Question

示例数据：

Name       | Domain    | Count | datetime
----------------------------------------------------
John       | google    | 5     | 2019-07-09 01:00:01
John       | apple     | 6     | 2019-07-09 01:02:01
John       | apple     | 8     | 2019-07-09 01:03:01
John       | google    | 10    | 2019-07-09 01:11:01
John       | dos       | 1     | 2019-07-09 01:11:01
John       | dos       | 3     | 2019-07-09 01:11:01

预期输出：

Name       | max_Domain1| max_Count1 | max_Domain2 | max_Count2 | datetime
--------------------------------------------------------------------------
John       | google     | 10          | apple      | 8         |   2019-07-09

我尝试在分区上使用row_number（），但它给我的日期值与前2个最大值相同。

select t.*,
             row_number() over (partition by name, date(datetime) order by count desc) as seqnum
      from table t
      where datetime >= '2019-07-08' and
            datetime < '2019-07-09'

Answer 1

您只需要按年，月和日进行分区，而不要按整个日期时间进行分区：

with main as
    (
    select name,domain,count,datetime
    ,row_number() over (partition by 
    name,domain,count,year(datetime)*10000+month(datetime)*100+day(datetime) order by 
    datetime desc) as row_number 
    from table
    )
select *
from main
where row_number = 1

Answer 2

尝试以下操作：

select name, domain, count,  CONVERT(varchar(10),datetime, 126) as datetime
from
(
select top 2 t.*
from tab t
where datetime >= CONVERT(datetime,'2019-07-08', 101) and
      datetime < CONVERT(datetime,'2019-07-10', 101)
order by row_number() over (partition by name, domain, cast(datetime as date) order by count desc) asc, count desc
) as t2
order by t2.datetime asc

示例here

Answer 3

您似乎想要每天最多的计数：

select t.*
from (select t.*,
              row_number() over (partition by name, date(datetime) order by count desc) as seqnum
      from tab t
      where datetime >= '2019-07-08' and
            datetime < '2019-07-10'
     ) t
where t.seqnum = 1
order by count desc

前2个最大值（不包括重复值）

3 个答案: