SQL Server:查询分层和引用的数据

时间:2013-04-28 07:45:26

标签: sql sql-server hierarchical

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->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< 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:

  1. 将整个系统层次结构选择到临时表(#treeIDs)中,该表由每个整个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(来自#temp)
  3. 获取来自heirarchy的资产所指向的所有参考资产
  4. 解析所有参考资产的层次结构
  5. 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!

2 个答案:

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。