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How to use ajax to update mysql db when checkbox condition is changed?
<script type="text/javascript" src="jquery-1.2.1.min.js"></script>
<script type="text/javascript">

    function checkbox_click (id, favorite)
    {
        // see if checkbox is checked
        if(favorite==1)
        {
            $.ajax({
                type:'POST',
                url:'check_favorite.php', // this external php file isn't connecting to mysql db
                data:'id= ' + id + '&favorite=1',
            });
        }// if
        // the checkbox was unchecked
        else
        {
            $.ajax({
                type:'POST',
                url:'check_favorite.php', // this external php file isn't connecting to mysql db
                data:'id= ' + id + '&favorite=0',
            });
        }//else
    }
</script>

echo "<input type='checkbox' id='$rowid;' name='favorite' checked='checked' onclick='checkbox_click('id','favorite',this();' />";


else
echo "<input type='checkbox' id='$rowid;' name='favorite'  onclick='checkbox_click('id','favorite',this.checked);' />";

<?php
//Database Variables - with the variables entered it doesn't connect
$dbhost = 'localhost';   // usually localhost
$dbuser = 'username';      // database username
$dbpass = 'password';      // database password

//Establish connection to MySQL database

$con = @mysql_connect($dbhost, $dbuser, $dbpass);
if (!$con)
die('Unable to connect.' . mysql_error());

mysql_select_db('devcake', $con);

// Get the variables.
$query = "UPDATE mytable SET favorite=".$_POST['favorite'] . "
WHERE id=".$_POST['id'] . ";";

mysql_query($query);
mysql_close($con);
?>

$dbhost='localhost';
$dbuser='root';
$dbpass='';

$id=$_POST['id'];
$favorite=$_POST['favorite'];
$query="UPDATE mytable SET favorite='$favorite' WHERE id='$id'";

echo "<input type='checkbox' id='$rowid;' name='favorite'";
if(//check in database if 'favorite' column is set to 'on'){
print "checked='checked'";
 }
    echo "/>";

$('input[name=favorite]').click(function(){
    //if a checkbox with name 'favorite' is clicked, do the following.
    //grab the id from the clicked box
    var id=$(this).attr('id');
    //grab the value from the checkbox (remember, if it is checked it will be 'on', else ''
    var favorite=$(this).val();
    //setup the ajax call
    $.ajax({
            type:'POST',
            url:'check_favorite.php',
            data:'id= ' + id + '&favorite='+favorite
        });
    }
    });

echo "<input type='checkbox' id='$rowid;' name='favorite' checked='checked' onclick='checkbox_click('$rowid','favorite',this();' />";

else
echo "<input type='checkbox' id='$rowid;' name='favorite'  onclick='checkbox_click('$rowid','favorite',this.checked);' />";