MySQL:两个结果集的差异

问题描述

如何获得两个结果集的集合差异?

How can I get the set difference of two result sets?

说我有一个结果集(每个结果中只有一列):

Say I have a result set (just one column in each):

result1:
'a'
'b'
'c'

result2:
'b'
'c'

我想通过result2减去result1中的内容:result1-result2等于:

I want to minus what is in result1 by result2: result1 - result2 such that it equals:

 difference of result1 - result2:
 'a'

推荐答案

要执行result1-result2，可以将result1与result2连接起来，并且只输出result1中存在的项目.例如:

To perform result1 - result2, you can join result1 with result2, and only output items that exist in result1. For example:

SELECT DISTINCT result1.column
FROM result1 LEFT JOIN result2 ON result1.column = result2.column
WHERE result2.column IS NULL

请注意，这不是集合 difference ，并且不会在result2中输出在result1中不存在的项目.设置为减法.

Note that is not a set difference, and won't output items in result2 that don't exist in result1. It's set subtraction.

另请参见:相关博客文章的网络存档版本.

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