Question

I have a page that displays the data of a mysql entry, depending on the link the user clicked ($pagename).. Im wondering how I can create a very basic rating system, that will consist of a form with drop down options of 1 to 5, and when the user submits this value, it posts the data to the corresponding ID of the entry thats currently on the page.

   <?php
$pagename = $_GET['name'];
$sql = "SELECT * FROM tblCocktail WHERE name = '$pagename' LIMIT 1";
/*$sql = sprintf(%sql, mysql_real_escape_string($pagename));*/
$result = mysql_query($sql);
if(!$result) {
    // error occured
}
$data = mysql_fetch_assoc($result);
echo "<p class=\"paratitle\">".$data["name"]."&nbsp;&nbsp;&nbsp;&nbsp;</p>";
echo "<p class=\"paratitle3\">".$data["howto"]."&nbsp;&nbsp;&nbsp;&nbsp;</p>";
echo "<p class=\"paratitle2\">".$data["ingredient1"]."&nbsp;&nbsp;&nbsp;&nbsp;</p>";
echo "<p class=\"paratitle3\">".$data["quantity1"]."&nbsp;&nbsp;&nbsp;&nbsp;</p>";
echo "<p class=\"paratitle2\">".$data["ingredient2"]."&nbsp;&nbsp;&nbsp;&nbsp;</p>";
echo "<p class=\"paratitle3\">".$data["quantity2"]."&nbsp;&nbsp;&nbsp;&nbsp;</p>";
echo "<p class=\"paratitle2\">".$data["ingredient3"]."&nbsp;&nbsp;&nbsp;&nbsp;</p>";
echo "<p class=\"paratitle3\">".$data["quantity3"]."&nbsp;&nbsp;&nbsp;&nbsp;</p>";
echo "<p class=\"dateadded\">".$data["dateadded"]."&nbsp;&nbsp;&nbsp;&nbsp;</p>";
?>
                        </div>
                    </div>
                    <div id="cont2">
                        <div id="contentwrap">
                    <form method="POST" action="addrating.php" >
           <input type="hidden" name="cocktailID" value="<?=$data["id"]?>">
           <select id="ratinglevel" name="ratinglevel">
              <option></option>
              <option>1</option>
              <option>2</option>
              <option>3</option>
              <option>4</option>
              <option>5</option>
            </select>
            <input type="submit" value="submit" />
            </form>

addrating.php:

<?php
mysql_select_db("mwheywood", $con);

//insert cocktail details
$sql="INSERT INTO tblRating (cocktailID, value, counter)
VALUES
('$_POST[id]','$_POST[ratinglevel]','1'";

if (!mysql_query($sql,$con))
  {
  die('Error: you fail at life' . mysql_error());
  }
echo "<p>Thanks for voting</p>"

?>

the table I want to save the rating into is linked via “cocktailID” to the data that is being echo’d in the above code.

and the table structure of “tblRating” is: ratingID, cocktailID, value, counter..

I therefore want the option value to save to the corresponding “cocktailID”, in the “value” field, and a “1” posted to the counter field.

-any help is appreciated -matt

Answer 1

Just include the current ID when you send teh form like this

           <form method="POST" action="addrating.php" >
           <input type="hidden" name="cocktailID" value="<?=$data["id"]?>">
           <select id="ratinglevel" name="ratinglevel">
              <option></option>
              <option>1</option>
              <option>2</option>
              <option>3</option>
              <option>4</option>
              <option>5</option>
            </select>
            </form>

This will then return the ID along with the form results at which time you will have the cocktailID available

    if (isset($_POST['cocktailID']) && isset($_POST['ratinglevel']))
        $sql = "INSERT INTO tblRating SET cocktailID = ".mysql_real_escape_string($_POST['cocktailID']).", value = ".mysql_real_escape_string($_POST['ratinglevel']).",counter = 1";
        $result = mysql_query($sql);
        if(!$result) {
            // error occured
        }
    }