查询获取树下的所有节点

--构造测试数据: 只作演示用
CREATE TABLE [dbo].[Tim_LinqTable](
[Id] int PRIMARY KEY IDENTITY(1,1) NOT NULL,
[Name] [varchar](50) NOT NULL,
[Parent] int NOT NULL,
)
GO

INSERT INTO [Tim_LinqTable]
SELECT 'A',0 UNION ALL
SELECT 'A1',1 UNION ALL
SELECT 'A2',1 UNION ALL
SELECT 'B1',2 UNION ALL
SELECT 'B2',3 UNION ALL
SELECT 'C1',4 UNION ALL
SELECT 'C2',4 UNION ALL
SELECT 'D1',5 UNION ALL
SELECT 'D2',5 UNION ALL
SELECT 'D3',5
GO

WITH temp
AS
(
SELECT * FROM [Tim_LinqTable] WHERE Parent = 3
UNION ALL
SELECT m.* FROM [Tim_LinqTable] AS m
INNER JOIN temp AS child ON m.Parent = child.Id
)
SELECT * FROM temp
GO

--查询 Parent=3 的所有子数据结果如下:
Id Name Parent
----------- -------------------------------------------------- -----------
5 B2 3
8 D1 5
9 D2 5
10 D3 5

(4 row(s) affected)

原文地址:https://www.cnblogs.com/yfann/p/2578116.html