Question

我需要找到将图书归还图书馆的平均失踪天数最高的学生。我有2张桌子：

用户 - ＆gt; id，fname，lname，professional
借出 - ＆gt; isbn（书号），id，自（日期），到期日（日期），实际（日期）

这是我使用子查询

执行此操作的方法

select concat(fname, '', lname) as name 
  from users 
 where id in ( select id 
                 from loaned 
                group by id 
               having avg(datediff(due, actual))
                      = ( select min(m) 
                            from ( select avg(datediff(due, actual)) as m 
                                     from loaned
                                    group by id 
                                          ) as minavg
                                  )
                      );

我怎样才能用JOIN解决这个问题？

Answer 1

select concat(fname, '', lname) as name 
from users u
join loaned l on l.id = u.id and due < actual
group by 1
order by avg(actual - due) desc
limit 1

此查询仅取得延迟的图书的平均值，而不是整体平均值。对于全部平均值，请移除and due < actual

返回所有平均分享最高的学生：

select concat(fname, '', lname) as name 
from users u
join loaned l on l.id = u.id and due < actual
group by 1
having avg(actual - due) = (
    select avg(actual - due)
    from loaned
    where due < actual
    group by id
    order by 1 desc
    limit 1)

选择平均失踪天数最多的学生

1 个答案: