使用SQL实现小计,合计以及排序

--说明：个人学习笔记，实现小计合计显示，分组按BANK_ID+OP_DATE升序排序
复制代码 代码如下:
--测试数据
CREATE TABLE #TB(ID VARCHAR(10),BANK_ID VARCHAR(10),OP_DATE VARCHAR(10),OPERATOR_NO VARCHAR(20),AMT DECIMAL(10,2))
INSERT #TB SELECT '1111','001','20121210',1234567,111.00
UNION ALL SELECT '2222','002','20121210',1234567,222.00
UNION ALL SELECT '3333','001','20121112',1234567,250.00
UNION ALL SELECT '4444','002','20121110',1234567,330.00
UNION ALL SELECT '5555','001','20121210',1234567,300.00
UNION ALL SELECT '6666','002','20121112',1234567,150.00
GO

--查询
SELECT A.ID
,A.BANK_ID
,A.AMT
,B.OP_DATE
,B.OPERATOR_NO
FROM
(
(SELECT
ID=CASE
WHEN GROUPING(BANK_ID)=1 THEN '合计'
WHEN GROUPING(ID)=1 THEN '小计'
ELSE ID END
,BANK_ID
,SUM(AMT) AMT
,ORDER_SIGN1=GROUPING(BANK_ID),ORDER_SIGN2=BANK_ID
,ORDER_SIGN3=GROUPING(ID)
FROM #TB
GROUP BY BANK_ID,ID WITH ROLLUP
HAVING GROUPING(ID)=1
UNION ALL--先得出表的统计，再加上表中的数据
SELECT ID
,BANK_ID
,AMT
,ORDER_SIGN1=0,ORDER_SIGN2=BANK_ID
,ORDER_SIGN3=0
FROM #TB) A
LEFT JOIN--为了显示出OP_DATE、OPERATOR_NO
(SELECT ID
,OP_DATE
,OPERATOR_NO
FROM #TB) B ON A.ID = B.ID
) ORDER BY ORDER_SIGN1,ORDER_SIGN2,ORDER_SIGN3,OP_DATE

GO

--删除测试
DROP TABLE #TB

/*--测试结果
IDBANK_IDAMTOP_DATEOPERATOR_NO
3333001250.00201211121234567
5555001300.00201212101234567
1111001111.00201212101234567
小计001661.00NULLNULL
4444002330.00201211101234567
6666002150.00201211121234567
2222002222.00201212101234567
小计002702.00NULLNULL
合计NULL1363.00NULLNULL
--*/