Question

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

客户：

╔════╦═════════════╗
║ 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 ║
╚═══════╩══════════╩═══════╝

如何选择以下结果：

╔════╦═════════════╦═════════════╦═════════════╗
║ 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 Demo

OR

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

要进一步了解联接，请访问以下链接：

Visual Representation of SQL Joins

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deriderated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，然后通过分组用户ID来使用总和。

SQL连接用于通过等式点合并表，您应该看一下这个http://dev.mysql.com/doc/refman/5.0/en/join.html

抱歉，我现在没有mysql服务器，就像这样：

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

几乎可以肯定不是很正确，但在某个地方很近。

MySQL - 选择所有客户和每个客户的总订单和总价值

3 个答案: