Syntax error
Posted: Tue Aug 28, 2007 11:18 pm
Code: Select all
echo "<form action='pages.php?page=2 method='post'><select name='rates'><option value='x' selected>Rate me</option><option value='10'>10</option><option value='9'>9</option><option value='8'>8</option><option value='7'>7</option><option value='6'>6</option><option value='5'>5</option><option value='4'>4</option><option value='3'>3</option><option value='2'>2</option><option value='1'>1</option></select><input type='hidden' name='rateit_id' value='100'><input type='hidden' name='action' value='doit'> <input type='submit' value='Rate!'></form>";
if($_POST['action'] == "doit")
{
if($_POST['rates']>0 && $_POST['rates']<11 && !empty($_POST['rateit_id']))
{
$new_count = ($row['votes'] + 1);
$tut_rating2 = ($row['rates'] * $row['votes']);
$new_rating = (($_POST['rates'] + $tut_rating2) / ($new_count));
$new_rating2 = number_format($new_rating, 2, '.', '');
$edit = "UPDATE wow SET
rating='$new_rating2',votes='$new_count' WHERE id = $id";
$result = mysql_query($edit) or die (mysql_error());
// did it work?
if ($result) {
echo 'Thank you for rating this article.<br /><br />';
} else {
echo 'Error! The values sent are below.<br /><br />';
}
echo $row['votes'];
}
}and when i click on rate it gives
Undefined variable: id in c:\program files\easyphp1-8\www\pages.php on line 95
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 2