Return 当前行之前的最新记录

Question

我有 table 条消息。每条消息都有时间戳 created_ts、类型（purchase 或 support）、user_id 和主键 msg_id。例如，

msg_id	user_id	created_ts	type
1	1	1	purchase
2	1	2	support
3	2	3	purchase
4	2	4	support
5	2	5	support

我想生成 table 支持消息，这样每一行都将包含支持消息的详细信息，以及最后一次购买的详细信息（如果存在），即purchase 类型的最新消息，发生在该行的支持消息之前。例如，

support_msg_id	user_id	support_created_ts	type	last_purchase_msg_id	last_purchase_created_ts
2	1	2	support	1	1
4	2	4	support	3	3
5	2	5	support	3	3

我试过以下查询：

select * from
(select msg_id as support_msg_id, user_id, created_ts as support_created_ts, type
  (select msg_id as last_purchase_msg_id
   from messages
   where type = 'purchase' and created_ts < support_created_ts
 order by created_ts desc
limit 1)
  
 from messages
 where type = 'support'
) as supports
inner join
(
 select msg_id as last_purchase_msg_id, created_ts as last_purchase_ts
   from messages
   where type = 'purchase'
) as purchases
on supports.last_purchase_msg_id = purchases.last_purchase_msg_id

然而这太慢了。

Answer 1

您的问题的解决方案是双重使用 window 函数。首先使用 count 函数来识别组的边界，其中每个组以 purchase 开头并继续所有后续的 support。然后使用 max 函数在每个组中获取（实际上是单个非空）值。

Dbfiddle 适用于 Postgres，但仅使用标准 SQL，我希望它也可用于 Redshift。

with messages(msg_id,   user_id,    created_ts,     type) as (values
(1,     1,  1,  'purchase'),
(2,     1,  2,  'support'),
(3,     2,  3,  'purchase'),
(4,     2,  4,  'support'),
(5,     2,  5,  'support')
)
, precomputed as (
  select m.*
       , case m.type when 'purchase' then m.msg_id end as last_purchase_msg_id
       , case m.type when 'purchase' then m.created_ts end as last_purchase_created_ts
       , count(case m.type when 'purchase' then 1 end) over (order by m.created_ts) as grp
  from messages m
)
, lasts as (
  select g.msg_id, g.user_id, g.created_ts, g.type
       , max(g.last_purchase_msg_id) over (partition by g.grp) as last_purchase_msg_id
       , max(g.last_purchase_created_ts) over (partition by g.grp) as last_purchase_created_ts
  from precomputed g
)
select * from lasts
where type = 'support'

Answer 2

with p as (
    select *, lead(created_ts) over (partition by user_id order by created_ts) as nextts
    from messages where type = 'purchase'
)
select *
from messages s
left outer join p on p.user_id = s.user_id
    and s.created_ts >= p.created_ts and (s.created_ts < p.nextts or p.nextts is null)
where s.type = 'support'

https://dbfiddle.uk/?rdbms=postgres_10&fiddle=aea646e85dfa35896f8cacfabec439d6

Return 当前行之前的最新记录

Return the latest record which was before the current row

sql

amazon-redshift