Question

我为拥有2011年约会的网站的客户提供客户ID

SELECT customers.id
FROM customers 
INNER JOIN sites ON customers.id = sites.customer_id
INNER JOIN appointments ON appointments.site_id = sites.id
WHERE YEAR(appointments.day) = 2011

我希望得到2011年没有任何约会的客户。对于如何去做而感到困惑。

Answer 1

有两种方式：

SELECT
    C.id
FROM
    Customers C
WHERE
    NOT EXISTS (
        SELECT *
        FROM
            Sites S
        INNER JOIN Appointments A ON
            A.site_id = S.id AND
            YEAR(A.day) = 2011
        WHERE
            S.customer_id = C.id
    )

或：

SELECT DISTINCT
    C.id
FROM
    Customers C
INNER JOIN Sites S ON
    S.customer_id = C.id
LEFT OUTER JOIN Appointments A ON
    A.site_id = S.id AND
    YEAR(A.day) = 2011
WHERE
    A.id IS NULL

Answer 2

......年纪不到（appointmentments.day）= 2011

Answer 3

应该做一个答案而不是编辑。由于缩放，不想真正寻找NULL。

SELECT id
FROM customers 
WHERE NOT EXISTS (SELECT * FROM sites 
                      INNER JOIN appointments ON appointments.site_id = sites.id
                  WHERE sites.customer_id = customers.id AND YEAR(appointments.day) = 2011)

Answer 4

 SELECT customers.id
 FROM customers, sites, appointments
 WHERE (customers.id = sites.customer_id and appointments.id = sites.id) 
 and  YEAR(appointments.day) <> 2011

请检查您的特定数据库服务器“＆lt;＆gt;”这是“不平等”。

MySQL Join不存在关联的地方

4 个答案: