X-Original-To: pgsql-sql-postgresql.org@localhost.postgresql.org Received: from localhost (unknown [200.46.204.2]) by svr1.postgresql.org (Postfix) with ESMTP id 5D595D1B4FE for ; Sun, 9 Nov 2003 20:59:54 +0000 (GMT) Received: from svr1.postgresql.org ([200.46.204.71]) by localhost (neptune.hub.org [200.46.204.2]) (amavisd-new, port 10024) with ESMTP id 49304-10 for ; Sun, 9 Nov 2003 16:59:25 -0400 (AST) Received: from guinness.cs.stevens-tech.edu (guinness.cs.stevens-tech.edu [155.246.89.8]) by svr1.postgresql.org (Postfix) with ESMTP id 83269D1B515 for ; Sun, 9 Nov 2003 16:59:23 -0400 (AST) Received: by guinness.cs.stevens-tech.edu (Postfix, from userid 4405) id DAE7314187B; Sun, 9 Nov 2003 15:59:19 -0500 (EST) Received: from localhost (localhost [127.0.0.1]) by guinness.cs.stevens-tech.edu (Postfix) with ESMTP id CDE1BC2452 for ; Sun, 9 Nov 2003 15:59:19 -0500 (EST) Date: Sun, 9 Nov 2003 15:59:19 -0500 From: Yasir Malik To: pgsql-sql@postgresql.org Subject: Re: help me... Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII X-Virus-Scanned: by amavisd-new at postgresql.org X-Archive-Number: 200311/75 X-Sequence-Number: 15890 Hi, I'm usually wrong, but you can create a view for Query1 and Query2. That is do create view Query1 [your query for query1]; create view Query2 [your query for query2]; Then what you are doing should work. Are views offered in Access? Regards, Yasir On Sat, 8 Nov 2003, [iso-8859-1] ron_tabada wrote: > Date: Sat, 8 Nov 2003 06:02:15 +0000 (GMT) > From: "[iso-8859-1] ron_tabada" > Reply-To: ronald@my.msuiit.edu.ph > To: pgsql-sql@postgresql.org > Subject: [SQL] help me... > > Hello everyone, Good day! Could anyone help me translate this query in Microsoft Access to Postgresql. I'm having a difficulty. Pls... > Query1: > SELECT items.description, Sum(supplieditems.qty) AS SumOfqty > FROM items INNER JOIN supplieditems ON items.itemno = supplieditems.itemno > GROUP BY items.description; > Query2: > SELECT [items].[description], Sum([customer].[qty]) AS SumOfqty > FROM (items INNER JOIN OtherItem ON [items].[itemno]=[OtherItem].[Itemno]) INNER JOIN customer ON [OtherItem].[Itemno]=[customer].[itemcode] > GROUP BY [items].[description]; > Query3: > SELECT [Query1].[SumOfqty], [Query2].[SumOfqty], [Query1]![SumOfqty]-[Query2]![SumOfqty] AS remain > FROM Query1, Query2; > I have translated Query1 and Query2 in POSTGRESQL but I don't know how to implement Query3. > > > --------------------------------- > Want to chat instantly with your online friends?Get the FREE Yahoo!Messenger