oracle--聚合函数和case when结合使用

--创建测试表 客户信息表
CREATE TABLE CUST_INFO_TEST(
CUST_NO VARCHAR2(5),
AUM_EDU NUMBER,
DT VARCHAR(10)
)

--在表中插入测试数据
SELECT * FROM CUST_INFO_TEST FOR UPDATE 

--测试数据如下
CUST_NO       AUM_EDU DT
---------- ---------- -----------
1                  11 20180610
2                  22 20180610
3                  33 20180610
1               11.11 20180611
2               22.22 20180611
3               33.33 20180611
1               11.21 20180612
2               22.32 20180612
3               33.33 20180612
1               11.44 20180613
2               22.44 20180613
3               33.44 20180613

--使用表的自关联实现
SELECT A1.CUST_NO
      ,A3.AUM_EDU - A2.AUM_EDU AS BSR_11
      ,A4.AUM_EDU - A3.AUM_EDU AS BSR_12
      ,A1.AUM_EDU - A4.AUM_EDU AS BSR_13
  FROM CUST_INFO_TEST A1
LEFT JOIN CUST_INFO_TEST A2
  ON A1.CUST_NO = A2.CUST_NO 
 AND A2.DT = ‘20180610‘ 
 
LEFT JOIN CUST_INFO_TEST A3
  ON A1.CUST_NO = A3.CUST_NO 
 AND A3.DT = ‘20180611‘
 
LEFT JOIN CUST_INFO_TEST A4
  ON A1.CUST_NO = A4.CUST_NO 
 AND A4.DT = ‘20180612‘
 
WHERE A1.DT= ‘20180613‘;

--得到如下结果
CUST_NO      BSR_11     BSR_12     BSR_13
-------- ---------- ---------- ----------
1              0.11        0.1       0.23
2              0.22        0.1       0.12
3              0.33          0       0.11

--优化
SELECT 
       CUST_NO
      ,SUM(CASE WHEN DT=‘20180611‘ THEN AUM_EDU ELSE 0 END - CASE WHEN DT=‘20180610‘ THEN AUM_EDU ELSE 0 END) AS BSR_11
      ,SUM(CASE WHEN DT=‘20180612‘ THEN AUM_EDU ELSE 0 END - CASE WHEN DT=‘20180611‘ THEN AUM_EDU ELSE 0 END) AS BSR_12
      ,SUM(CASE WHEN DT=‘20180613‘ THEN AUM_EDU ELSE 0 END - CASE WHEN DT=‘20180612‘ THEN AUM_EDU ELSE 0 END) AS BSR_13
  FROM CUST_INFO_TEST
 WHERE DT IN (‘20180610‘,‘20180611‘,‘20180612‘,‘20180613‘)
 GROUP BY CUST_NO 
 
--得到结果和上面一样
CUST_NO      BSR_11     BSR_12     BSR_13
-------- ---------- ---------- ----------
1              0.11        0.1       0.23
2              0.22        0.1       0.12
3              0.33          0       0.11