Updated values are not being added to server database using ajax? [duplicate] - php

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Database values are not updating?
Closed 8 years ago.
I am using this code to call ajaxvote.php
$('.vote').click(function(){
$.ajax({
url: 'ajaxvote.php',
type: 'POST',
cache: 'false',
success: function () { alert("Success!"); } ,
error: function () { alert("Error!"); }});
var self = $(this);
var action = self.data('action');
var parent = self.parent().parent();
var postid = parent.data('postid');
var score = parent.data('score');
if (!parent.hasClass('.disabled')) {
if (action == 'up') {
parent.find('.vote-score').html(++score).css({'color':'orange'});
self.css({'color':'orange'});
$.ajax({data: {'postid' : postid, 'action' : 'up'}});
}
else if (action == 'down'){
parent.find('.vote-score').html(--score).css({'color':'red'});
self.css({'color':'red'});
$.ajax({data: {'postid' : postid, 'action' : 'down'}});
};
parent.addClass('.disabled');
This is the code from my webpage
<div class="item" data-postid="<?php echo $rows['ID'] ?>" data-score="<?php echo $rows['VOTE'] ?>">
<div class="vote-span">
<div class="vote" data-action="up" title="Vote up"><i class="fa fa-camera-retro"></i></div>
<div class="vote-score"><?php echo $rows['VOTE'] ?></div>
<div class="vote" data-action="down" title="Vote down"><i class="fa fa-camera-retro"></i></div>
</div>
This is my php code
if ($_SERVER['HTTP_X_REQUESTED_WITH']) {
if (isset($_POST['postid']) && (isset($_POST['action']))) {
$postId = $_POST['postid'];
if (isset($_SESSION['vote'][$postId])) return;
$query = $mysqli - > query("SELECT VOTE from stories WHERE ID = $postId LIMIT 1");
while ($rows = mysqli_fetch_array($query)) {
if ($_POST['action'] === 'up') {
$vote = ++$rows['VOTE'];
} else {
$vote = --$rows['VOTE'];
}
$mysqli - > query("UPDATE stories SET VOTE = $vote WHERE ID = $postId ");
$_SESSION['vote'][$postId] = true;
}
}
}
I know I can connect to database because I can login. I also get the alert success I have set up above, However, the values are not updating in Database.
EDIT
I have added more Ajax code that I had already written.

Answer 1

When posting via ajax, you need to send through the data you actually want to post.
var postData = {name:"Mister",lastName:"Frodo"}; //Array
$.ajax({
url : "ajaxvote.php",
type: "POST",
data : postData,
success: function(data, textStatus, jqXHR)
{
//Handle response
},
error: function (e) {
// Handle error
}
});
In this case, the post ID and score needs to be grabbed. You also need to grab what kind of action is clicked (typically through a click event bind on the divs with class="vote". For example purposes, let's just set it to "up" for now:
var postId = $('div.item').attr('data-postid').val();
var score = $('div.item').attr('data-score').val();
var postData = {postId: postId, score: score, action: 'up'}
You can now post that "postData" to your ajaxvote.php.
Also, you can use jQuery's $.POST method
$.post("ajaxvote.php", { name: "Mister", lastName: "Frodo" } );
Now for parsing your form, have a look at jQuery's serialize which goes through your form takes each input's [name] attribute along with the value to create a data-string.
Example
name=Mister&lastName=Frodo
This is ideal for sending through with the "data" attribute in $.ajax. Have a look at this answer for more regarding jQuery's serialize.