【Sql Server】SQL SERVER 递归查询

　　SQL SERVER 2005之前的版本只能用函数方法实现，SQL SERVER 2005之后新增了CTE功能，可以利用CTE实现递归查询；

　　CTE：公用表达式Common Table Expression 是SQL SERVER 2005版本之后引入的一个特性；

#填充测试数据

1、sql

 1 Create table GroupInfo1([Id] int,[GroupName] nvarchar(50),[ParentGroupId] int)
 2
 3 Insert GroupInfo1
 4
 5 select 0,'某某大学',null union all
 6
 7 select 1,'外语学院',0 union all
 8 select 2,'英语专业',1 union all
 9 select 3,'日语专业',1 union all
10 select 4,'英语专业一班',2 union all
11 select 5,'英语专业二班',2 union all
12 select 6,'日语专业一班',3 union all
13 select 7,'日语专业二班',3 union all
14
15 select 8, '法学院',0 union all
16 select 9, '刑法学专业',8 union all
17 select 10,'经济法学专业',8 union all
18 select 11,'刑法学专业一班',9 union all
19 select 12,'刑法学专业二班',9 union all
20 select 13,'经济法学专业一班',10 union all
21 select 14,'经济法学专业二班',10 

2、效果图

 

#递归实现Demo

1、根据指定的节点向上获取所有父节点,向下获取所有子节点

 1 --根据指定的节点向下获取所有子节点
 2 with
 3 CTE
 4 as
 5 (
 6     select * from GroupInfo where Id=1
 7     union all
 8     select G.* from CTE inner join GroupInfo as G
 9     on CTE.Id=G.ParentGroupId
10 )
11 select * from CTE order by Id

 1 --根据指定的节点向上获取所有父节点
 2 with
 3 CTE
 4 as
 5 (
 6     select * from GroupInfo where Id=14
 7     union all
 8     select G.* from CTE inner join GroupInfo as G
 9     on CTE.ParentGroupId=G.Id
10 )
11 select * from CTE order by Id

2、构造递归路径

 1 --构造递归路径
 2 with
 3 CTE
 4 as
 5 (
 6     select Id,GroupName,ParentGroupId,GroupPath=CAST( GroupName as nvarchar(max)) from GroupInfo where Id=1
 7     union all
 8     select G.*,CAST(CTE.GroupPath+'//'+G.GroupName as nvarchar(max)) as GroupPath from CTE
 9     inner join GroupInfo as G
10     on CTE.Id=G.ParentGroupId
11 )
12 select * from CTE

 3、分组递归，将同一条分支上节点放到一起

 1 --通过id字段的字符串的拼接，形成sort字段，再通过sort排序，来实现同一分支上的节点放到一起
 2 WITH
 3 CTE
 4 AS
 5 (
 6     SELECT * ,CAST(RIGHT('000' + CAST([Id] AS VARCHAR), 3) AS VARCHAR(MAX)) AS sort FROM GroupInfo
 7     WHERE ParentGroupId = 0
 8     UNION ALL
 9     SELECT   GroupInfo.* ,CAST(sort + RIGHT('000' + CAST(GroupInfo.[Id] AS VARCHAR),3) AS VARCHAR(MAX)) AS sort
10     FROM CTE
11     INNER JOIN GroupInfo ON CTE.Id = GroupInfo.ParentGroupId
12 )
13 SELECT * FROM CTE ORDER BY sort 

 

4、递归层级查询（查询出节点所属的层级）

1 --查询节点层级
2 WITH CTE AS (
3     SELECT *,1 AS [Level] FROM GroupInfo WHERE ParentGroupId=0
4     UNION ALL
5     SELECT G.*,CTE.Level+1 FROM GroupInfo as G
6     JOIN CTE ON CTE.Id =G.ParentGroupId
7 )
8 SELECT * FROM CTE