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up vote 1 down vote favorite

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?

share|improve this question
1  
What error are you getting? – Joshua Dwire Nov 1 '12 at 18:38
2  
Always escape your SQL INSERTS!!!!!! – Svetlio Nov 1 '12 at 18:47
i'm just trying things first, don't worry :) – Sándor Tamás Nov 1 '12 at 18:52

2 Answers

up vote 0 down vote

when you request using http request the params from browser and get params on your server must be same, but in your code i don't see params from your client

$.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);
        }
    });

this is the point data: result, and you get params result on your server with this $_POST['result']

try to add result params on your client here will be

$.ajax({
        url: "http://localhost:8888/snake/addScore.php",
        type: "POST",
        data: 'result=' + result,
        //dataType: "jsonp",
        success: function(data){
            alert("siker");

        },
        error: function(jqXHR, textStatus, errorThrown) {
            alert("bukta " + textStatus);
            //console.log(data);
        }
    });
share|improve this answer
up vote 4 down vote

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 } };
share|improve this answer
1  
and there is no need to use json_decode 2 times :) – Svetlio Nov 1 '12 at 18:48
Excellent point and one that I missed. Except in very rare cases, you should never have to json_decode your input twice – user1026361 Nov 1 '12 at 18:50
okay so, with nick and score separated from post,it's working, but i get still error back on ajax, but it inserts into database. on second approach with json_decode, nothing works, even inserting failed. why? – Sándor Tamás Nov 1 '12 at 18:54
Can you edit your OP to reflect the new code? – user1026361 Nov 1 '12 at 19:52
okay, u updated it. – Sándor Tamás Nov 1 '12 at 19:59
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