how to resolve loops

More Tricks of the Universe Masters

More Tricks of the Universe Masters

David G. Rathbun, Integra Solutions

IntroductionTricks of the MastersWrap UpQ&A

Please hold questions until the end of the presentation

Topics

Introduction 1/3

Focused on BusinessObjects Solutions since 1995Consulting / mentoring / trouble shootingCertified BusinessObjects trainerPrimary focus on knowledge transfer and client education

Selected to present at 1996 - 2005 User Conference10 consecutive years as a presenter

Charter member of BOBhttp://busobj.forumtopics.com

A few words about me…

Introduction 2/3

Designer has changed very little since 4.0Initially released in 1996Introduced contexts, aggregate awareness, multiple SQL statementsCore technology has been stable for many years

Most important component of your implementationA bad universe = a bad implementationThink ZEN

• Zero Defect Coding• Efficient SQL Generation• Narrow Project Scope

Know the rules and when to break them

A few words about Designer…

Introduction 3/3

ANSI 92 join syntax now availableDerived tablesIndex awareness

6.5

Parameter settings as part of the universeSave as PDF

6.1

Find and replaceIntegrity check does not check cardinalities by defaultList of aliases available

6.0FeatureVersion

Designer Evolution

What has changed over the years?

IntroductionTricks of the MastersWrap UpQ&A

Topics

Intelligent SQL generationEffective use of derived tablesWorking with hierarchical data

Tricks of the Masters

Demonstrations today will use Designer 6.5 although many techniques are valid in 5.x

Intelligent SQL Generation

The pitfalls of the all-in-one solutionJoin loopsFan trapsChasm traps

Not generally present in a “one-of” solutionImpact on success

Cause incorrect SQL to be generated

Techniques demonstrated in this section are essential to the success of your project

Join Loops 1/4

Loops are often present in logical data diagrams

Join Loops 2/4

SQL Generator uses all available paths

Why are loops bad?

SELECTEmployees.EMP_LASTNAME,Salary.SALARY_VALUE

FROMSalary,Employees,Absenses,Lookups

WHERE( Absenses.EMP_ID=Employees.EMP_ID )AND ( Salary.EMP_ID=Employees.EMP_ID )AND ( Lookups.LOOKUP_ID=Salary.LOOKUP_ID )AND ( Lookups.LOOKUP_ID=Absenses.LOOKUP_ID )

Join Loops 3/4

Join loops may be addressed using several techniquesAliasesContextsShortcutsMultiple universes

What is the “best” solution for join loops?

Join Loops 4/4

There is no “best” solution!Each loop must be examined independentlyMake sure you evaluate all of your options

When Do I Use an Alias? 1/3

More than one logical purpose for the same tableRequire different objects

Types of salaryTypes of absences

When Do I Use an Alias? 2/3

Always set cardinality on every joinOne – ManyOne – One

A table on the one side is a candidate for an alias

Note: You should never have a Many – Many relationship in a production database!

When Do I Use an Alias? 3/3Fixed the loop!

Demonstration

When Do I Use a Context? 1/3

More than one logical path between two or more tables


A table on the many side helps identify a context


Each context identifies a path through the structure

Alias versus Context

There is no “better” answerDo you have paths or purposes?

AliasesRequire additional objects for each purposeGenerally used on tables on the outside of the universe structureEliminate the loop

ContextsDo not require additional objectsDo not actually eliminate the loopRequire ongoing maintenance

Are loops the only issue?

Chasm Trap DefinitionA chasm spans a many – one – many relationship

Each showroom has many employeesEach showroom has many annual salary figuresThe number of employee records should not impact the total salary on the report

Chasm Trap Example 1/3

SELECTShowrooms.SHOWROOM_NAME,sum(Salary_Summary_Annual.ANNUAL_SAL_TOTAL)

FROMSalary_Summary_Annual,Showrooms

WHERE( Salary_Summary_Annual.SHOWROOM_ID =

Showrooms.SHOWROOM_ID )GROUP BY

Showrooms.SHOWROOM_NAME

Showroom Name Annual SalarayPrestige Cars $316,720.00Prestige Motors $565,730.00Prestige Sports Cars $461,380.00

Sum: $1,343,830.00


SELECTShowrooms.SHOWROOM_NAME,Employees.EMP_LASTNAME,sum(Salary_Summary_Annual.ANNUAL_SAL_TOTAL)

FROMEmployees,Salary_Summary_Annual,Showrooms

WHERE( Employees.SHOWROOM_ID=Showrooms.SHOWROOM_ID )AND (

Salary_Summary_Annual.SHOWROOM_ID=Showrooms.SHOWROOM_ID )GROUP BY

Showrooms.SHOWROOM_NAME, Employees.EMP_LASTNAME


A chasm trap inflates the number of rowsX rows on the leftY rows on the rightX * Y rows in the combined set

Can you prevent a query writer from using both objects?No

Can you still get the correct answer?Yes

Context Chasm Trap Solution 1/2

Set up contexts for each side of the chasmBusinessObjects automatically generates separate SQL

Context Chasm Trap Solution 2/2

You can prevent users from spanning a chasm

Intelligent SQL Generation Review

The pitfalls of the all-in-one solutionJoin loopsFan traps (not discussed today)Chasm traps

Not generally present in a “one-of” solutionImpact on success

Cause incorrect SQL to be generated

Designer gives us the tools to handle these issues

Effective Use of Derived Tables

Another option for aliasesReplace database viewsDenormalize dataCreate summary dataEliminate intermediate joins… all this and more

Insert + Derived Tables…

Enter desired SQL code

Creating A Derived Table 1/2

Select Models.MODEL_ID, Models.MODEL_NAME, Models.MODEL_PRICE, Models.MODEL_COST, Makers.MAKER_NAME, Styles.STYLE_NAMEFROM Models, Makers, StylesWHERE Models.STYLE_ID = Styles.STYLE_IDAND Models.MAKER_ID = Makers.MAKER_ID

Creating A Derived Table 2/2

This example shows denormalized dataCould be done with a viewAre there better uses for this technique?

Summary Data

Database views are staticDerived tables can include promptsSELECT Departments.DEPT_ID, Departments.DEPT_NAME, Lookups.LOOKUP_TEXT AS Salary_type, sum(Salary.SALARY_VALUE) as Total_salaryFROM Departments, Employees, Salary, LookupsWHERE Departments.DEPT_ID=Employees.DEPT_IDAND Employees.EMP_ID=Salary.EMP_IDAND Salary.LOOKUP_ID=Lookups.LOOKUP_IDAND Lookups.LOOKUP_TYPE='SAL'GROUP BY Departments.DEPT_ID, Departments.DEPT_NAME, Lookups.LOOKUP_TEXTHAVING sum(Salary.SALARY_VALUE) >= @Prompt('EnterDepartment Salary Threshhold','N',,,)

Demonstration

Hierarchical Data 1/2

Hierarchical data is not uncommonCompany organizational chartManufacturing BOM (Bill of Materials)Referral tracking

Hierarchical data is recursiveA table joins to itselfDepth is unlimitedDepth is unknown

SQL is not recursiveHow can we solve this?

Employee ID Last Name Manager ID101 Noakes102 Ferrerez 101103 Field 102104 Fraser 101105 Snow 101106 Speed 105

Hierarchical Data 2/2

Who do I work for?Who are my direct reports?

Who reports directly to me in the organization

Who are my indirect reports?Who works for someone that works for me

What is my level in the hierarchy?Show my parents in the treeShow my children in the tree

Some typical questions

Alias Solution 1/2

Use aliases to represent the hierarchical relationships

One solution

Demonstration

Alias Solution 2/2

Unknown depthEach new level requires an aliasAdding new employees should not require universe maintenance

Varying depthAn organization tree is not always balancedDifferent employees have different depthsOuter joins are used to allow for missing dataOuter joins can cause performance issuesNot all databases support outer join chains (sequences of outer joins)

The problems

Creating New Tables 1/2

Create a new table to flatten out the structureEmployee IDManager IDTop Level ID

Each person exists as a top level IDWho works for me?

Each person exists as an Employee IDWho do I work for?What is my level?

Another solution

Creating New Tables 2/2

This table has to be created / populated / updatedStores each potential manager as a top level

Inflates the amount of data

Have to change objects if I change questionsWho works for me?Who do I work for?What is my level in the hierarchy?

The problems

Tree Traversal Data 1/5

Assign each person (node) a left and right valueValue is determined by tree traversal order

Tree traversal means walk around the outside edgeNumber each node each time you pass the left or right sideStore those values in the original source table

Store the tree traversal path in the original table

Noakes

Ferrerez Fraser Snow Pickworth

Forest Porter Duckworth

5 6

9 10

7 8 15

11 12 13 14

161

2 3 4

Note: This is not the complete tree


ID Full Name Left Tree Right Tree101 Noakes, Nicholas 1 54102 Ferrerez, Ferdinand 2 5103 Field, Felicity 3 4104 Fraser, Frank 6 7105 Snow, Sara 8 13106 Speed, Sonya 9 10107 Spencer, Steve 11 12108 Helen, Harrison 14 15109 Thomas, Tom 16 19110 Thatcher, Terry 17 18111 Davis, Diana 20 21201 Pickworth, Paul 22 39202 Forest, Florence 23 28203 Brown, Bella 24 25204 Porter, Pete 29 34205 Irving, Ira 30 31206 Bailey, Ben 32 33207 Duckworth, Dave 35 38208 Ince, Ian 36 37209 Hilary, Hibbs 26 27

Name Descendants Parents LevelNoakes, Nicholas 26 0 1 Ferrerez, Ferdinand 1 1 2 Field, Felicity 0 2 3 Fraser, Frank 0 1 2 Snow, Sara 2 1 2 Speed, Sonya 0 2 3 Spencer, Steve 0 2 3 Helen, Harrison 0 1 2 Thomas, Tom 1 1 2 Thatcher, Terry 0 2 3 Davis, Diana 0 1 2 Pickworth, Paul 8 1 2 Forest, Florence 2 2 3 Brown, Bella 0 3 4 Hilary, Hibbs 0 3 4 Porter, Pete 2 2 3 Irving, Ira 0 3 4 Bailey, Ben 0 3 4 Duckworth, Dave 1 2 3 Ince, Ian 0 3 4

Using left and right data to build the hierarchy

Demonstration


Inserts and updates must adjust left and right values

Noakes

Ferrerez Fraser Snow Pickworth

Forest Porter Duckworth

5 6

9

10

7 8 17

11

14 15 16

181

2 3

New12 13

Update table set left = left + 2 where left >= 10;

Update table set right = right + 2 where right >= 10;

Insert (name, left, right) values (‘New’, 10, 11);


Including the following derived table in our universe…

… enables us to answer the required questionsWho do I work for?Who are my direct reports?Who are my indirect reports?What is my level in the hierarchy?

Revisiting derived tables

SELECT A.EMP_ID,(A.RIGHT_TREE-A.LEFT_TREE-1)/2 AS Number_of_descendants, COUNT(B.EMP_ID) AS Employee_Level,COUNT(B.EMP_ID)-1 AS Number_of_parents

FROM Employees A, Employees BWHERE B.LEFT_TREE <= A.LEFT_TREE AND B.RIGHT_TREE >= A.RIGHT_TREEGROUP BY A.EMP_ID, (A.RIGHT_TREE-A.LEFT_TREE-1)/2


Who do I work for?Left < My Left and Right > My Right and Level = My Level - 1

Who are my direct descendants?Left > My Left and Right < My Right and Level = My Level + 1

Who are my indirect descendants?Left > My Left and Right < My Right and Level > My Level + 1

Who are all of my descendants?Left > My Left and Right < My Right

What is my level in the hierarchy?Provided directly by the derived table

Create a complete treeDisplay all records order by LEFT indent by LEVEL

Hierarchical Data Recap

There are various solutions for handling hierarchiesAliasesExtra tablesTree traversal

Tree traversal data seems to provide the best solutionSample data included one hierarchy onlyIf multiple hierarchies are present add the TREE_ID to each nodeAssign TREE_ID during the initial tree traversal process

Database views may be substituted for derived tablesReminder: Derived tables were added in Designer 6.5

Intelligent SQL generationResolve loopsResolve chasm trapsResolve fan traps

Effective use of derived tablesDenormalize dataSummarize dataInclude prompts

Working with hierarchical dataNumber each node according to its position in the tree pathUse that information to answer typical hierarchical questions



Questions?Contact information

David G. RathbunIntegra Solutionswww.IntegraSolutions.net

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