Question

我要做的是将查询优化95％，作为学习psql的任务的一部分。我想要的是所有拥有与第三大借书相同或更大的客户。

explain analyze
select clb.f_name, clb.l_name, noofbooks 
from 
      (select f_name, l_name, count(*) as noofbooks 
      from customer left join loaned_book on customer.customerid = loaned_book.customerid 
      group by f_name, l_name) as clb,
      (select f_name as fname, l_name as lname, count(*) as noofbooksf 
      from customer left join loaned_book on customer.customerid = loaned_book.customerid
      group by fname, lname order by noofbooksf  desc  LIMIT 1 OFFSET 2) as clb1
where noofbooks  >= noofbooksf 
order by noofbooks desc;

这就是我到目前为止，它已经优化到高80但是这两个连接增加了成本。因此我要做的就是做一次加入。然后使用结果获取它的第三个元素。我想创建一个临时视图，但我确信会有更好的方法。

感谢。

Answer 1

如果你想要拥有第一或第二大借书的人，那就去做：

select c.f_name, c.l_name, count(lb.customerid) as noofbooks 
from customer c left join
     loaned_book lb
     on c.customerid = lb.customerid 
group by c.f_name, c.l_name
order by noofbooks desc
limit 3;

如果你担心关系：

select f_name, l_name, noofbooks
from (select c.f_name, c.l_name, count(lb.customerid) as noofbooks,
             rank() over (order by noofbooks desc) as seqnum
      from customer c left join
           loaned_book lb
           on c.customerid = lb.customerid 
      group by c.f_name, c.l_name
     ) clb
where seqnum <= 3;

Postgresql - 在另一个select语句中使用From中的变量

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