Here is the code
$parent_select = "<TR><TD BGCOLOR=\"#F7F7F7\"><FONT FACE=\"Verdana, Arial, Tahoma\"
SIZE=\"2\"
COLOR=\"#000000\">Parent Category:</font></td><TD BGCOLOR=\"#F7F7F7\"
WIDTH=\"80%\"><select
name=\"parent_id\">";
$parent_sql = "SELECT * FROM categories ORDER BY sort_order ASC";
$parent_result = mysql_query($parent_sql);
while ($parent = mysql_fetch_array($parent_result) ) { <-- the problem line
$parent_select .= "<option
value=\"".$parent['id']."\">".$parent['cat_name']."</option>";
}
$parent_select = "</select></td></tr>\n";
print $parent_select;
this is similar to code that I use to get the result for several other places
--- Andrey Hristov <[EMAIL PROTECTED]> wrote:
> Possibilities:
> 1)There was an error when mysql_query(). Possibly the SQL statement is not correct.
>Trace with
> var_dump()/mysql_errno().
> 2)You give as parameter to some of mysql_* functions a variable not returned from
>mysql_query().
>
> HTH
>
> Regards,
> Andrey Hristov
>
> ----- Original Message -----
> From: "Dan McCullough" <[EMAIL PROTECTED]>
> To: "PHP General List" <[EMAIL PROTECTED]>
> Sent: Tuesday, January 08, 2002 6:47 PM
> Subject: [PHP] Supplied argument is not a valid MySQL result resource
>
>
> > I know this is a common error, but what does it mean. This snippet of code is
>something I use
> all
> > the time.
> >
> > help please.
> >
> > dan
> >
> > __________________________________________________
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