CODEWITHSUNDEEP
phpjqueryjsonajax
Nikos Takiris
Nikos Takiris

Reputation: 149

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

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?

Upvotes: 0

Views: 75

Answers (1)

Barmar
Barmar

Reputation: 780724

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.

Upvotes: 4

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