info refuses to go into table

hellz

For some reason this data is not transferring into the data, the data transfers but i dont want it just out on the screen, i want it in a table. At the moment the page shows the table headers but no data. I cant see any problem with this code but its not working, anyone see any errors?

 
<?php
$query = "SELECT * FROM item WHERE ProductId='$id'"; 

$result=mysql_query($query); 
$row= mysql_fetch_array($result); ?>

<table>
  <tr>
    <td ><strong>Artist Name</strong></td>
    <td ><strong>Album Name </strong></td>
    <td ><strong>Price </strong></td>
   </tr>
</table>

<?php

while($row=mysql_fetch_array($result))

echo "<tr><td>$row[Artist]</td> <td>$row[Album]</td><td>$row[Price]</td></tr> ";
 mysql_close();


?>

coldwerturkey

$id isn't defined. adding or die([man]mysql_error/man); to the end of your query will help you when troubleshooting mysql_query()'s

the while() statement is unnecessary as you've already defined $row and I'm assuming you don't intend on showing more than one row.

bradgrafelman

To expand on what coldwerturkey is saying:

$id is indeed undefined; did you perhaps mean to reference some external data? If so, try reading this section in the manual: [man]variables.external[/man].
If the data is indeed coming from an external source, then your code is vulnerable to SQL injections. User-supplied data should never be placed directly into a SQL query string. Instead, it must first be sanitized with a function such as [man]mysql_real_escape_string/man. Note that in this case, since $id is presumed to be an integer, you could (and should) instead simply cast the external data to an (int).
Even if the while() loop was necessary (which it isn't, as coldwerturkey pointed out), you'd be throwing away the first row returned. Each time you call one of the mysql_fetch_*() functions, the data from a row is returned and MySQL then examines the next row. Since you called [man]mysql_fetch_array/man once before the while() loop, you're throwing away the first (and only, it would seem) row of data.
Try to break the habit of doing SELECT * queries; instead, only SELECT the columns that you actually need data from.