ONY*_*NYX 6 sql sql-server sql-server-2008
是否可以将参数传递给选择节点的CTE,然后选择其父节点到parentId空的根节点?
在我的下面的代码中,如果我传入一个选择Rain Coats的参数,然后将树递归到男士服装,其中它parentId为null并选择该分支中的所有节点,包括子节点.请有人帮我这个.我的例子只是递归并显示深度
SQL示例:
DECLARE @Department TABLE
(
Id INT NOT NULL,
Name varchar(50) NOT NULL,
ParentId int NULL
)
INSERT INTO @Department SELECT 1, 'Toys', null
INSERT INTO @Department SELECT 2, 'Computers', null
INSERT INTO @Department SELECT 3, 'Consoles', 2
INSERT INTO @Department SELECT 4, 'PlayStation 3', 3
INSERT INTO @Department SELECT 5, 'Xbox 360', 2
INSERT INTO @Department SELECT 6, 'Games', 1
INSERT INTO @Department SELECT 7, 'Puzzles', 6
INSERT INTO @Department SELECT 8, 'Mens Wear', null
INSERT INTO @Department SELECT 9, 'Mens Clothing', 8
INSERT INTO @Department SELECT 10, 'Jackets', 9
INSERT INTO @Department SELECT 11, 'Rain Coats', 10
;WITH c
AS
(
SELECT Id, Name,1 AS Depth
FROM @Department
WHERE ParentId is null
UNION ALL
SELECT t.Id, t.Name, c.Depth + 1 AS 'Level'
FROM @Department T
JOIN c ON t.ParentId = c.Id
)
SELECT * FROM c WHERE c.Id = 3
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mar*_*c_s 14
您当前的CTE只显示树中的所有项目,包括它们Depth和所有其他属性.因此,它工作正常.
要做你正在寻找的东西,你必须几乎"反转"CTE - 首先抓住你感兴趣的项目,作为你的CTE的"锚",然后"递归"到根:
DECLARE @StartID INT = 11
;WITH c
AS
(
SELECT Id, ParentId, Name, 1 AS Depth
FROM @Department
WHERE Id = @startID
UNION ALL
SELECT t.Id, t.ParentId, t.Name, c.Depth + 1 AS 'Level'
FROM @Department T
INNER JOIN c ON t.Id = c.ParentId
)
SELECT *
FROM c
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这将完成您正在寻找和输出的内容:
Id ParentId Name Depth
11 10 Rain Coats 1
10 9 Jackets 2
9 8 Mens Clothing 3
8 NULL Mens Wear 4
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更新
对于深度的反向顺序,您可以使用:
;WITH c
AS
(
SELECT Id, ParentId, Name, 1 AS Depth
FROM @Department
WHERE Id = @startID
UNION ALL
SELECT t.Id, t.ParentId, t.Name, c.Depth + 1 AS 'Level'
FROM @Department T
INNER JOIN c ON t.Id = c.ParentId
)
SELECT Id,
ParentID,
Name,
MAX(Depth) OVER() - Depth + 1 AS InverseDepth
FROM c
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输出来自:
Id ParentId Name InverseDepth
11 10 Rain Coats 4
10 9 Jackets 3
9 8 Mens Clothing 2
8 NULL Mens Wear 1
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