--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

ID BANK_ID AMT OP_DATE OPERATOR_NO

3333 001 250.00 20121112 1234567

5555 001 300.00 20121210 1234567

1111 001 111.00 20121210 1234567

小计 001 661.00 NULL NULL

4444 002 330.00 20121110 1234567

6666 002 150.00 20121112 1234567

2222 002 222.00 20121210 1234567

小计 002 702.00 NULL NULL

合计 NULL 1363.00 NULL NULL

--*/

时间： 2013-06-09