最近,在论坛中,遇到了不少比较难的sql问题,虽然自己都能解决,但发现过几天后,就记不起来了,也忘记解决的方法了。
所以,觉得有必要记录下来,这样以后再次碰到这类问题,也能从中获取解答的思路。
sql2008 树形结构分组
http://bbs.csdn.net/topics/390634930
ID DeprtID DeprtName
1 0 1
2 1 2
3 1 3
4 2 4
5 3 5
6 4 6
7 5 7
分组后效果
ID DeprtID DeprtName
1 0 1
2 1 2
4 2 4
6 4 6
3 1 3
5 3 5
7 5 7
我的解法:
--drop table tb create table tb(ID int, DeprtID int, DeprtName varchar(10)) insert into tb select 1, 0, '1' union all select 2 , 1 , '2' union all select 3 , 1 , '3' union all select 4 , 2 , '4' union all select 5 , 3 , '5' union all select 6 , 4 , '6' union all select 7 , 5, '7' go ;with t as ( select id,DeprtID,DeprtName,1 as level, cast(right('000'+cast(id as varchar),3) as varchar(max)) as sort from tb where DeprtID =0 union all select tb.id,tb.DeprtID,tb.DeprtName,level + 1 , cast(sort+right('000'+cast(tb.id as varchar),3) as varchar(max)) from t inner join tb on t.id = tb.DeprtID ) select id,deprtid,deprtname from t order by sort /* id deprtid deprtname 1 0 1 2 1 2 4 2 4 6 4 6 3 1 3 5 3 5 7 5 7 */
这里还有个例子,就是递归查询后,按照树形来排序:
drop table tb create table tb ( id int, pid int, name varchar(20) ) insert into tb select 1,null,'x' union all select 2,1,'a' union all select 3,1,'b' union all select 4,2,'aa' union all select 5,3,'bb' go ;with t as ( select id,pid,name,1 as level, cast(right('000'+cast(id as varchar),3) as varchar(max)) as sort from tb where pid is null union all select tb.id,tb.pid,tb.name,level + 1 , cast(sort+right('000'+cast(tb.id as varchar),3) as varchar(max)) from t inner join tb on t.id = tb.pid ) select * from t order by sort /* id pid name level sort 1 NULL x 1 001 2 1 a 2 001002 4 2 aa 3 001002004 3 1 b 2 001003 5 3 bb 3 001003005 */