SQL最早的记录

问题描述 投票:2回答:3

假设我有一个20列的表格订单。我只对前4列感兴趣:id,department_id,region_id,datetime,其中id是客户ID,datetime是客户下订单的时间。其他列更具体于产品详细信息(例如product_id),因此在给定订单上,您可能有多行。我很难写一个查询来让每个客户获得最早的部门和区域,因为同一个客户可以有多个department_id和region_id的组合。

SELECT a.*
FROM (
    SELECT id,
        department_id,
        region_id,
        min(DATETIME) AS ts
    FROM orders
    GROUP BY id,
        department_id,
        region_id
    ) a
INNER JOIN (
    SELECT id,
        min(DATETIME) AS ts
    FROM orders
    GROUP BY id
    ) b
    ON a.id = b.id
        AND a.ts = b.ts

这似乎有效,但它看起来效率不高,写得不好。有没有更好的方法来写这个?表本身相当大,因此查询速度很慢。

sql hive hiveql
3个回答
0
投票

我会这样做:

SELECT id, department_id, region_id, datetime
FROM (SELECT o.*
             row_number() over (partition by id order by datetime) as seqnum
      FROM orders o
     ) o
where seqnum = 1;

编辑:

您可以尝试使用此版本以查看它是否更好:

select o.*
from orders o join
     (select id, min(datetime) as min_datetime
      from orders
      group by id
     ) oo
     on oo.id = o.id and oo.datetime = o.datetime;

在大多数数据库中,row_number()版本可能具有更好的性能。但是,Hive可以做出神秘的优化决策,这可能会更好。


0
投票

我想你可能会喜欢这样:

SELECT id, department_id, region_id, min(datetime) AS ts 
FROM orders 
GROUP BY id, department_id, region_id 
HAVING ts=min(datetime)

0
投票

使用dense_rank()分析函数:

SELECT 
        id,
        department_id,
        region_id,
        min(DATETIME) AS ts
 FROM
(
SELECT  id,
        department_id,
        region_id,
        DATETIME,
        dense_rank() over(partition by id order by DATETIME) AS rnk
  FROM orders
)s 
WHERE rnk=1 --records with minimal date by id
GROUP BY id,
         department_id,
         region_id;

此查询与您的查询相同,但该表将被扫描一次,无需加入。

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