在PostgreSQL中选择每个客户的最后两条记录

Question

你们能否给我看看以下情况的查询？ 我在表access_log这些列：

customer_id | service_name | accessed_time 

一位顾客可能会访问该服务的任何数量的时间。但是我只需要列出每个customer_id的最后两条记录。

Answer 1

东西沿着这些路线应该工作：

select * from access_log a1 
where 2 > (select count(*) from access_log a2 
      where a1.customer_id = a2.customer_id 
      and a1.accessed_time < a2.accessed_time) 

这写着：获得其中存在0或1，其他的访问日志为同一客户与后来的“accessed_time”的所有日志。确保在相关列上有适当的索引。

Answer 2

Window functions救援（再次）：

select customer_id, service_name, accessed_time 
from (
    select customer_id, service_name, accessed_time, 
      rank() over (partition by customer_id order by accessed_time desc) as rank 
    from access_log 
) dt 
where dt.rank <= 2 

这假定 “过去两年” 是指 “两个最近期的”。取决于您想如何处理重复项，row_number窗口函数可能更合适。

这样的假设数据（对不起，我不觉得有想象力今晚）：

=> select * from access_log order by customer_id, accessed_time; 
customer_id | service_name | accessed_time  
-------------+--------------+--------------------- 
      1 | one   | 2011-01-01 00:00:00 
      1 | two   | 2011-01-02 00:00:00 
      1 | three  | 2011-01-03 00:00:00 
      2 | two   | 2011-01-02 00:00:00 
      2 | one   | 2011-04-01 00:00:00 
      2 | three  | 2011-05-03 00:00:00 

上面的查询率：

customer_id | service_name | accessed_time  
-------------+--------------+--------------------- 
      1 | three  | 2011-01-03 00:00:00 
      1 | two   | 2011-01-02 00:00:00 
      2 | three  | 2011-05-03 00:00:00 
      2 | one   | 2011-04-01 00:00:00