Question

我有一种情况，我想提取在特定期间内每天仅存在一次且仅一次的设备ID（DID）。我尝试了不同的方法和分区，但似乎每天只能单独获取该数据（日期= X，但是我需要查询可以在X和Y之间放置日期的地方）

示例，这是数据：

DID date     
A   2019-01-01
A   2019-01-01
A   2019-01-02
A   2019-01-03
B   2019-01-01
B   2019-01-02
B   2019-01-03
C   2019-01-01
C   2019-01-02
C   2019-01-02
C   2019-01-03
D   2019-01-01
D   2019-01-02
D   2019-01-03

查询应仅返回B＆D（因为B＆D从01到03每天存在一次）我也希望得到计数，在这种情况下为2

谢谢！

Answer 1

您希望设备在该时段的每个日仅一次存在，因此，如果您group by did需要返回did count(date)和count(distinct date)等于该时间段的天数：

select did
from tablename
where date between cast('2019-01-01' as date) and cast('2019-01-03' as date)
group by did
having 
  count(distinct date) = cast('2019-01-03' as date) - cast('2019-01-01' as date) + 1
  and
  count(date) = cast('2019-01-03' as date) - cast('2019-01-01' as date) + 1

请参见demo。
或者：

select t.did
from (
  select did, date
  from tablename
  where date between cast('2019-01-01' as date) and cast('2019-01-03' as date)
  group by did, date
  having count(*) = 1
)t  
group by t.did
having count(*) = cast('2019-01-03' as date) - cast('2019-01-01' as date) + 1

请参见demo。
结果：

| did |
| --- |
| B   |
| D   |

Answer 2

一个选择是用DID进行聚合，并断言总计数等于不同日期的计数。如果此断言通过，则意味着给定的DID仅具有不同的日期。

SELECT DID
FROM yourTable
GROUP BY DID
HAVING COUNT(date) = COUNT(DISTINCT date);

Demo

如果要获取匹配的DID的总数，则可以对上面的子查询进行取COUNT(*)。或者，如果您想使用相同的查询，则可以尝试：

SELECT DID, COUNT(*) OVER () AS total_cnt
FROM yourTable
GROUP BY DID
HAVING COUNT(date) = COUNT(DISTINCT date);

获取日期范围内每天仅存在一次的唯一记录

2 个答案:

Demo