error event. i read that if i set the dataType to jsonp it will go trough , but then i get parse error, how do i get past that?

Answer 1

Wehn you access the $_POST variables in your php script, you need to refer to them the way they are packaged with the JSON object:
$_POST['nick']
$_POST['score']
If you want to refer to the items as $_POST['result'] and use your json decoding approach, youll need to package it this way:
var result = { result : { "nick":nickName, "score":gameScore } };