(HELP!!!!!!!!) Creating a 1 - 5 Star rating script

Orc · 12-21-2007, 02:31 PM

For now I just made it say Rating Successful and Unsuccessful, and query.

PHP Code:


			
<style type="text/css">
 input.ratingsnstuff {
   background:url('rate/default.png') no-repeat;
   border:none;
   width:16px;
   height:16px;
   margin:1px;
   cursor: pointer;
   }
   
    input:hover.ratingsnstuff {
   background:url('rate/default_hover.png') no-repeat;
   border:none;
   width:16px;
   height:16px;
   margin:1px;
   cursor: pointer;
   }
   </style>
   
<form class="ratingsnstuff" action="rating.php" method="post">
<input class="ratingsnstuff" name="star1"  type="button" value="">
<input class="ratingsnstuff" name="star2"  type="button" value="">
<input class="ratingsnstuff" name="star3"  type="button" value="">
<input class="ratingsnstuff" name="star4"  type="button" value="">
<input class="ratingsnstuff" name="star5"  type="button" value="">
</form>
<?php

$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$database = "rates";

 mysql_connect($dbhost, $dbuser, $dbpass) or die(mysql_error());
 mysql_select_db($database) or die(mysql_error());


 function ratingInsert($ratings) {
$rateid =  intval($_REQUEST['rateid']);
   $rating = array($_POST['star1'], $_POST['star2'],$_POST['star3'], $_POST['star4'], $_POST['star5']);
 foreach($rating as $ratings) {
   $query = mysql_query("Insert into db_ratings WHERE rateid=".$rateid." and rate=".$ratings."");

    if($query!=null) {
       echo "Rated successfully!";
   } 
   elseif(!$query) {
     echo 'Rated Unsuccessfully!';
     }
    }
  }
           



$ratingInsert = ratingInsert($ratings);
return $ratingInsert;


?>

This outputted:

Code:

Rated Unsuccessfully!Rated Unsuccessfully!Rated Unsuccessfully!Rated Unsuccessfully!Rated Unsuccessfully!

Please help!

UPDATE!!!!

Code:

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 'WHERE rateid=0 and rate=' at line 1

No idea what this means :[