You forgot to select a database.
mysql_select_db("database",$link_id);
That should work.
> --
> From: paul morgan[SMTP:[EMAIL PROTECTED]]
> Sent: Saturday, March 31, 2001 1:53 AM
> To: [EMAIL PROTECTED]
> Subject: [PHP-WIN] unhappy camper
>
> hi,
> I'm new to thi
");
$result = mysql_list_tables ("MYSQL")
or die ("Invalid query");
print ("Connected successfully");
while ($i < mysql_num_rows ($result)) {
$tb_names[$i] = mysql_tablename ($result, $i);
echo $tb_names[$i] . "";
$i++;
}
?>
this is from the php manual
""paul morgan"" <[EMAIL PROT
Warning: Supplied argument is not a valid MySQL result resource in
e:/apache/apache/htdocs/tr.php on line 7
huum this is the error I got. you need to read a bit more mysql
documentation
and perhaps the mysql functions in php.
what data base are you trying to show_tables on ?
""paul morgan"" <[
