oper_id 收银员
0174 刘
0174 刘
0494 王英
0910 何
0910 何
查询出来的结果:
oper_id 收银员
0174 刘
0494 王英
0910 何
------解决方案--------------------
----------------------------------------------------------------
-- Author :DBA_HuangZJ(发粪涂墙)
-- Date :2014-06-16 15:52:53
-- Version:
-- Microsoft SQL Server 2008 R2 (RTM) - 10.50.1600.1 (X64)
-- Apr 2 2010 15:48:46
-- Copyright (c) Microsoft Corporation
-- Enterprise Edition (64-bit) on Windows NT 6.2 <X64> (Build 9200: ) (Hypervisor)
--
----------------------------------------------------------------
--> 测试数据[huang]
if object_id('[huang]') is not null drop table [huang]
go
create table [huang]([收银员] nvarchar(2),[sheet_qty] int,[sale_amt] int,[AA] int,[结算方式] nvarchar(6),[现金] int)
insert [huang]
select N'刘',392,123,16,'DSY',10 union all
select N'刘',392,123,16,'RMB',1 union all
select N'何',561,456,28,'DSY',3 union all
select N'何',561,456,28,'RMB',4
--------------生成数据--------------------------
SELECT *
FROM HUANG
UNION ALL
SELECT DISTINCT N'合计',(SELECT SUM(DISTINCT [sheet_qty]) FROM HUANG ),(SELECT SUM(DISTINCT[sale_amt]) FROM HUANG ),(SELECT SUM(DISTINCT aa) FROM huang ),NULL ,(select SUM( DISTINCT [现金]) FROM huang )
from [huang]
----------------结果----------------------------
/*
收银员 sheet_qty sale_amt AA 结算方式 现金
---- ----------- ----------- ----------- ------ -----------
刘 392 123 16 DSY 10
刘 392 123 16 RMB 1
何 561 456 28 DSY 3
何 561 456 28 RMB 4
合计 953 579 44 NULL 18
*/