mysql - MySQL - 选择所有客户和每个客户的总订单和总价值

Question

对于每个客户，我想返回：id、name、total_orders、total_value

顾客：

╔════╦═════════════╗
║ ID ║    NAME     ║
╠════╬═════════════╣
║  1 ║ John Smith  ║
║  2 ║ Jim Jimmers ║
╚════╩═════════════╝

订单：

╔═══════╦══════════╦═══════╗
║  ID   ║ CUSTOMER ║ VALUE ║
╠═══════╬══════════╬═══════╣
║ 34656 ║        1 ║    20 ║
║ 37345 ║        2 ║    25 ║
║ 38220 ║        1 ║    15 ║
║ 39496 ║        1 ║    38 ║
║ 41752 ║        1 ║     9 ║
║ 43734 ║        2 ║    20 ║
╚═══════╩══════════╩═══════╝

如何选择如下结果：

╔════╦═════════════╦═════════════╦═════════════╗
║ ID ║    NAME     ║ TOTALORDERS ║ TOTAL_VALUE ║
╠════╬═════════════╬═════════════╬═════════════╣
║  1 ║ John Smith  ║           4 ║          82 ║
║  2 ║ Jim Jimmers ║           2 ║          45 ║
╚════╩═════════════╩═════════════╩═════════════╝

Answer 1

SELECT  a.ID,
        a.Name,
        COUNT(b.Customer) totalOrders,
        SUM(b.value) total_value
FROM    Customers a
        LEFT JOIN Orders b
            ON a.ID = b.Customer
GROUP   BY a.ID,
        a.Name

• SQLFiddle 演示

或者

SELECT  a.ID,
        a.Name,
        COUNT(b.Customer) totalOrders,
        COALESCE(SUM(b.value), 0) total_value
FROM    Customers a
        LEFT JOIN Orders b
            ON a.ID = b.Customer
GROUP   BY a.ID,
        a.Name

要进一步了解有关联接的更多信息，请访问以下链接：

• SQL 连接的可视化表示

结果，

╔════╦═════════════╦═════════════╦═════════════╗
║ ID ║    NAME     ║ TOTALORDERS ║ TOTAL_VALUE ║
╠════╬═════════════╬═════════════╬═════════════╣
║  1 ║ John Smith  ║           4 ║          82 ║
║  2 ║ Jim Jimmers ║           2 ║          45 ║
╚════╩═════════════╩═════════════╩═════════════╝

Answer 2

此查询应提供您的贬低输出：

SELECT c.id, c.name, count(*) AS total_orders, sum(o.value) AS total_value
FROM Customers AS c
     LEFT JOIN Orders AS o ON c.id = o.customer
GROUP BY c.id

Answer 3

使用内部联接，然后通过对用户 ID 进行分组来求和。

SQL join 用于通过相等点合并表，你应该看看这个http://dev.mysql.com/doc/refman/5.0/en/join.html

抱歉，我现在没有 mysql 服务器 bu 是这样的：

select id, name, sum(value) from (select customers.id as id, customers.name as name, orders.value as value from customers inner join orders on customers.id = orders.customer) group by id

几乎可以肯定不是很正确，但在某个接近的地方。