See Below
On Sun, 2004-07-04 at 05:54, Ronald "The Newbie" Allen wrote:
> Here is my code.... I keep on getting a Resource id #2 ......whatever that
> means....any help would be appreciated!
>
> <html>
>
> <head>
>
> <title>Top Frame</title>
>
> </head>
>
> <body bgcolor="#ffffff" text="#000000" link="#0000ff" vlink="#800080"
> alink="#ff0000">
> <EMBED SRC="" autostart="true" hidden="true" loop="true" volume="100%">
> <?
>
> $db_host = '######;
> $db_user = '######';
> $db_pass = '######';
> $db_name = '######';
>
> $conn = mysql_connect($db_host,$db_user,$db_pass);
>
> if ($conn == true) {
>
>
> // Select the database.
> mysql_select_db($db_name,$conn);
>
>
> $db_table = 'music';
> $music = mysql_query ("Select * from $db_table ORDER BY RAND() Limit 1");
>
> if(!$music) die("Query Failed.");
>
> while($row = mysql_fetch_row($music))
>
> print "$music";
you need something like print $row
however, that still will not work.
print_r( $row ) will give you all the values.
print( $row->the_table_name would yield each value.
NOTE: the syntax above is for a call to $row = mysql_fetch_object(
$music )
lookup the correct syntax for mysql_fetch_row( $table ). You might want
to check mysql_fetch_assoc and mysql_fetch_array to have even more fun!
If you really wish to know more check out mysql_result( ) quite useful
when you only seek a couple values from the resultset.
hope that helps
Alvaro Zuniga
(337) 654 6515
www.zunitek.com
Zunitek Solutions
> }
> ?>
>
>
> </body>
>
> </html>
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