BJ,
A simple "while" command should do it. Try something
like this:
print("<select name=artist>\n");
$qry = "SELECT artist_id, artist FROM $db";
$res = mysql_query($qry) or die(mysql_error());
while($data = mysql_fetch_array($res))
{
print("<option
value=\"$data[0]\">$data[1]</option>\n");
}
print("</option>");
That should do the trick for ya!
Best,
John
>I have two fields artist_id, artist. How do I put
the contents of
"artist" into a dropdown list.
__________________________________________________
Do You Yahoo!?
Great stuff seeking new owners in Yahoo! Auctions!
http://auctions.yahoo.com
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
