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MySQL比较运算符IN问题求解
2022-07-02 09:34:00 【CSDN问答】
关于比较运算符IN
我看书上、教程视频上、网上，都是这么介绍的：
判断给定值是否是IN列表中的一个值。如果是，返回1，否则返回0。如果给定值为NULL，或者IN列表存在NULL，则结果为NULL。
可是 SELECT 1 IN (1, NULL) 的结果是1，而不是NULL。
这是为什么？不是说 或者IN列表存在NULL，则结果为NULL嘛？？
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