A typical performance case of SQL Optimization - sql-server

Question

I have an engineering practice of SQL Optimization problem, which I think is a typical case ,and will help a lot of guys.
SQL SERVER 2005,
Firstly, create the main table. This is a person info table.
CREATE TABLE [dbo].[OLAPAgentDim](
[RoleID] [varchar](50) NULL CONSTRAINT [DF_OLAPAgentDim_RoleID] DEFAULT ((1)),
[OLAPKey] [bigint] IDENTITY(1,1) NOT NULL,
[FatherKey] [bigint] NULL,
[FatherKeyValue] [nvarchar](100) NULL,
[System] [varchar](6) NULL,
[Level] [int] NULL,
[IfLeaf] [real] NULL,
[IfDel] [real] NULL CONSTRAINT [DF_OLAPAgentDim_IfDel] DEFAULT ((0)),
[SourceKey] [varchar](50) NULL,
[MainDemoName] [nvarchar](100) NULL,
[FastCode] [varchar](50) NULL,
[TagValue] [varchar](50) NULL,
[Script] [nvarchar](max) NULL,
[Birthday] [datetime] NULL,
[EarlyStartTime] [datetime] NULL,
[StartTime] [datetime] NULL,
[EndTime] [datetime] NULL,
[EditTime] [datetime] NULL,
[BecomesTime] [datetime] NULL,
[ContractTime] [datetime] NULL,
[ContractEndTime] [datetime] NULL,
[XMLIcon] [nvarchar](max) NULL,
[PassKey] [varchar](50) NULL CONSTRAINT [DF_OLAPAgentDim_PassKey] DEFAULT ('N3pkY3RHaeZXA9mGJdfm8A=='),
[Address] [nvarchar](100) NULL,
[HomeTel] [varchar](50) NULL,
[Mobile] [varchar](50) NULL,
[Email] [varchar](100) NULL,
[IDCard] [varchar](50) NULL,
[IDSecu] [varchar](50) NULL,
[IDEndowment] [varchar](50) NULL,
[IDAccumulation] [varchar](50) NULL,
[ContactPerson] [nvarchar](100) NULL,
[ContactPersonTel] [varchar](50) NULL,
[Others1] [varchar](50) NULL,
[SexKey] [varchar](2) NULL CONSTRAINT [DF_OLAPAgentDim_SexKey] DEFAULT ((1)),
[SexKeyValue] [nvarchar](100) NULL,
[MarrageKey] [varchar](2) NULL CONSTRAINT [DF_OLAPAgentDim_MarrageKey] DEFAULT ((1)),
[MarrageKeyValue] [nvarchar](100) NULL,
[Nation] [nvarchar](50) NULL,
[Race] [nvarchar](50) NULL,
[PartyMemberKey] [varchar](2) NULL CONSTRAINT [DF_OLAPAgentDim_PartyMemberKey] DEFAULT ((1)),
[PartyMemberKeyValue] [nvarchar](100) NULL,
[RegionKey] [bigint] NULL CONSTRAINT [DF_OLAPAgentDim_RegionKey] DEFAULT ((1)),
[RegionKeyValue] [nvarchar](100) NULL,
[LeaveResonKey] [bigint] NULL CONSTRAINT [DF_OLAPAgentDim_LeaveResonKey] DEFAULT ((1)),
[LeaveResonKeyValue] [nvarchar](100) NULL,
[RoleStr] [varchar](max) NULL,
[RoleStrValue] [nvarchar](max) NULL,
[LeaderKey] [bigint] NULL CONSTRAINT [DF_OLAPAgentDim_LeaderKey] DEFAULT ((1)),
[LeaderKeyValue] [nvarchar](100) NULL,
[FastCode2] [varchar](50) NULL,
[FastCode3] [varchar](50) NULL,
[FastCode4] [varchar](50) NULL,
[FastCode5] [varchar](50) NULL,
[OtherAddress] [nvarchar](100) NULL,
[ShowOrder] [int] NULL,
[RaceKey] [bigint] NULL DEFAULT ((1)),
[RaceKeyValue] [nvarchar](100) NULL,
[DepartLevelKey] [bigint] NULL DEFAULT ((1)),
[DepartLevelKeyValue] [nvarchar](100) NULL,
[forumname] [nvarchar](100) NULL,
[IfCloseKey] [bigint] NULL DEFAULT ((1)),
[IfCloseKeyValue] [nvarchar](100) NULL,
[InsureStartTime] [datetime] NULL,
[AccumulationStartTime] [datetime] NULL,
[Rate] [varchar](50) NULL,
[DirectLeaderKey] [bigint] NULL CONSTRAINT [DF_OLAPAgentDim_DirectLeaderKey] DEFAULT ((1)),
[DirectLeaderAttriKey] [bigint] NULL CONSTRAINT [DF_OLAPAgentDim_DirectLeaderAttriKey] DEFAULT ((1)),
[DirectLeaderKeyValue] [nvarchar](100) NULL,
[DirectLeaderSourceKey] [varchar](50) NULL,
[DirectLeaderPartName] [nvarchar](100) NULL,
[DirectLeaderPositionName] [nvarchar](100) NULL,
[NOTSync] [int] NULL,
[FatherPath] [nvarchar](max) NULL,
[SaleDiscount] [real] NULL,
CONSTRAINT [PK_OLAPAgent Dim] PRIMARY KEY CLUSTERED
(
[OLAPKey] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
Secondly, insert abount 10,000 record into the table. I think 10,000 record is not a very big number to SQL SERVER. You can see this is a father and children dimention table in fact. The records with ifleaf=0 means the person's department structure node, the records with ifleaf=1 means the person. You can define father-children relationship using FahterKey column. For Example:
OLAPKey IfLeaf FatherKey DepartLevelKey MainDemoName
2 0 0 1 IBM Company
3 0 2 2 Sales Depart
4 0 2 2 Service Depart
5 0 3 3 Sales Team1
6 1 5 NULL John Smith
7 1 4 NULL Mary
......
DepartLevelKey Column means the depart node's level.
So in this table, we can save the whole HR tree info.
Thirdly, we see the problem SQL:
create table #t
(
TableID int IDENTITY(1,1),
OLAPKey bigint,
MainDemoName nvarchar(max)
)
declare #t4 table
(
TableID int IDENTITY(1,1),
MainDemoName nvarchar(max),
OLAPKeystr varchar(100)
)
declare #agentkey bigint
set #agentkey ='2'
--Part A
--DepartLevelKey=2, to get #agentkey node's all level=2 department
;WITH Result AS(
SELECT OLAPKey,DepartLevelKey,maindemoname FROM OLAPAgentDim WHERE OLAPKey =#agentkey
UNION ALL
SELECT a.OLAPKey,a.DepartLevelKey,a.maindemoname FROM OLAPAgentDim AS a,Result AS b WHERE a.FatherKey = b.OLAPKey
)
insert #t select OLAPKey,maindemoname from Result where DepartLevelKey=4
--Part B
;with One as
(
select *,convert(varchar(50),OLAPKey) as Re from #t
)
insert #t4 select maindemoname,stuff((select ','+Re from One where One.maindemoname=#t.maindemoname for xml path('')),1,1,'') as Two
from #t
group by maindemoname
drop table #t
The SQL above is divided into Part A and Part B.
Part A SQL get all the childrens below a root node(and filtered those belong to the specified DepartLevelKey). For example, to get all persons in Sales Department's child-department with level=3.
Part B SQL change the rows to column, For example:
Change:
TableID OLAPKey MainDemoName
1 6 Sales Team1
2 10 Sales Team1
3 12 Sales Team1
to:
TableID MainDemoName OLAPKeystr
1 Sales Team1 6,10,12
Thus we get each goal department's persons, for further processing(omited here).
The Problem:
The Part A is very slow, cost about 5 minutes. The Part B is slow too.
I wonder how to optimize it basing the table struc existed.
yours,
Ivan

Answer 1

Try:
(i) Adding this index to OLAPAgentDim:
create index IX_OLAPAgentDim_FatherKey on OLAPAgentDim (FatherKey) include (DepartLevelKey, MainDemoName)
(ii) Changing MainDemoName in #t from nvarchar(max) to nvarchar(100). This matches the column definition in OLAPAgentDim.
(iii) Between Part A and Part B, i.e. after Part A and before Part B, adding this index to #t:
create clustered index IX on #t (MainDemoName)