oracle--聚合函数和case when结合使用

Posted 2020-11-19 WANGZIHONG

需求：

1 根据客户信息表中信息比较出相应余额。

2 表中有客户号，余额，各个时间点的分区信息，每个分区中客户号唯一。

3 客户信息中分区有10、11、12、13、14、15。

4 得到每个比上日余额信息，如下面表样。

--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

 case when与聚合函数结合使用示例

CREATE TABLE TEST(
CUST_NO NUMBER,
ACT_TYPE VARCHAR2(50),
ACT_TIME DATE);

SELECT * FROM TEST FOR UPDATE; 

   CUST_NO ACT_TYPE    ACT_TIME
---------- ----------- -----------
         1 IN          2017/8/1
         1 MSG         2018/8/10
         2 MSG         2017/8/10
         3 IN          2018/8/23
         4 MSG         2018/8/17
         5 IN          2018/8/10
         5 MSG         2018/8/15
         5 MSG         2017/8/1
         5 IN          2018/8/6

SELECT CUST_NO
      ,SUM(CASE WHEN ACT_TYPE=\'IN\' AND ACT_TIME>=DATE\'2018-08-01\' THEN 1 ELSE 0 END) AS IN_FLAG
      ,SUM(CASE WHEN ACT_TYPE=\'MSG\' AND ACT_TIME>=DATE\'2018-08-01\' THEN 1 ELSE 0 END) AS MSG_FLAG
  FROM TEST
 GROUP BY CUST_NO 

--最后大于1就是对应的客户标签
   CUST_NO    IN_FLAG   MSG_FLAG
---------- ---------- ----------
         1          0          1
         2          0          0
         4          0          1
         5          2          1
         3          1          0