SQL SERVER CTE递归查询
SQL SERVER CTE递归查询
正文
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(Common Table Expression)功能,可以利用CTE实现递归查询;
一、建表
1、sql
1 Create table test_GroupInfo([Id] int,[GroupName] nvarchar(50),[ParentGroupId] int) 2 3 Insert test_GroupInfo 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
二、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 test_GroupInfo where Id=1 7 union all 8 select G.* from CTE inner join test_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 test_GroupInfo where Id=14 7 union all 8 select G.* from CTE inner join test_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 test_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 test_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 test_GroupInfo 7 WHERE ParentGroupId = 0 8 UNION ALL 9 SELECT test_GroupInfo.* ,CAST(sort + RIGHT('000' + CAST(test_GroupInfo.[Id] AS VARCHAR),3) AS VARCHAR(MAX)) AS sort 10 FROM CTE 11 INNER JOIN test_GroupInfo ON CTE.Id = test_GroupInfo.ParentGroupId 12 ) 13 SELECT * FROM CTE ORDER BY sort 14 15 16 select id,CAST(RIGHT('000' + CAST([Id] AS VARCHAR), 3) AS VARCHAR(MAX)) from test_GroupInfo
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 test_GroupInfo WHERE ParentGroupId=0 4 UNION ALL 5 SELECT G.*,CTE.Level+1 FROM test_GroupInfo as G 6 JOIN CTE ON CTE.Id =G.ParentGroupId 7 ) 8 SELECT * FROM CTE