Question

我的问题是找到所有那些已经达到最低和最高无订单的客户。

我能想出的是：

select customer_id , count(order_id) as num 
from bab_customer right outer join bab_order_details using(customer_id)
group by customer_id
having count(order_id) >= 
    all(select count(order_id) 
        from bab_customer right outer join bab_order_details using(customer_id)
        group by customer_id)
or count(order_id) <= 
    all(select count(order_id) 
        from bab_customer right outer join bab_order_details using(customer_id)
        group by customer_id);

它给了我正确的输出，但我必须做同样的连接3次。有更好的方法吗？

Answer 1

这样更好吗？我不知道......

SELECT x.* 
  FROM 
     ( SELECT customer_id
            , COUNT(*) cnt 
         FROM orders 
        GROUP  
           BY customer_id
     ) x 
  JOIN 
     ( SELECT MIN(cnt) min_cnt
            , MAX(cnt) max_cnt 
         FROM 
            ( SELECT customer_id
                   , COUNT(*) cnt 
                FROM orders 
               GROUP 
                  BY customer_id
            ) n
     ) y 
    ON y.min_cnt = x.cnt 
    OR y.max_cnt = x.cnt;

Answer 2

我意识到，如果不多次调用相同的子查询，就无法在单个查询中运行它。当您引用min（numOrders）和max（numOrders）时，您必须将numOrders存储为时态表或计算两次。如果请求相同，DBMS很可能会缓存您的请求，因此您不必担心浪费资源。

所以你的查询必须有点重写并且足够好。我的变种是

select customer_id , count(order_id) as num 
from customers right outer join orders using(customer_id)
group by customer_id
having 
  num = (SELECT min(num) FROM (select count(order_id) as num 
    from customers right outer join orders using(customer_id)
    group by customer_id) numOrders)
or  
  num = (SELECT max(num) FROM (select count(order_id) as num 
    from customers right outer join orders using(customer_id)
    group by customer_id) numOrders)

您可以使用此sqlfiddle

进行检查

Answer 3

如果我绝对觉得这必须是一个查询，我可能只使用union all。

(
select
  customer_id,
  count(*) as num
from bab_customer right outer join bab_order_details using(customer_id)
group by customer_id
order by num desc limit 1
)
union all
(
select
  customer_id,
  count(*) as num
from bab_customer right outer join bab_order_details using(customer_id)
group by customer_id
order by num asc limit 1
)

最初，我一直认为这会因为避免连接中间值而表现更好。但是，我没有看到任何性能提升。（如果它更快，那么限制可以提升到某个任意数字，并且应用程序可以减少一些额外的值而不会损失大部分增加的性能）

但是，没有骰子。

获得最大和最小订单数量的客户的更好方法

3 个答案: