It looks like your sql query failed, so the result is invalid.
Assuming that this is really the whole script, then you're counting on
the mysql_db_query function to open a connection to the database "db"
using the default connection values which (according to the manual) are
host: localhost, user: (whatever user is running the script, probably
'www', 'apache', or 'nobody' depending on your system), and password:
(blank). Do you really have your mysql database setup to allow access
with these defaults?
I'd recommend trying mysql_connect() first, then if that works continue
on to the mysql_query(). At least you'll know better which aspect failed.
-Steve
On Friday, March 15, 2002, at 05:14 PM, cosmin laslau wrote:
> <?
> $query = "SELECT * from mytable";
> $result = mysql_db_query("db", $query);
>
> while ($myarray = mysql_fetch_array($result))
> {
> $title = $myarray[title];
> echo "$title<br>";
> }
> ?>
>
> Can someone PLEASE tell me why the coding above gives the following
> error:
>
>
> Warning: Supplied argument is not a valid MySQL result resource in
> /var/web/somesite.com/html/index.php on line 5
>
>
> It's absurd. It's driving me nuts. I'm about to introduce my computer
> to the pavement 40 feet below.
>
> Thanks in advance for whoever sees what I am sure is a glaring and
> obvious flaw in the coding. I've been looking at it for an hours and
> just can't get anything from where I'm standing, maybe a different
> perpective will help.
>
>
>
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