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

您似乎想要条件聚合：

select name,
       max(case when seqnum = 1 then domain end) as domain_1,
       max(case when seqnum = 1 then count end) as count_1,
       max(case when seqnum = 2 then domain end) as domain_2,
       max(case when seqnum = 2 then count end) as count_2,
       date(datetime)
from (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'
     ) t
group by name, date(datetime);

从查询中获取前两个域

1 个答案: