Question

我有一张桌子可以在任何特定日期存储一个人的茶水偏好。 ;）

customer_id, tea_preference,     date
     1        'yes please'    2011-01-01
     1        'no thanks'     2011-01-02

我需要在给定的日期范围内找到喜欢茶的客户，例如：

SELECT * FROM customers
WHERE tea_preference = 'yes please'
AND date BETWEEN '2011-01-01' AND '2011-01-03'

现在，该查询将提供一个匹配，但如果客户不喜欢该范围内某个日期的茶，我不希望该客户有任何匹配...

我尝试过自我加入，但要么他们不是答案，要么我做得不对，因为我似乎无法得到我追求的结果集。帮助？！

Answer 1

也许：

SELECT c1.* FROM customers c1
WHERE c1.tea_preference = 'yes please' 
AND c1.date BETWEEN '2011-01-01' AND '2011-01-03'
AND NOT EXISTS 
    (SELECT c2.* FROM customers c2 
    WHERE c2.tea_preference = 'no thanks'
    AND c2.date BETWEEN '2011-01-01' AND '2011-01-03' 
    AND c1.custom_id = c2.customer_id)

排除其中一行与标准不匹配 - 自联接？

1 个答案: