[英]Simulate ring (algebra) in SQL
我正在创建小型配置文件查看器,我想在SQL中模拟铃声。 这个怎么运作? 简单来说:我的桌子上有5位用户。 当我从中间到实验(如用户号2、3、4)中的一个时,下一个和上一个用户为CURRENT_USER_ID + -1。但是,当我选择用户号1或5时,它并不像以前那样简单。 因此,如果我有一些像代数环这样的工具,那很容易。 我可以模拟吗?
哦,我在用MySQL。
像这样: SQLFIDDLE
set @curr_user = 1;
set @maxid = (select max(u1.id) maxid from users u1);
set @minid = (select min(u2.id) minid from users u2);
set @next_user = (if(@curr_user = @maxid
,@minid
,@curr_user +1));
set @prev_user = (if(@curr_user = @minid
,@maxid
,@curr_user - 1));
SELECT t1.name as prevuser, t2.name as curruser, t3.name as nextuser
FROM users t1
join users t2
on 1=1
JOIN users t3
on 1=1
WHERE t1.id = @prev_user
AND t2.id = @curr_user
and t3.id = @next_user
编辑:我做了一点调整,所以即使缺少ID,它也可以工作,假设您有用户1,3,4,5,7,如果当前为5, 则将获得用户4,5,7 SQLFIDDLE
set @curr_user = 3;
set @maxid = (select max(u1.id) maxid from users u1);
set @minid = (select min(u2.id) minid from users u2);
set @nextid = (select min(id) from users where id > @curr_user );
set @previd = (select max(id) from users where id < @curr_user );
set @next_user = (if(@curr_user = @maxid
,@minid
,@nextid));
set @prev_user = (if(@curr_user = @minid
,@maxid
,@previd));
SELECT t1.name as prevuser, t2.name as curruser, t3.name as nextuser
FROM users t1
join users t2
on 1=1
JOIN users t3
on 1=1
WHERE t1.id = @prev_user
AND t2.id = @curr_user
and t3.id = @next_user
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