help a newbie with an error

[buckeyejack]

Hey guys here's the deal. I'm trying to display results from a column, but having some difficulty. Here's the code.
<?
$db_name = "testDB";
$connection = mysql_connect("host", "user", "password") or die..
$query = "SELECT name FROM testDB";
$result = mysql_db_query($db_name, $query,$connection) or die..
// here I have a table
echo "<table width="450" border=1>";
echo "<tr><td>name</td></tr>";
while($field= mysql_fetch_array($result))
{
$name = $field["name"];
echo "<tr><td>$name</td></tr>
}
?>

[buckeyejack]

oopps- I forgot to say what is wrong. I just doesn't render. I have a parse error which is actually the end of the script.
Thanks matt

[[deleted]]

Here's your error.

If you want to print double quotes around the 450, you must escape them with a backslash.

echo "<table width="450" border=1>";

[Anon]

Hmmm....

Why are you selecting field name from a database instead of a table?

first you connect - you did that
second - select your database <b>$db=mysql_select_db("$db_name", $conn);</b>
Once you did that you need to query it
$sql="Select name from $table_name(not $db_name)";

Also die is a function, I don't know if it'll matter, but write it like this - die("comments..");

Di

[buckeyejack]

Thanks for the help. I was having difficulty with the design which was causing all the problems, got fixed now.
matt