是否可以执行按位组功能？

本文介绍了是否可以执行按位组功能？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我在包含按位标志的表中有一个字段。比方说，为了举例，有三个标志： 4 =>阅读，2 =>写，1 =>执行，表看起来像这样 * ： user_id |文件|权限 ----------- + -------- + --------------- 1 | a.txt | 6（ 1 | b.txt | 4（ 2 | a.txt | 4 2 | c.exe | 1（ 我有兴趣找到所有拥有在任何记录上设置特定标志（例如：写入）。为了在一个查询中做到这一点，我认为如果你将所有用户的权限组合在一起，你会得到一个单一的值，这是他们权限的总和： user_id | all_perms ----------- + ------------- 1 | 6（ 2 | 5（ * 我的实际表格不是用来处理文件或文件权限，而是一个例子 有一种方法可以在一个声明中执行此操作吗？我看到它的方式，它与使用GROUP BY的普通聚合函数非常相似： SELECT user_id，SUM（permissions）as all_perms FROM权限 GROUP BY user_id ...但显然，有些神奇的按位或功能，而不是SUM。任何人都知道这样的事情？ （对于奖励点，它在oracle中是否有效？） 解决方案 MySQL： SELECT user_id，BIT_OR（permissions）as all_perms FROM权限 GROUP BY user_id I have a field in a table which contains bitwise flags. Let's say for the sake of example there are three flags: 4 => read, 2 => write, 1 => execute and the table looks like this*: user_id | file | permissions-----------+--------+--------------- 1 | a.txt | 6 ( <-- 6 = 4 + 2 = read + write) 1 | b.txt | 4 ( <-- 4 = 4 = read) 2 | a.txt | 4 2 | c.exe | 1 ( <-- 1 = execute)I'm interested to find all users who have a particular flag set (eg: write) on ANY record. To do this in one query, I figured that if you OR'd all the user's permissions together you'd get a single value which is the "sum total" of their permissions: user_id | all_perms-----------+------------- 1 | 6 (<-- 6 | 4 = 6) 2 | 5 (<-- 4 | 1 = 5)*My actual table isn't to do with files or file permissions, 'tis but an exampleIs there a way I could perform this in one statement? The way I see it, it's very similar to a normal aggregate function with GROUP BY:SELECT user_id, SUM(permissions) as all_permsFROM permissionsGROUP BY user_id...but obviously, some magical "bitwise-or" function instead of SUM. Anyone know of anything like that?(And for bonus points, does it work in oracle?) 解决方案 MySQL:SELECT user_id, BIT_OR(permissions) as all_permsFROM permissionsGROUP BY user_id 这篇关于是否可以执行按位组功能？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！