从一个维度中选择所有值,该维度在所有其他维度中都有事实

作为第一次迭代,您可以尝试以下操作:

假设

• 客户(所有客户的列表)
• 邮政编码(所有邮政编码的列表)
• 天(所有天的列表)

我相信你必须把它分解

1) 让客户和邮政编码每天都有事实

SELECT zip_id, customer_id, COUNT(DISTINCT days)
FROM facts
GROUP BY zip_id, customer_id
HAVING COUNT(DISTINCT days) = (SELECT COUNT(*) FROM days) 

2) 从这个结果集中检查所有客户的拉链

SELECT zip_id, COUNT(DISTINCT customer_id)
FROM ( 
    SELECT zip_id, customer_id, COUNT(DISTINCT days)
    FROM facts
    GROUP BY zip_id, customer_id
    HAVING COUNT(DISTINCT days) = (SELECT COUNT(*) FROM days) 
    ) S
GROUP BY zip_id
HAVING COUNT(DISTINCT customer_id) = (SELECT COUNT(*) FROM customers)

在你得到邮政编码后,它应该很容易加入它回到美国。