Question

我有一个包含2列usersID及其siblingID

的表格

查找给定用户的所有兄弟姐妹的最佳方法是什么？

问题很复杂。这是一个例子。

用户1有5个兄弟姐妹（2,3,4,5,6）

表格如下所示

userID|siblingID
1     | 2
1     | 3
6     | 5
5     | 3
3     | 1
4     | 6

Answer 1

ANSI SQL：

with recursive tree (userid, siblingid) as
(
   select userid, 
          siblingid
   from users
   where userId = 1
   union all 
   select c.userid,
          c.siblingid
   from users c
     join tree p on p.userid c.siblingId
)
select *
from tree;

对于Oracle 11.2和SQL Server - 显然没有仔细查看ANSI规范 - 您需要删除recursive关键字（根据标准是强制性的）

Answer 2

即使使用多个SQL语句，答案也比我想象的要难得多。

您问题的简单答案是：创建一个包含所有兄弟关系的表格。然后您可以将其查询为：

select siblingid
from @allsiblings sa
where sa.userid = 3

一个注意事项。我正在使用SQL Server的语法，因为它恰好是最方便的数据库。我只在MySQL中使用功能，所以它应该很容易翻译。

如何创建表@AllSiblings？好吧，继续添加不存在的兄弟对，直到没有更多要添加。我们通过自我加入获得对。

以下是代码（受先前警告的影响）：

declare @allsiblings table (userid integer, siblingid integer);

declare @siblings table (userId int, siblingID int);

-- Initialize the @siblings table    
insert into @siblings(userId, siblingID)
    select 1 as userID, 2 as siblingID union all
    select 1 as userID, 3 as siblingID union all
    select 6 as userID, 5 as siblingID union all
    select 5 as userID, 3 as siblingID union all
    select 3 as userID, 1 as siblingID union all
    select 4 as userID, 6 as siblingID;

-- Initialize all siblings.  Note that both pairs are going in here    
insert into @allsiblings(userid, siblingid)
    select userId, siblingid from @siblings union
    select siblingID, userid from @siblings

-- select * from @allsiblings

while (1=1)
begin
    -- Add in new siblings, that don't exist by doing a self-join to traverse the links
    insert into @allsiblings
        select distinct sa.userid, sa2.siblingid
        from @allsiblings sa join
             @allsiblings sa2
             on sa.siblingid = sa2.userid
        where not exists (select * from @allsiblings sa3 where sa3.userid = sa.userid and sa3.siblingid = sa2.siblingid)

    -- If nothing was added, we are done        
    if (@@ROWCOUNT = 0) break;

    select * from @allsiblings;
end;

Answer 3

您可以使用循环和临时表来模拟递归。首先在临时表中插入起始节点。然后在临时表中有行，获取第一个，从临时表中删除它，并在其中插入所有兄弟姐妹...

Answer 4

http://sqlfiddle.com/#!4/0ef0c/5是一个例子，有人必须获得某些条目的所有亲属。

相应的堆栈溢出问题在这里： Hierarchical Query Needs to Pull Children, Parents and Siblings

棘手的递归MySQL查询

4 个答案: