如何查找列表/集合是否正好在另一个列表中 [英] How to find if a list/set is exactly within another list

问题描述

我的问题基本上是这个问题的扩展 - 如何查找列表/集合是否包含在另一个列表中

My question is basically an extension of this question - How to find if a list/set is contained within another list

我有相同的数据，但有更多完全相同的产品 -

I have the same data, but with more exact same products -

order_id | product_id
----------------------
1        | 222
1        | 555
2        | 333
2        | 222
2        | 555

我想查找订单完全指定了产品 ID 的订单.

I want to find the order(s) where the order has EXACTLY given product ids.

例如:如果我搜索 222 &555，我应该得到 order_id 1 作为输出，因为它只有那 2 个确切的产品 ID.并且不应将订单 ID 2 作为输出，因为它有一个额外的 333 产品 ID.

For example: If I search for 222 & 555, I should get order_id 1 as output since it only have those 2 exact product ids. And should not get order id 2 as output since it has an additional 333 product id.

推荐答案

您仍然可以使用 have，但要对每个产品进行测试——以及是否没有任何其他产品:

You can still use having, but test for each product -- and for the absence of any other product:

SELECT o.order_id
FROM orders o
GROUP BY o.order_id
HAVING SUM( product_id = 222 ) > 0 AND
       SUM( product_id = 555 ) > 0 AND
       SUM( product_id NOT IN (222, 555) ) = 0 ;

请注意，这里使用 MySQL 简写，其中布尔表达式被视为数字，1 表示真，0 表示假.标准语法是:

Note that this uses the MySQL shorthand, where boolean expressions are treated as numbers with 1 for true and 0 for false. The standard syntax is:

HAVING SUM( CASE WHEN product_id = 222 THEN 1 ELSE 0 END) > 0 AND
       SUM( CASE WHEN product_id = 555 THEN 1 ELSE 0 END ) > 0 AND
       SUM( CASE WHEN product_id NOT IN (222, 555) THEN 1 ELSE 0 END ) = 0 ;

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