On Mon, Mar 19, 2001 at 10:06:52PM -0800, Toke Herkild wrote:
> I've written this login rutine
>
> But that doesn't work... I've tried to use mysql_num_rows too but I still
> get :
> <b>Warning</b>: Supplied argument is not a valid MySQL result resource in
> <b>g://web/herkild/html/loginout.php</b> on line <b>113</b><br>
>
> if ($ACTION == 'Login') {
> $query = "SELECT user.ID FROM user WHERE ";
> $query .= "user.brugerID = $LoginBrugerID AND ";
> $query .= "user.kodeord = $LoginKodeord;";
>
> $mysql_result = mysql_query($query, $mysql_link);
>
> (line 113 >>) if ($row = mysql_fetch_row($mysql_result)){
> setcookie("Herkild_Domain", $row[0]);
> Logged_in();
> }
> else {
> wronguser();
> }
> }
It looks like you might be missing some quotes in your query. After you
call mysql_query(), call mysql_error($mysql_link); It will show you
your error in your query.
--
Jason Stechschulte
[EMAIL PROTECTED]
--
There are still some other things to do, so don't think if I didn't fix
your favorite bug that your bug report is in the bit bucket. (It may be,
but don't think it. :-) Larry Wall in <[EMAIL PROTECTED]>
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