Mysql SQL-通过不同的条件聚合同一列

假设我有一张这样的桌子

|country| customer_id | order_id |
| CA    | 5           |     3    |
| CA    | 5           |     4    |
| CA    | 6           |     5    |
| CA    | 6           |     6    |
| US    | 2           |     7    |
| US    | 7           |     8    |
| US    | 7           |     9    |
| US    | 7           |    10    |
| US    | 2           |    11    |

我想写一个查询来填充一个表

| country | customers_w_2_orders | customers_w_2_plus_orders |
| CA      | 2                    | 0                         |
| US      | 1                    | 1                         |

其中，按国家/地区汇总2份订单的客户数量和3份订单的客户数量

这就是我所做的，它没有给我想要的结果

SELECT country, count(*) as cnt1, count(*) as cnt2 
FROM Orders 
GROUP BY country 
HAVING cnt1=2 AND cnt2>2;

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dbfiddle

首先构建一个表，其中包含每个客户及其在每个国家/地区的订单数量，其中每行是国家/地区、客户id、订单数量

现在，您可以通过在派生表上分组来计算\u订单数为2或大于2的频率

select country, sum(num_orders = 2), sum(num_orders > 2)
from (
    select country, customer_id, count(*) as num_orders
    from Orders
    group by country, customer_id
) t group by country

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将无法获得OP想要的格式。您将生成一个稀疏表。您提供的查询在0.5毫秒内完成，这比其他答案要好。谢谢我很高兴能帮上忙。 country | cust_w_2_orders | cust_2_plus_orders :------ | --------------: | -----------------: CA | 2 | 0 US | 1 | 1

select country, sum(num_orders = 2), sum(num_orders > 2)
from (
    select country, customer_id, count(*) as num_orders
    from Orders
    group by country, customer_id
) t group by country

SELECT country,
       (select count(distinct(customer_id)) from Orders o where o.country = Orders.country and (select count(*) from Orders o2 where o2.country = orders.country and o2.customer_id = o.customer_id) = 2) as customers_w_2_orders,
       (select count(distinct(customer_id)) from Orders o where o.country = Orders.country and (select count(*) from Orders o2 where o2.country = orders.country and o2.customer_id = o.customer_id) > 2) as customers_w_2_plus_orders
  FROM Orders 
 GROUP BY country;