SQL 树向下遍历到根并备份一个

Question

我有下表(ObjectStates)并且想要获得根以及根的第一个 child :

ID    Title    ParentID
1     Draft    null
2     Green    null
3     Red      null
4     Foo      1
5     Bar      4
6     Some1    1
7     Some2    6
8     XYZ      2
9     Some3    7

我想要以下输出:

GetState(5)
-- returns root: 1, first-child: 4
GetState(6)
-- returns root: 1, first-child: 6
GetState(7)
-- returns root: 1, first-child: 6
GetState(9)
-- returns root: 1, first-child: 6
GetState(8)
-- returns root: 2, first-child: 8

所以无论我查询的层次结构有多深 - 我总是想要根元素以及第一个子元素。如果你考虑这棵树，我总是想要红色和蓝色元素，无论我在树中有多深。



我可以像这样获得“根”状态:

WITH CTEHierarchy
AS (
    SELECT 
    ID
        ,0 AS LEVEL
        ,ID AS root

    FROM ObjectStates
    WHERE  ParentID IS NULL

    UNION ALL

    SELECT 
    ObjectStates.ID
        ,LEVEL + 1 AS LEVEL
        ,[root]

    FROM ObjectStates
    INNER JOIN CTEHierarchy uh ON uh.id = ObjectStates.ParentID
    )    
    SELECT [root]
    FROM CTEHierarchy
    WHERE ID = @ObjectStateID 

这给了我想要的根结果:

GetState(5)
-- returns root: 1
GetState(9)
-- returns root: 1
GetState(2)
-- returns root: 2

我怎样才能从那里向上穿越？那么从根获取树中的下一个 child ？或者反过来 - 获得根以及第一级。递归让我头疼。

Answer 1

我今天正在使用这个查询来遍历主键外键关系以跟踪整个路径，这个问题似乎很相似。所以只需粘贴相同的代码。您可以直接运行此代码并检查这是否是您需要的。

此查询在 CTE 中添加了另外两列，即 Hops 和 Path，其中 Hops 是元素级别，Path 是从头到尾遍历的节点。

WITH cte
AS
(
    SELECT 
        fk.create_date
        , fk.modify_date
        , fkc.constraint_object_id AS ConstraintId
        , OBJECT_NAME(fkc.constraint_object_id) AS ConstraintName
        , OBJECT_NAME(fkc.referenced_object_id) AS PrimaryKeyTableName
        , rc.name AS PrimaryKeyColumnName
        , OBJECT_NAME(fk.parent_object_id) AS ForeignKeyTableName
        , lc.name AS ForeignKeyColumnName
    FROM sys.foreign_key_columns fkc
    INNER JOIN sys.columns rc 
        ON  rc.OBJECT_ID = fkc.referenced_object_id 
        AND fkc.referenced_column_id = rc.column_id
    INNER JOIN sys.foreign_keys fk 
        ON  fk.OBJECT_ID = fkc.constraint_object_id
    INNER JOIN sys.columns lc 
        ON  lc.OBJECT_ID = fk.parent_object_id
        AND fkc.parent_column_id = lc.column_id
)
, cte2(create_date, modify_date, ConstraintName
        , PrimaryKeyTableName, PrimaryKeyColumnName
        , ForeignKeyTableName, ForeignKeyColumnName
        , Hops, path ) AS 
    (
        SELECT
            create_date, modify_date, ConstraintName
            , PrimaryKeyTableName, PrimaryKeyColumnName
            , ForeignKeyTableName, ForeignKeyColumnName 
            , 1 , CAST(QUOTENAME(PrimaryKeyTableName + '.' + PrimaryKeyColumnName) AS VARCHAR(4000))
        FROM cte
    UNION ALL
        SELECT 
            cte.create_date, cte.modify_date, cte.ConstraintName
            , cte.PrimaryKeyTableName, cte.PrimaryKeyColumnName
            , cte.ForeignKeyTableName, cte.ForeignKeyColumnName
            , cte2.Hops +1, CAST(cte2.path + '-> ' +QUOTENAME(cte.PrimaryKeyTableName+ '.' + cte.PrimaryKeyColumnName) AS VARCHAR(4000))
        FROM cte2 INNER JOIN cte ON cte2.ForeignKeyTableName = cte.PrimaryKeyTableName
        AND cte2.ForeignKeyColumnName != cte.PrimaryKeyColumnName
    )
SELECT 
ConstraintName
        , PrimaryKeyTableName, PrimaryKeyColumnName
        , ForeignKeyTableName, ForeignKeyColumnName
        , Hops, path + '-> ' + QUOTENAME(ForeignKeyTableName + '.' + ForeignKeyColumnName) AS Path
FROM cte2

仅当数据库中物理存在外键约束时，此查询才有效。