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我们通常被要求在报告中包含某些层次结构的级别，我正在寻找一种使用 hierarchyid 提高查询性能的方法。我做了一些测试，并且我有一些有效的查询，但我认为性能可能会更好。下面是一个为给定条目提取级别 2 到 4 的示例。我想出了一种方法来使用 GetAncestor() 函数结合hierarchyid的级别。第一个查询看起来很快，但它被硬编码为只返回某个级别的行，以避免用负值破坏 GetAncestor 查询。第二个示例解决了该问题，但速度要慢得多。理想情况下，第二个选项是我想要使用的，但它不够快。

--drop table #hier

CREATE TABLE #hier
    (
    rec_ID int NOT NULL,
    rec_NAME varchar(6),
    nodeID hierarchyid NULL,
    lvl  AS [nodeid].[GetLevel]() PERSISTED 
    )  ON [PRIMARY]
GO
ALTER TABLE #hier ADD CONSTRAINT
    rec_ID PRIMARY KEY CLUSTERED 
    (
    rec_ID
    ) WITH( STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
GO
CREATE NONCLUSTERED INDEX IX_hier_nodeID ON #hier
    (
    nodeID
    ) WITH( STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
GO
ALTER TABLE #hier SET (LOCK_ESCALATION = TABLE)
GO

insert into #hier (rec_ID, rec_NAME, nodeID)
    SELECT 1, 'CEO', cast('/' as hierarchyid)
    union
    SELECT 2, 'VP1', cast('/1/' as hierarchyid)
    union
    SELECT 3, 'VP2', cast('/2/' as hierarchyid)
    union
    SELECT 4, 'VP3', cast('/3/' as hierarchyid)
    union
    SELECT 5, 'Mgr1', cast('/1/1/' as hierarchyid)
    union
    SELECT 6, 'Mgr2', cast('/1/2/' as hierarchyid)
    union
    SELECT 7, 'Super1', cast('/1/2/1/' as hierarchyid)
    union
    SELECT 8, 'Ldr1', cast('/1/2/1/1/' as hierarchyid)
    union
    SELECT 9, 'Work1', cast('/1/2/1/1/1/' as hierarchyid)
    union
    SELECT 10, 'Work2', cast('/1/2/1/1/2/' as hierarchyid)
    union
    SELECT 11, 'Work3', cast('/1/2/1/1/3/' as hierarchyid)
GO

-- this runs fast but is hard coded to a level

declare @recname varchar(6)
set @recname = 'Work3'

select 
    x.rec_name
    ,x.lvl
    ,(select rec_name from  #hier where nodeid = x.nodeid.GetAncestor(x.lvl - 2) ) as l2
    ,(select rec_name from  #hier where nodeid = x.nodeid.GetAncestor(x.lvl - 3) ) as l3
    ,(select rec_name from  #hier where nodeid = x.nodeid.GetAncestor(x.lvl - 4) ) as l4
from #hier x 
where x.rec_name = @recname
    and x.lvl >= 4


-- this works for all levels but runs too slow

set @recname = 'Mgr2'
select 
    x.rec_name
    ,x.lvl
    ,case
        when x.lvl >=2 then (select rec_name from  #hier where nodeid = x.nodeid.GetAncestor(x.lvl - 2) )
        else '*N/A' end as l2
    ,case
        when x.lvl >=3 then (select rec_name from  #hier where nodeid = x.nodeid.GetAncestor(x.lvl - 3) )
        else '*N/A' end as l2
    ,case
        when x.lvl >=4 then (select rec_name from  #hier where nodeid = x.nodeid.GetAncestor(x.lvl - 4) )
        else '*N/A' end as l2
from #hier x 
where x.rec_name = @recname
4

1 回答 1

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在我看来，您应该创建以下索引：

1)

CREATE INDEX IX_hier_rec_name_#_lvl_nodeid 
ON #hier (rec_name) 
INCLUDE (lvl, nodeid)

满足查询#2

select 
    x.rec_name
    ,x.lvl
    ,case
        when x.lvl >=2 then (... = x.nodeid.GetAncestor(x.lvl - 2) )
        else '*N/A' end as l2
    ,...
from #hier x 
where x.rec_name = @recname

注意：对于查询 #1，您可以使用

CREATE INDEX IX_hier_rec_name_lvl_#_nodeid 
ON #hier (rec_name, lvl) 
INCLUDE (nodeid)

和

2）你应该改变这个IX_hier_nodeID索引

CREATE INDEX IX_hier_nodeID_#_rec_name 
ON #hier (nodeid) 
INCLUDE (rec_name)

或（更好）与

CREATE UNIQUE INDEX IUN_hier_nodeID_#_rec_name 
ON #hier (nodeid) 
INCLUDE (rec_name)

如果nodeid值不允许重复。

于 2014-10-07T19:59:59.970 回答