how to update PHP variable with Jquery Ajax?

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";
}

Please tell me that this code is a joke. You are using the most inefficient way of counting rows. Using mysql_num_rows() on the resultset would be inefficient, but your code iterating over every row is worse. Anyway, use SELECT COUNT(*) FROM and then use mysql_result($resultset, 0, 0) to get the row count. — ThiefMaster
– ThiefMaster, Commented Jul 9, 2011 at 9:13
^ SELECT COUNT(*) FROM voter WHERE like=0 would be much better — endy
– endy, Commented Jul 9, 2011 at 9:17
select like, count(*) from voter group by like would even be better: one query to get both the likes and dislikes — JB Nizet
– JB Nizet, Commented Jul 9, 2011 at 9:32
where is the javascript code that sends the vote and the PHP code that reads it and put it in database? — JB Nizet
– JB Nizet, Commented Jul 9, 2011 at 9:34

Hamid · Accepted Answer · 2011-07-09 10:55:38Z

2

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 !

............
............
............
$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;
    });

edited Jul 9, 2011 at 10:55

answered Jul 9, 2011 at 9:59

Hamid

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5 Comments

user701510 Over a year ago

I don't have a html section for the output. Can you give an example what it should look like?

Hamid Over a year ago

Hmmm! I mean where does your like or dislike count show ? div ? input ? Where you display this number to user ? you could change it's value!

user701510 Over a year ago

In the first PHP code section I posted, I am just using "echo $numlike" and "echo $numdislike" to show the vote count. I am not using any divs to show vote count. How do I show the vote count without submitting a vote?

Hamid Over a year ago

So What was your problem ? You said that when i update my vote but the number of votes do not update after I put in a vote! So how you figure out that vote not updated ?

user701510 Over a year ago

For example, when I click the "like" button, the jquery Ajax uploads the "like" vote into the SQL table but on the website itself, I do not see an increase in the number of "like" votes based on "echo $numlike" result

endy · Accepted Answer · 2011-07-09 09:16:51Z

1

You can send a $_GET or $_POST variable to the file that you are calling with AJAX.

.load("google.com", "foo=bar", function(){ 

 });

answered Jul 9, 2011 at 9:16

endy

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1 Comment

user701510 Over a year ago

can you please explain in terms of my sample code? I'm quite new to Jquery and Ajax.

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