Select 每个客户的第一个和第 n 个订单

Question

我正在尝试为每个用户显示第一个和第五个 order_id。

Table 看起来像这样：

+----------+-----------+-------------------+
| Order_ID | Client_ID | Datetime          |
+----------+-----------+-------------------+
| 1        | 1         | YYYYMMDD HH:MM:SS |
+----------+-----------+-------------------+
| 2        | 1         | YYYYMMDD HH:MM:SS |
+----------+-----------+-------------------+
| 3        | 2         | YYYYMMDD HH:MM:SS |
+----------+-----------+-------------------+

我写了这样的东西：

select 
  t.client_id, 
  t.order_id as first_order, 
  t2.order_id as fifth_order,
  t.datetime as first_dt,
  t2.datetime as fifth_dt,
from
(
  select o.client_id, o.order_id, o.datetime,
  row_number() over(partition by o.client_id order by o.datetime) as rn
  from "OhMyTable" as o
) as t
  left join
    (
  select o.client_id, o.order_id, o.datetime,
  row_number() over(partition by o.client_id order by o.order_id) as rn
  from "OhMyTable" as o
      order by o.order_id
  ) as t2
  on t.client_id = t2.client_id
where t.rn = 1 and t2.rn = 5

但我想看到下了第一个订单但没有第五个订单的客户。应该有NULL，但我不明白如何制作它。 :c

Answer 1

将where t2.rn = 5条件放入left join:

select 
  t.client_id, 
  t.order_id as first_order, 
  t2.order_id as fifth_order,
  t.datetime as first_dt,
  t2.datetime as fifth_dt,
from
(
  select o.client_id, o.order_id, o.datetime,
  row_number() over(partition by o.client_id order by o.datetime) as rn
  from "OhMyTable" as o
) as t
  left join
    (
  select o.client_id, o.order_id, o.datetime,
  row_number() over(partition by o.client_id order by o.order_id) as rn
  from "OhMyTable" as o
      order by o.order_id
  ) as t2
  on t.client_id = t2.client_id
  and t2.rn = 5
where t.rn = 1 and t2.datetime is null

Answer 2

如果发现条件聚合比两个子查询更简单。您可以使用 having 子句进行过滤：

select client_id,
    max(order_id) filter(where rn = 1) first_order,
    max(order_id) filter(where rn = 5) fifth_order,
    min(datetime) first_dt,
    max(datetime) fifth_dt
from (
    select o.client_id, o.order_id, o.datetime,
        row_number() over(partition by o.client_id order by o.datetime) as rn
    from "OhMyTable" as o
) t
where rn in (1, 5)
group by client_id
having count(*) = 2

Answer 3

使用条件聚合：

select o.client_id,
       min(datetime) as first_datetime,
       max(datetime) as fifth_datime,
       max(case when seqnum = 1 then order_id end) as first_orderid,
       max(case when seqnum = 5 then order_id end) as fifth_orderid
from (select o.*
             row_number() over(partition by o.client_id order by o.datetime) as seqnum
      from "OhMyTable" o
     ) o
where seqnum in (1, 5);

Select 每个客户的第一个和第 n 个订单

Select first and nth order for each client

sql

postgresql

pivot

left-join

greatest-n-per-group