how to update PHP variable with Jquery Ajax? - php

Question

I'm making a simple voter that takes either a "like" vote or "dislike" vote. Then, I count the total number of likes and dislikes and output the total numbers. I figured out how to put in the votes using Jquery Ajax, but the number of votes do not update after I put in a vote. I would like to update the $numlike and $numdislike variables using Jquery Ajax.
Here is the PHP script pertaining to the output:
$like = mysql_query("SELECT * FROM voter WHERE likes = 1 ");
$numlike = 0;
while($row = mysql_fetch_assoc($like)){
$numlike++;
}
$dislike = mysql_query("SELECT * FROM voter WHERE likes = 0 ");
$numdislike = 0;
while($row = mysql_fetch_assoc($dislike)){
$numdislike++;
}
echo "$numlike like";
echo "<br>";
echo "$numdislike dislike";
UPDATE:
Jquery Ajax for uploading vote
<script>
$(document).ready(function(){
$("#voter").submit(function() {
var like = $('#like').attr('value');
var dislike = $('#dislike').attr('value');
$.ajax({
type: "POST",
url: "vote.php",
data: "like=" + like +"& dislike="+ dislike,
success: submitFinished
});
function submitFinished( response ) {
response = $.trim( response );
if ( response == "success" ) {
jAlert("Thanks for voting!", "Thank you!");
}
return false;
});
});
</script>
<form id="voter" method="post">
<input type='image' name='like' id='like' value='like' src='like.png'/>
<input type='image' name='dislike' id='dislike' value='dislike' src='dislike.png'/>
</form>
vote.php:
if ($_POST['like'])
{
$likeqry = "INSERT INTO test VALUES('','1')";
mysql_query($likeqry) or die(mysql_error());
echo "success";
}
if ($_POST['dislike'])
{
$dislikeqry = "INSERT INTO test VALUES('','0')";
mysql_query($dislikeqry) or die(mysql_error());
echo "success";
}

Answer 1

If you want to change current like or dislike number after clicking it you must return result instead of printing it ! return json result and echo this and change div innerHTML to see new result !
............
............
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$dislike = mysql_query("SELECT * FROM voter WHERE likes = 0 ");
$numdislike = 0;
while($row = mysql_fetch_assoc($dislike)){
$numdislike++;
}
echo json_encode( array( $numlike, $numdislike ) ) ;
exit();
Now in your html code :
$.ajax({
type: "POST",
url: "vote.php",
context:$(this)
data: "like=" + like +"& dislike="+ dislike,
success: submitFinished(data)
});
function submitFinished( response ) {
response = $.parseJSON( response );
//Now change number of like and dilike but i don't know where are shown in your html
$('#like').attr('value',response[0]);
$('#dislike').attr('value',response[1]);
return false;
});

Answer 2

You can send a $_GET or $_POST variable to the file that you are calling with AJAX.
.load("google.com", "foo=bar", function(){
});