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-＆gt;资产），parent_id（始终为资产）
---编辑5月27日----
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＆lt; 0（与最后一行一样）表示引用的资产。引用的资产有一些被覆盖的列（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
你看，id：-5的父亲在那里，但缺少的是需要的，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：
这适用于现在，因为参考资产始终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，而且我知道它不健全（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.获取分支机构中的所有参考资产 （参考资产具有负的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。我非常感谢能帮助我找到更好解决方案的人！
答案 0 :(得分：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
答案 1 :(得分：0)
了解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。