Dysan Posted November 16, 2007 Share Posted November 16, 2007 On executing the code below, an error message appears. Why is this? Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\User\Desktop\Xampp\htdocs\cart.php on line 51 $total = 0; echo '<table>'; foreach ($contents as $id=>$qty) { $sql = 'SELECT * FROM books WHERE id = '.$id; $result = mysql_query($sql); $row = mysql_fetch_array($result); echo '<tr>'; echo '<td><a href="cart.php?action=delete&id='.$id.'" class="r">Remove</a></td>'; echo '<td>'.$row['title'].' by '.$author.'</td>'; echo '<td>£'.$price.'</td>'; echo '<td><input type="text" name="qty'.$id.'" value="'.$qty.'" size="3" maxlength="3" /></td>'; echo '<td>£'.($price * $qty).'</td>'; $total += $price * $qty; echo '</tr>'; } echo '</table>'; echo '<p>Grand total: £'.$total.'</p>'; echo writeShoppingCart(); echo '<a href="cart.php?action=add&id=1">Add to cart</a>'; Link to comment https://forums.phpfreaks.com/topic/77673-shopping-cart-not-working/ Share on other sites More sharing options...
alecks Posted November 16, 2007 Share Posted November 16, 2007 print $result and see what it comes out with, it is probably returning (I think your query is incorrectly formatted) $sql = "SELECT * FROM `books` WHERE `id` = '$id'"; $result = mysql_query($sql); Link to comment https://forums.phpfreaks.com/topic/77673-shopping-cart-not-working/#findComment-393217 Share on other sites More sharing options...
Dysan Posted November 16, 2007 Author Share Posted November 16, 2007 Can data in the database be displayed without using a while loop. Obviously only one record will get displayed. mysql_select_db("db", $con); $result = mysql_query("SELECT * FROM details"); while($row = mysql_fetch_array($result)) { echo $row['FirstName'] . " " . $row['LastName']; echo "<br />"; } Link to comment https://forums.phpfreaks.com/topic/77673-shopping-cart-not-working/#findComment-393223 Share on other sites More sharing options...
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