如何从两个不同的日期比较中找到结果集？

Question

如何找到导致从两个不同的日期比较

如何找到period1_label_cost设置，period2_label_cost在此查询，而无需使用此查询？

只有period1_label_cost列的值越来越

select distinct customers.id as customer_id, customers.first_name as first_name, customers.last_name as last_name, SUM(orders.total_cost) as period1_label_cost 

from customers inner join orders 
      on customers.id= orders.customer_id 

      where 
      date(orders.created_at) between 'start_date1' and 'end_date1' 

     group by customers.id , customers.first_name, customers.last_name, customers.preferred 
      having(SUM(orders.total_cost) > sales_exceeded 
intersect 

     select distinct customers.id as customer_id, customers.first_name as first_name, customers.last_name as last_name, 
     customers.preferred as preferred, SUM(orders.total_cost) as period2_label_cost 

     from customers inner join orders 
     on customers.id= orders.customer_id 


     where 
      date(orders.created_at) between start_date2 and end_date2 


     group by customers.id , customers.first_name, customers.last_name, customers.preferred 
     having(SUM(orders.total_cost) < sales_below) order by first_name asc 

Answer 1

您是否在寻找纯粹的加入？将你的两个查询放在你的FROM子句中，并加入customer_id，然后加入customers表并显示结果。

select 
    c.id as customer_id, 
    c.first_name, 
    c.last_name, 
    c.preferred, 
    period1.label_cost as period1_label_cost, 
    period2.label_cost as period2_label_cost 
from 
(
    select 
    customer_id, 
    sum(total_cost) as label_cost 
    from orders 
    where date(created_at) between <start_date1> and <end_date1> 
    group by customer_id 
    having sum(total_cost) > <sales_exceeded> 
) period1 
join 
(
    select 
    customer_id, 
    sum(total_cost) as label_cost 
    from orders 
    where date(created_at) between <start_date2> and <end_date2> 
    group by customer_id 
    having sum(total_cost) < <sales_below> 
) period2 on period2.customer_id = period1.customer_id 
join customers c on c.id = period1.customer_id;