选择按日期记录的最高发生次数

Question

表结构:

ScheduledDate Date
Commodity varchar    

示例数据

ScheduledDate | Commodity
--------------+------------
 2017-11-27   | KIDNEY BEANS
 2017-11-27   | KIDNEY BEANS
 2017-11-27   | RED BEANS
 2017-11-28   | LARGE GREEN LENTIL
 2017-11-28   | SMALL GREEN LENTIL
 2017-11-29   | LARGE GREEN LENTIL

我需要的东西:表格中每日最高出现商品的一条记录,按日期排序.所以期望的输出将是:

ScheduledDate | Commodity
--------------+------------
 2017-11-27   | KIDNEY BEANS
 2017-11-28   | LARGE GREEN LENTIL
 2017-11-29   | LARGE GREEN LENTIL

...如果多个商品在某个日期出现的次数相同,那么我只需要退回一次.没有真正的偏好.

我认为我非常接近但只需要最后一块拼图来解决这个问题......

SELECT
     ScheduledDate
    ,Commodity
    ,OCCURANCE_COUNT = MAX(OCCURANCES)
FROM (
    SELECT 
        ScheduledDate
        ,Commodity
        ,OCCURANCES = COUNT(Commodity)
    FROM 
        vwScheduledLoads 
    GROUP BY
        ScheduledDate,
        Commodity
) qc
GROUP BY
    ScheduledDate
    ,Commodity
ORDER BY
    ScheduledDate

...内部SQL做了我想要的,我想要对结果进行分组的部分是我难以理解的.该查询吐出以下内容:

ScheduledDate | Commodity          | OCCURANCE_COUNT
--------------+--------------------+-----------------
 2017-11-27   | KIDNEY BEANS       | 2
 2017-11-27   | RED BEANS          | 1
 2017-11-28   | LARGE GREEN LENTIL | 1
 2017-11-28   | SMALL GREEN LENTIL | 1
 2017-11-29   | LARGE GREEN LENTIL | 1

Answer 1

您想要一个聚合查询row_number():

select ScheduledDate, Commodity
from (select ScheduledDate, Commodity, count(*) as cnt,
             row_number() over (partition by ScheduledDate order by count(*) desc) as seqnum
      from t
      group by ScheduledDate, Commodity
     ) sc
where seqnum = 1;