关于各个数据库递归（start with connect by

查找根结点（自下而上）

Oracle：

select a.id from hrmsubcompany a start with a.id=2 connect by prior a.id = a.supsubcomid；

SQL Server／DB2:

WITH allsub(id,subcompanyname,supsubcomid) as (
SELECT id,subcompanyname ,supsubcomid FROM HrmSubCompany where id=2 
UNION ALL SELECT a.id,a.subcompanyname,a.supsubcomid FROM HrmSubCompany a,allsub b where a.supsubcomid = b.id
) 
select * from allsub

Mysql:

-- 起始值为2，获得所有父节点（包括自己）
select t.id from (
    select @id idlist,
    (select @id:=group_concat(supsubcomid separator ',') from hrmsubcompany where find_in_set(id,@id)) sub
    from hrmsubcompany,(select @id:=2) vars
    where @id is not null) tl,hrmsubcompany t
where find_in_set(t.id,tl.idlist);

-- 起始值为2，获得所有父节点（包括自己）
select tl.lv,t.* from (
    select @id idlist, @lv:=@lv+1 lv,
    (select @id:=group_concat(pid separator ',') from treenodes where find_in_set(id,@id)) sub
    from treenodes,(select @id:=2,@lv:=0) vars
    where @id is not null) tl,treenodes t
where find_in_set(t.id,tl.idlist);

查找叶子结点，自上而下

Oracle：

select a.id from hrmsubcompany a start with a.id=2 connect by prior a.supsubcomid = a.id；

SQL Server／DB2:

WITH allsub(id,subcompanyname,supsubcomid) as (
SELECT id,subcompanyname ,supsubcomid FROM HrmSubCompany where id=2 
UNION ALL SELECT a.id,a.subcompanyname,a.supsubcomid FROM HrmSubCompany a,allsub b where a.id = b.supsubcomid
) 
select * from allsub

Mysql:

-- 起始值为2，获得所有父节点（包括自己）
select t.id from (
    select @id idlist,
    (select @id:=group_concat(id separator ',') from hrmsubcompany where find_in_set(pid,@id)) sub
    from hrmsubcompany,(select @id:=2) vars
    where @id is not null) tl,hrmsubcompany t
where find_in_set(t.id,tl.idlist);

-- 起始值为2，获得所有子节点（包括自己）
select tl.lv,t.* from (
    select @id idlist, @lv:=@lv+1 lv,
    (select @id:=group_concat(id separator ',') from treenodes where find_in_set(pid,@id)) sub
    from treenodes,(select @id:=2,@lv:=0) vars
    where @id is not null) tl,treenodes t
where find_in_set(t.id,tl.idlist);