Question

所以说我们有行

ID | Date 
1 1/20 
2 1/20 
1 1/21 
7 1/21 
4 1/22 
5 1/22

所以说我只想看到出现在1/20和1/21的ID。这应该只给我ID 1，因为这是1/20和1/21中出现的唯一行。实现这一目标的最简单方法是什么？

我试过这样做：

Select ID, [date]
FROM t1
INNER JOIN (
SELECT ID, Count(*) countRow
FROM t1
Where [date] in(1/20, 1/21)
GROUP BY ID
having count(DISTINCT [Date]) > 1
) aa on t1.id = aa.id

我觉得这是实现这一目标的简单方法。有什么想法吗？

Answer 1

你现在拥有的方式非常简单。使用exists()的替代方法：

select 
    t.id
  , t.date
from t
where t.date in (1/20, 1/21)
  and exists (
    select 1
    from t as i
    where i.id = t.id
      and i.date <> t.date
      and i.date in (1/20, 1/21)
  )

Answer 2

一种方法使用group by和having：

select id
from t1
where date in ('1/20', '1/21')
group by id
having count(distinct date) = 2;

当然，如果date确实存储为日期，您应该将日期常量的格式修改为YYYY-MM-DD。

Answer 3

SELECT ID FROM T1 WHERE Date = '1/20' AND EXISTS (
SELECT ID FROM T1 AS T2 WHERE T2.Date = '1/21' AND 
T1.ID = T2.ID)

Answer 4

不确定是否更好郎你这样做是非常标准的不确定为什么你需要列出那个条件

的日期

如果ID，日期是唯一的，那么您可以使用count（*）

Select ID, [date]
FROM t1
INNER JOIN ( SELECT ID
             FROM t1
             Where [date] = 1/20
             intersection
             SELECT ID
             FROM t1 
             Where [date] = 1/21
           ) aa 
  on t1.id = aa.id

Select ID, [date]
FROM t1
INNER JOIN ( SELECT ID
             FROM t1 t1a 
             join t1 t1b 
               on t1a.ID = t1b.ID
              and t1a.[date] = 1/20
              and t1b.[date] = 1/21
           ) aa 
  on t1.id = aa.id

如果是id，则值是唯一的

select id, value 
from (select id, value
           , count(*) over (partition by id order by value) as cnt 
       from t 
       where value in ('a', 'b')
     ) td 
where cnt = 2 
order by id, value

TSQL仅查询在指定的多个日期中出现的行？

4 个答案: