SQL Server：查询分层和引用的数据

Question

我正在开发一个具有层次结构的资产数据库。 此外，还有一个“ReferenceAsset”表，可以有效地指回资产。 参考资产基本上用作覆盖，但它被选择为好像它是一个独特的新资产。 设置的覆盖之一是parent_id。

与选择层次结构相关的列：
资产：id（主要），parent_id
资产参考：id（主要），asset_id（foreignkey-> Asset），parent_id（始终是资产）
---编辑5/27 ----

样本Relevent表数据（连接后）：

  id | asset_id | name | parent_id | milestone | type 3 3 suit null march shape 4 4 suit_banker 3 april texture 5 5 tie null march shape 6 6 tie_red 5 march texture 7 7 tie_diamond 5 june texture -5 6 tie_red 4 march texture 

id <0（如最后一行）表示引用的资产。 引用的资产有几个被覆盖的列（在这种情况下，只有parent_id很重要）。

期望是如果我从四月选择所有资产，我应该进行二次选择以获得匹配查询的整个树分支：

所以最初查询匹配会导致：

  4 4 suit_banker 3 april texture 

然后在CTE之后，我们得到完整的层次结构，我们的结果应该是这个（到目前为止这是有效的）

  3 3 suit null march shape 4 4 suit_banker 3 april texture -5 6 tie_red 4 march texture 

并且你看，id：-5的父亲在那里，但缺少的是所需的，是引用的资产，以及引用的资产的父级：

  5 5 tie null march shape 6 6 tie_red 5 march texture 

目前我的解决方案适用于此，但它仅限于一个参考深度（我觉得实现非常难看）。

---编辑----这是我的主要选择功能。 这应该更好地展示真正的复杂性所在：AssetReference。

Select A.id  as id, A.id as asset_id, A.name,A.parent_id as parent_id, A.subPath, T.name as typeName, A2.name as parent_name,  B.name as batchName, 
L.name as locationName,AO.owner_name as ownerName, T.id as typeID,
M.name as milestoneName, A.deleted as bDeleted, 0 as reference, W.phase_name, W.status_name
FROM Asset as A Inner Join Type as T on A.type_id = T.id
Inner Join Batch as B on A.batch_id = B.id
Left Join Location L on A.location_id = L.id
Left Join Asset A2 on A.parent_id = A2.id   
Left Join AssetOwner AO on A.owner_id = AO.owner_id
Left Join Milestone M on A.milestone_id = M.milestone_id
Left Join Workflow as W on W.asset_id = A.id
where A.deleted <= @showDeleted

UNION 

Select -1*AR.id as id, AR.asset_id as asset_id, A.name, AR.parent_id as parent_id, A.subPath, T.name as typeName, A2.name as parent_name,  B.name as batchName, 
L.name as locationName,AO.owner_name as ownerName, T.id as typeID,
M.name as milestoneName, A.deleted as bDeleted, 1 as reference, NULL as phase_name, NULL as status_name
FROM Asset as A Inner Join Type as T on A.type_id = T.id
Inner Join Batch as B on A.batch_id = B.id
Left Join Location L on A.location_id = L.id
Left Join Asset A2 on AR.parent_id = A2.id  
Left Join AssetOwner AO on A.owner_id = AO.owner_id
Left Join Milestone M on A.milestone_id = M.milestone_id
Inner Join AssetReference AR on AR.asset_id = A.id
where A.deleted <= @showDeleted

我有一个存储过程，它接受临时表（#temp）并查找层次结构的所有元素。 我采用的策略是：

1. 选择整个系统heirarchy到临时表（#treeIDs）中，由每个整个树分支的逗号分隔列表表示
2. 获取匹配查询的资产的整个层次（来自#temp）
3. 获取来自heirarchy的Assets指向的所有参考资产
4. 解析所有参考资产的层次结构

这是有效的，因为参考资产总是分支上的最后一项，但如果它们不是，我想我会遇到麻烦。 我觉得我需要一些更好的递归形式。

这是我当前的代码，它正在运行，但我并不为此感到自豪，而且我知道它不健壮（因为它仅在引用位于底部时才有效）：

步骤1.构建整个层次结构

;WITH Recursive_CTE AS (
 SELECT Cast(id as varchar(100)) as Hierarchy, parent_id, id
 FROM #assetIDs
Where parent_id is Null

UNION ALL

 SELECT
 CAST(parent.Hierarchy + ',' + CAST(t.id as varchar(100)) as varchar(100)) as Hierarchy, t.parent_id, t.id
 FROM Recursive_CTE parent
 INNER JOIN #assetIDs t ON t.parent_id = parent.id
)



Select Distinct h.id, Hierarchy as idList into #treeIDs
FROM ( Select Hierarchy, id FROM Recursive_CTE ) parent 
CROSS APPLY dbo.SplitIDs(Hierarchy) as h

步骤2.选择与查询匹配的所有资产的分支

Select DISTINCT L.id into #RelativeIDs FROM #treeIDs
CROSS APPLY dbo.SplitIDs(idList) as L
WHERE #treeIDs.id in (Select id FROM #temp)

步骤3.获取分支中的所有参考资产（参考资产具有负id值，因此id <0部分）

Select asset_id  INTO #REFLinks FROM #AllAssets WHERE id in 
(Select #AllAssets.asset_id FROM #AllAssets Inner Join #RelativeIDs
 on #AllAssets.id = #RelativeIDs.id  Where #RelativeIDs.id < 0)

步骤4.获取步骤3中找到的任何分支

Select DISTINCT L.id into #extraRelativeIDs FROM #treeIDs
CROSS APPLY dbo.SplitIDs(idList) as L
WHERE 
exists (Select #REFLinks.asset_id FROM #REFLinks WHERE #REFLinks.asset_id = #treeIDs.id) 
and Not Exists (select id FROM #RelativeIDs Where id = #treeIDs.id)

我试图只显示相关代码。 我非常感谢能帮助我找到更好解决方案的人！

Answer 1

--getting all of the children of a root node  ( could be > 1 ) and it would require revising the query a bit

DECLARE @AssetID int = (select AssetId from Asset where AssetID is null);


--algorithm is relational recursion
--gets the top level in hierarchy we want. The hierarchy column
--will show the row's place in the hierarchy from this query only
--not in the overall reality of the row's place in the table

WITH Hierarchy(Asset_ID, AssetID, Levelcode, Asset_hierarchy)
AS
(
 SELECT AssetID, Asset_ID,
        1 as levelcode, CAST(Assetid as varchar(max)) as Asset_hierarchy
 FROM   Asset 
 WHERE AssetID=@AssetID

 UNION ALL

 --joins back to the CTE to recursively retrieve the rows 
 --note that treelevel is incremented on each iteration

 SELECT  A.Parent_ID, B.AssetID,
        Levelcode + 1 as LevelCode,
        A.assetID + '\' + cast(A.Asset_id as varchar(20)) as Asset_Hierarchy
 FROM   Asset AS a
          INNER JOIN dbo.Batch AS Hierarchy
            --use to get children, since the parentId of the child will be set the value
            --of the current row
            on a.assetId= b.assetID 
--use to get parents, since the parent of the Asset_Hierarchy row will be the asset, 
            --not the parent.
            on Asset.AssetId= Asset_Hierarchy.parentID


SELECT  a.Assetid,a.name,
        Asset_Hierarchy.LevelCode, Asset_Hierarchy.hierarchy
FROM     Asset AS a
         INNER JOIN Asset_Hierarchy
              ON A.AssetID= Asset_Hierarchy.AssetID
ORDER BY    Hierarchy ;
--return results from the CTE, joining to the Asset data to get the asset name
---that is the structure you will want. I would need a little more clarification of your table structure  

Answer 2

了解您的基础表结构会有所帮助。 根据您的环境，有两种方法可以工作：SQL了解XML，因此您可以将SQL作为xml结构，或者只使用一个表，每个行项都具有唯一的主键ID和parentid。 id是parentid的fk。 节点的数据只是标准列。 您可以使用cte或为计算列提供动力的函数来确定每个节点的嵌套程度。 限制是节点只能有一个父节点。