mysql - 如何从左连接中获取前 n 个不同的列值？

Question

我想要得到的是“获得前 5 个不同客户的订单 ID 值”。当我将以下查询限制为 5 时，它只会按预期返回前 4 个客户的订单 ID 值。(Around the Horn 有两个不同的订单 ID)有没有办法获得前 5 个不同客户的订单 ID 值？

SELECT Customers.CustomerName, Orders.OrderID
FROM Customers
LEFT JOIN Orders
ON Customers.CustomerID=Orders.CustomerID
ORDER BY Customers.CustomerName;


CustomerName                        OrderID
Alfreds Futterkiste                 10001
Ana Trujillo Emparedados y helados  10308
Antonio Moreno Taquería             10365
Around the Horn                     10355
Around the Horn                     10383
B's Beverages                       10289
Antonio baa                         10444
....                                ....

编辑:我想要得到的是前 5 个不同的客户；

Alfreds Futterkiste                 10001 (first customer)
Ana Trujillo Emparedados y helados  10308 (second customer)
Antonio Moreno Taquería             10365 (third customer)
Around the Horn                     10355 (forth customer)
Around the Horn                     10383 (forth customer)
B's Beverages                       10289 (fifth customer)

在这里are我正在处理的表格。

Answer 1

试试这个。

            SELECT DISTINCT Orders.OrderID, CC.CustomerName

            FROM Customers CC
            LEFT JOIN Orders
            ON CC.CustomerID=Orders.CustomerID
            inner join  

            (
            SELECT DISTINCT Customers.CustomerName
            FROM Customers
            LEFT JOIN Orders
            ON Customers.CustomerID=Orders.CustomerID
            ORDER BY Customers.CustomerName
            LIMIT 5
            ) C 
            ON C.CustomerName=CC.CustomerName

            ORDER BY CC.CustomerName