problem with SQL statement

[Bastien]

select * from gulf_options where option_id in (1,2)

Warning: Supplied argument is not a valid MySQL result resource in C:\\Inetpub\\

http://wwwroot\\php\\gulfstream\\viewcart.php

on line 47

I am trying to find some data based on values from another query. The first query provides the \"(1,2)\" as a set of options to be found, where each number is an option_id.

What am i doing wrong?

TIA

Bastien

There are many ways to skin this cat,
but it still tastes like chicken

 

[sleipnir214]

May we please see what you were trying to do in line 47? ______________________________________________________________________
Perfection in engineering does not happen when there is nothing more to add.
Rather it happens when there is nothing more to take away.

 

[Bastien]

line 47 is the DB call

$Options = mysql_query($SQLGetOptions) or die ("Couldn't connect"

TIA

Bastien

There are many ways to skin this cat,
but it still tastes like chicken

 

[vcherubini]

Hmm, I still don't think you are giving enough information. You do realize that the mysql_query() function does not connect to the database, right?

I think you want something like:

$conn = mysql_pconnect("hostname", "username", "password") or die("Failed to connect");
@mysql_select_db("database", $conn);

$query = "SELECT * FROM gulf_options WHERE option_id IN (1,2)";
$result = mysql_query($query) or die("Failed to perform query");

What is the first query you are running? I am still somewhat unclear on what you are trying to do.

Hope this helps,
-Vic vic cherubini
krs-one@cnunited.com