Ajax cannot display json data from php. What's wrong with json format? - php

Question

I have read all the related questions that reference to this topic, but still cannot find answer here. So, php and ajax works great. The problem starts when i try to include json, between php and ajax, to passing data.
here is my ajax:
function likeButton(commentId, userId, sessionUserId) {
// check if the comment belong to the session userId
if(sessionUserId == userId) {
alert("You cannot like your own comment.");
}
else if(sessionUserId != userId) {
var like_upgrade = false;
$.ajax({
url: "requests.php",
type: "POST",
dataType: "json",
data: {
keyLike: "like",
commentId: commentId,
userId: userId,
sessionUserId: sessionUserId,
like_upgrade: like_upgrade
},
success: function(data) {
var data = $.parseJSON(data);
$("#comment_body td").find("#updRow #updComLike[data-id='" +commentId+ "']").html(data.gaming_comment_like);
if(data.like_upgrade == true) {
upgradeReputation(userId);
}
}
});
}
}
Note, that i try not to include this:
var data = $.parseJSON(data);
Also i tried with diferent variable like so:
var response = $.parseJSON(data);
and also tried this format:
var data = jQuery.parseJSON(data);
None of these worked.
here is requests.php file:
if(isset($_POST['keyLike'])) {
if($_POST['keyLike'] == "like") {
$commentId = $_POST['commentId'];
$userId = $_POST['userId'];
$sessionUserId = $_POST['sessionUserId'];
$sql_upgrade_like = "SELECT * FROM gaming_comments WHERE gaming_comment_id='$commentId'";
$result_upgrade_like = mysqli_query($conn, $sql_upgrade_like);
if($row_upgrade_like = mysqli_fetch_assoc($result_upgrade_like)) {
$gaming_comment_like = $row_upgrade_like['gaming_comment_like'];
}
$gaming_comment_like = $gaming_comment_like + 1;
$sql_update_like = "UPDATE gaming_comments SET gaming_comment_like='$gaming_comment_like' WHERE gaming_comment_id='$commentId'";
$result_update_like = mysqli_query($conn, $sql_update_like);
$sql_insert_like = "INSERT INTO gaming_comment_likes (gaming_comment_id, user_id, user_id_like) VALUES ('$commentId', '$userId', '$sessionUserId')";
$result_insert_like = mysqli_query($conn, $sql_insert_like);
$like_upgrade = true;
//json format
$data = array("gaming_comment_like" => $gaming_comment_like,
"like_upgrade" => $like_upgrade);
echo json_encode($data);
exit();
}
}
Note: i also try to include this to the top of my php file:
header('Content-type: json/application');
but still not worked.
What am i missing here?

Answer 1

Don't call $.parseJSON. jQuery does that automatically when you specify dataType: 'json', so data contains the object already.
You should also learn to use parametrized queries instead of substituting variables into the SQL. Your code is vulnerable to SQL injection.