Question

我有这些表格：

customer
--------
customer_id int
name        varchar(255)

order
-----
order_id    int
customer_id int
discount    boolean

我可以通过以下查询获得每位客户的订单数量：

select c.id, count(o.order_id)
from customer c 
left join order as o using c.customer_id = o.customer_id
group by 1

或者，我可以通过以下方式获得每位客户的折扣订单数量：

select c.id, count(o.order_id)
from customer c 
left join order as o using c.customer_id = o.customer_id and o.discount = true
group by 1

但我无法想出一种方法可以在一个查询中获得两者。我尝试了以下内容：

select c.id, count(o.order_id), count(o2.order_id)
from customer c 
left join order as o using c.customer_id = o.customer_id
left join order as o2 using c.customer_id = o2.customer_id and o2.discount = true
group by 1

但它没有用。是否可以在单个（MySql）查询中计算两者？

干杯，唐

Answer 1

像

这样的东西

select c.id, count(o.order_id),sum(if(o.discount,1,0))
from customer c 
left join order as o using c.customer_id = o.customer_id
group by c.id

Answer 2

你可以做像

这样的事情

select 
 c.id,
 sum(case o.discount when true then 1 else 0 end) as 'total discounted',
 count(o.order_id) as 'total orders'
from customer as c
 left join order as o using c.customer_id = o.customer_id 
group by c.id

Answer 3

其他答案很接近，但这是我写的方式：

SELECT c.id, COUNT(o.order_id) AS order_count, 
  SUM(o.discount = true) AS discount_order_count
FROM customer c 
  LEFT OUTER JOIN order AS o USING (customer_id)
GROUP BY c.id;

请注意USING的使用需要括号，并且只接受将与=进行比较的列列表。您无法使用USING语法使用ON语法提供完整的比较表达式。

此外，您可以简化SUM()内的表达式，因为相等比较返回1或0。

另请参阅“Query: count multiple aggregates per item”

按查询挑战分组

3 个答案: