Question

我有这个子查询结果：

DATE        REGCODE  REGCODE2    REGCODE 3   REGCODE4    
02/01/15    70771      5            6           8
02/01/15    null       5            21          8  
02/01/15    70855      4            5           13  
02/01/15    null       ...         ...          ...
03/01/15    70771
03/01/15    70772
05/01/15    null
05/01/15    70855
05/01/15    70826

我需要按日期计算不同的REGCODEs分组，但如果其中一个在上一个日期重复，则不再计算

所以，我需要这样的东西：

DATE          COUNT(REGCODE)
02/01/15           2
03/01/15           1
05/01/15           1

非常感谢你的帮助。

Answer 1

SELECT t.date, COUNT(DISTINCT t.regcode)
FROM yourtable t
WHERE NOT EXISTS (
   SELECT 1 
   FROM yourtable prev
   WHERE prev.regcode=t.regcode
   AND prev.date<t.date
)
GROUP BY t.date;

（可能需要注册码，日期索引）

Answer 2

试试这个。它使用ROW_NUMBER()和PARTITION BY使用REGCODE。

declare @table table ([DATE] date, REGCODE int)

insert into @table
select '02/01/15', 70771 union
select '02/01/15', null union
select '02/01/15', 70855 union
select '02/01/15', null union
select '03/01/15', 70771 union
select '03/01/15', 70772 union
select '03/01/15', null union
select '05/01/15', 70855 union
select '05/01/15', 70826

select [DATE], COUNT([DATE]) from 
(
    select [DATE], REGCODE, 
    row_number() over (partition by  REGCODE order by [DATE]) rn 
    from @table 
) e 
where rn = 1 and regcode is not null
group by [DATE]

按日计算不同的组

2 个答案: