计数许多列具有相同的值

Question

如何编写 SQL 查询以 return 计算一行中每一列的相似值？

我有这个：

emp_no	d1	d2	d3	d4	d5	d6	d7	d8	d9	d10	date
1002	2	2	2	26	26	4	4	53	53	53	2021-03-31
1003	4	4	4	26	26	2	26	26	26	26	2021-03-31
1002	2	2	2	26	26	4	4	26	26	26	2021-04-30

我想要这样的结果：

emp_no	2	4	26	53	date
1002	3	2	2	3	2021-03-31
1003	1	3	6	0	2021-03-31
1002	3	2	2	3	2021-04-30

我尝试了 UNPIVOT 数据，但我该如何调整它？

我是否使用逆透视数据创建视图，然后重新透视聚合数据？

SELECT EMP_NO, TS_MTH_YR, TSS_D
FROM (
    SELECT EMP_NO, TS_MTH_YR, [D1], [D2], [D3], [D4], [D5], [D6], [D7], [D8], [D9], [D10]
    FROM TSS_MONTHLY_TS
) AS TSS
UNPIVOT (
    TSS_D FOR TSS_DAYS IN ([D1], [D2], [D3], [D4], [D5], [D6], [D7], [D8], [D9], [D10])
) AS TS

Answer 1

正如我在评论中提到的，您需要在此处对数据进行逆透视和重新透视。因此，一种方法如下：

WITH YourTable AS(
    SELECT emp_no ,d1 ,d2 ,d3 ,d4 ,d5 ,d6 ,d7 ,d8 ,d9 ,d10 , CONVERT(date,date) AS date --That's not confusing
    FROM (VALUES(1002,2,2,2,26,26,4,4 ,53 ,53 ,53 ,'2021-03-31'),
                (1003,4,4,4,26,26,2,26, 26, 26, 26,'    2021-03-31'),
                (1002,2,2,2,26,26,4,4 ,26 ,26 ,26 ,'2021-04-30'))V(emp_no ,d1 ,d2 ,d3 ,d4 ,d5 ,d6 ,d7 ,d8 ,d9 ,d10 ,date))
SELECT YT.emp_no,
       COUNT(CASE V.Val WHEN 2 THEN 1 END) AS [2],
       COUNT(CASE V.Val WHEN 4 THEN 1 END) AS [4],
       COUNT(CASE V.Val WHEN 26 THEN 1 END) AS [26],
       COUNT(CASE V.Val WHEN 51 THEN 1 END) AS [51],
       COUNT(CASE V.Val WHEN 53 THEN 1 END) AS [53],
       YT.[date]
FROM YourTable YT
     CROSS APPLY (VALUES('d1',YT.d1),
                        ('d2',YT.d2),
                        ('d3',YT.d3),
                        ('d4',YT.d4),
                        ('d5',YT.d5),
                        ('d6',YT.d6),
                        ('d7',YT.d7),
                        ('d8',YT.d8),
                        ('d9',YT.d9),
                        ('d10',YT.d10))V(Col,Val)
GROUP BY YT.emp_no,
         YT.[date];

计数许多列具有相同的值

Count many columns have the same value

sql

tsql

sql-server

pivot

sql-server-2017