SQL Server：查询分层和引用数据

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，它有效地指向资产。参考资产基本上起到了覆盖的作用，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（外键->资产），parent_id（始终是资产）
---编辑于 5/27---

相关表数据示例（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

期望如果我选择了 4 月的所有资产，我应该进行二次选择以获取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，我们得到了完整的层次结构，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

你看，id:-5 的父级是存在的，但是缺少的是被引用的资产，以及被引用的资产的父级：

    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，但仅限于单一深度的引用（而且我觉得实现非常难看）。

---已编辑---- 这是我的主要选择功能。这应该更好地展示真正复杂的地方：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，它采用临时表 (#temp) 并查找层次结构的所有元素。我采用的策略是这样的：

1. 将整个系统层次结构选择到一个临时表 (#treeIDs) 中，该表由每个完整树分支的逗号分隔列表表示
2. 获取资产匹配查询的整个层次结构（来自#temp）
3. 从层次结构中获取 Assets 指向的所有参考资产
4. 解析所有参考资产的层次结构

这暂时有效，因为参考资产始终是分支上的最后一项，但如果不是，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，而且我知道它并不可靠（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 （参考资产的 id 值为负，因此 id

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)

第 4 步。获取第 3 步中找到的任何内容的分支

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。我非常感谢任何可以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

这将有助于了解您的基础表结构。有两种方法可以根据您的环境工作：SQL 理解 XML，因此您可以将 SQL 作为 xml 结构，或者简单地使用单个表，其中每个行项目具有唯一的主键 ID 和父 ID。 id 是 parentid 的 fk。节点的数据只是标准列。您可以使用 cte 或为计算列提供支持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。