12 things Oracle DBAs must know about SQL

12 THINGS EVERY
ORACLE DBA AND
DEVELOPER SHOULD
KNOW ABOUT SQL
IGGY FERNANDEZ
EDITOR,
NoCOUG JOURNAL

INTRODUCTION
» Editor of the NoCOUG Journal (Archive)
» Presentation based on The Twelve Days of SQL blog
series
» Also read The Twelve Days of NoSQL blog series
» New blog series: The Hitchhiker’s Guide to the
EXPLAIN PLAN (ToadWorld)

TOP TIPS
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desktop

“Those who are in love with practice without knowledge are
like the sailor who gets into a ship without rudder or
compass and who never can be certain where he is going.
Practice must always be founded on sound theory.”
—The Discourse on Painting
by Leonardo da Vinci

#1 SQL IS A NON-PROCEDURAL LANGUAGE

Charles Bachman
1973 Turing Award Winner
Dr. Edgar Codd
1981 Turing Award Winner
PATRON SAINTS OF DATABASES

NETWORK MODEL
CHAINS
OWNER
RECORD
MEMBER
RECORD 3
MEMBER
RECORD 2
MEMBER
RECORD 1

THE PROGRAMMER AS NAVIGATOR
1. Sequential access
2. ROWID access
3. Primary key access
4. Secondary key access
5. Starting from the owner record, get all the records in a
chain
6. Starting from any record, get the prior and next in a
chain
7. Starting from any record, get the owner record

“Each of these access methods is interesting in itself,
and all are very useful. However, it is the synergistic
usage of the entire collection which gives the
programmer great and expanded powers to come and
go within a large database while accessing only those
records of interest in responding to inquiries and
updating the database in anticipation of future inquiries.”
—Charles Bachman
ACM Turing Award Lecture, 1973

RELATIONAL MODEL
“In the choice of logical data structures that a system is to
support, there is one consideration of absolutely paramount
importance – and that is the convenience of the majority
of users. … To make formatted data bases readily
accessible to users (especially casual users) who have
little or no training in programming we must provide the
simplest possible data structures and almost natural
language. … What could be a simpler, more universally
needed, and more universally understood data structure
than a table?”
—Dr. Edgar Codd
Normalized Data Base Structure: A Brief Tutorial (1971)

THE INTENDED AUDIENCE FOR SQL
“There is a large class of users who, while they are not
computer specialists, would be willing to learn to interact
with a computer in a reasonably high-level, non-procedural
query language. Examples of such users are accountants,
engineers, architects, and urban planners. It is for this
class of users that SEQUEL is intended.”
—Donald Chamberlin and Raymond Boyce
Sequel: A Structured English Query Language (1974)

WHAT CAN YOU EXPECT FROM A NON-
PROCEDURAL QUERY LANGUAGE?
» Oracle is not using the indexes we created.
» The query plan changes without apparent reason.
» The query ran fast yesterday but is running slowly today.
» The query ran fast a few minutes ago but is running
slowly now.
» The query runs fast in the QA database but runs slowly
in the production database.
» The query runs slower after a database upgrade.

QUERY PLAN STABILITY
» Hints
» Stored Outlines
» SQL Profiles (the SQLT version)
» SQL Plan Management
» Don’t refresh statistics?
» Don’t use bind variable peeking?

#2 SQL IS BASED ON RELATIONAL
CALCULUS AND RELATIONAL
ALGEBRA

DEFINITION OF A RELATIONAL DATABASE
“A relational database is a database in which: The data is
perceived by the user as tables (and nothing but tables)
and the operators available to the user for (for example)
retrieval are operators that derive “new” tables from “old”
ones.”
—Chris Date
An Introduction to Database Systems

RELATIONAL OPERATORS
(FUNDAMENTAL AND SUFFICIENT)
» Selection
» Projection
» Union
» Difference (Minus)
» Join (Cross Join)

ADDITIONAL RELATIONAL OPERATIONS
(DERIVABLE FROM THE FUNDAMENTAL SET)
» Inner Join
» Outer Join
» Intersection
» Semi Join
» Anti Join
» Division

THE EMPLOYEES TABLE
Name Null? Type
----------------------------------------- -------- ------------------
EMPLOYEE_ID NOT NULL NUMBER(6)
FIRST_NAME VARCHAR2(20)
LAST_NAME NOT NULL VARCHAR2(25)
EMAIL NOT NULL VARCHAR2(25)
PHONE_NUMBER VARCHAR2(20)
HIRE_DATE NOT NULL DATE
JOB_ID NOT NULL VARCHAR2(10)
SALARY NUMBER(8,2)
COMMISSION_PCT NUMBER(2,2)
MANAGER_ID NUMBER(6)
DEPARTMENT_ID NUMBER(4

THE JOB HISTORY TABLE
Name Null? Type
----------------------------------------- -------- -----
EMPLOYEE_ID NOT NULL NUMBER(6)
START_DATE NOT NULL DATE
END_DATE NOT NULL DATE
DEPARTMENT_ID NUMBER(4)

THE JOBS TABLE
Name Null? Type
----------------------------------------- -------- ------------------
JOB_TITLE NOT NULL VARCHAR2(35)
MIN_SALARY NUMBER(6)
MAX_SALARY NUMBER(6)

EMPLOYEES WHO HAVE HELD ALL
ACCOUNTING JOBS (RELATIONAL
CALCULUS)
SELECT employee_id FROM employees e WHERE NOT EXISTS
(
SELECT job_id FROM jobs j WHERE job_id LIKE 'AC%‘ AND NOT EXISTS
(
SELECT * FROM
(
SELECT employee_id, job_id FROM job_history
UNION
SELECT employee_id, job_id FROM employees
)
WHERE employee_id = e.employee_id AND job_id = j.job_id
)
)

ACCOUNTING JOBS (RELATIONAL ALGEBRA)
SELECT employee_id FROM employees
MINUS
SELECT employee_id
FROM
(SELECT employees.employee_id,
jobs.job_id
FROM employees
CROSS JOIN jobs
WHERE jobs.job_id LIKE 'AC%'
MINUS
( SELECT employee_id, job_id FROM job_history
UNION
SELECT employee_id, job_id FROM employees
)
)

ACCOUNTING JOBS (SUBQUERY
FACTORING)
WITH
-- Step 1: Projection
all_employees_1 AS
( SELECT employee_id FROM employees
),
all_employees_2 AS
( SELECT employee_id FROM employees
),
all_jobs AS
( SELECT job_id FROM jobs
),

-- Step 4: Selection
selected_jobs AS
( SELECT * FROM all_jobs WHERE job_id LIKE 'AC%'
),
-- Step 5: Join
selected_pairings AS
( SELECT * FROM all_employees_2 CROSS JOIN
selected_jobs
),
current_job_titles AS
( SELECT employee_id, job_id FROM employees
),

previous_job_titles AS
( SELECT employee_id, job_id FROM job_history
),
-- Step 8: Union
complete_job_history AS
( SELECT * FROM current_job_titles
UNION
SELECT * FROM previous_job_titles
),
-- Step 9: Difference
nonexistent_pairings AS
( SELECT * FROM selected_pairings
MINUS
SELECT * FROM complete_job_history
),

ineligible_employees AS
( SELECT employee_id FROM nonexistent_pairings
)
-- Step 11: Difference
SELECT * FROM all_employees_1
MINUS
SELECT * FROM ineligible_employees

#3 THERE ISN’T ALWAYS A SINGLE OPTIMAL
QUERY PLAN FOR A SQL QUERY

IT DEPENDS ON WHAT THE MEANING OF THE
WORD “IS” IS
select * from employees
where first_name like 'Lex'
and last_name like 'De Haan'
select * from employees
where first_name like :b1
and last_name like :b2

#4 THE TWELVE DAYS OF SQL: THE WAY YOU
WRITE YOUR QUERY MATTERS

THE PERSONNEL TABLE
CREATE TABLE personnel
(
empid CHAR(9),
lname CHAR(15),
fname CHAR(12),
address CHAR(20),
city CHAR(20),
state CHAR(2),
ZIP CHAR(5)
);

THE PAYROLL TABLE
CREATE TABLE payroll
(
empid CHAR(9),
bonus INTEGER,
salary INTEGER
);

SOLUTION #1
RELATIONAL ALGEBRA METHOD
SELECT DISTINCT per.empid, per.lname
FROM personnel per JOIN payroll pay ON (per.empid = pay.empid)
WHERE pay.salary = 199170;
Plan hash value: 3901981856
-------------------------------------------------------
| Id | Operation | Name |
-------------------------------------------------------
| 0 | SELECT STATEMENT | |
| 1 | HASH UNIQUE | |
| 2 | NESTED LOOPS | |
| 4 | TABLE ACCESS BY INDEX ROWID| PAYROLL |
|* 5 | INDEX RANGE SCAN | PAYROLL_I1 |
|* 6 | INDEX UNIQUE SCAN | PERSONNEL_U1 |
| 7 | TABLE ACCESS BY INDEX ROWID | PERSONNEL |
-------------------------------------------------------

SOLUTION #2
UNCORRELATED SUBQUERY
SELECT per.empid, per.lname
FROM personnel per
WHERE per.empid IN (SELECT pay.empid
FROM payroll pay
WHERE pay.salary = 199170);
------------------------------------------------------
------------------------------------------------------
------------------------------------------------------

SOLUTION #3
CORRELATED SUBQUERY
FROM personnel per
WHERE EXISTS (SELECT *
FROM payroll pay
WHERE per.empid = pay.empid
AND pay.salary = 199170);
-------------------------------------------------------
-------------------------------------------------------
| 3 | SORT UNIQUE | |
-------------------------------------------------------

SOLUTION #4
SCALAR SUBQUERY IN THE WHERE CLAUSE
FROM personnel per
WHERE (SELECT pay.salary FROM payroll pay WHERE pay.empid = per.empid) =
199170;
---------------------------------------------------
---------------------------------------------------
|* 1 | FILTER | |
| 2 | TABLE ACCESS FULL | PERSONNEL |
|* 4 | INDEX UNIQUE SCAN | PAYROLL_U1 |
---------------------------------------------------
1 - filter(=199170)
4 - access("PAY"."EMPID"=:B1)

SOLUTION #5
SCALAR SUBQUERY IN THE SELECT CLAUSE
SELECT DISTINCT pay.empid, (SELECT lname FROM personnel per WHERE per.empid
= pay.empid)
FROM payroll pay
WHERE pay.salary = 199170;
-----------------------------------------------------
-----------------------------------------------------
| 3 | HASH UNIQUE | |
-----------------------------------------------------
2 - access("PER"."EMPID"=:B1)
5 - access("PAY"."SALARY"=199170)

SOLUTION #6
AGGREGATE FUNCTION TO CHECK
EXISTENCE
FROM personnel per
WHERE (SELECT count(*) FROM payroll pay WHERE pay.empid = per.empid AND
pay.salary = 199170) > 0;
----------------------------------------------------
----------------------------------------------------
|* 1 | FILTER | |
| 3 | SORT AGGREGATE | |
|* 4 | TABLE ACCESS BY INDEX ROWID| PAYROLL |
|* 5 | INDEX UNIQUE SCAN | PAYROLL_U1 |
----------------------------------------------------
1 - filter(>0)
4 - filter("PAY"."SALARY"=199170)
5 - access("PAY"."EMPID"=:B1)

SOLUTION #7
CORRELATED SUBQUERY (DOUBLE
NEGATIVE)
FROM personnel per
WHERE NOT EXISTS (SELECT *
FROM payroll pay
WHERE pay.empid = per.empid
AND pay.salary != 199170);
------------------------------------------
------------------------------------------
|* 1 | HASH JOIN RIGHT ANTI| |
|* 2 | TABLE ACCESS FULL | PAYROLL |
------------------------------------------
1 - access("PAY"."EMPID"="PER"."EMPID")
2 - filter("PAY"."SALARY"<>199170)

SOLUTION #8
UNCORRELATED SUBQUERY (DOUBLE
NEGATIVE)
FROM personnel per
WHERE per.empid NOT IN (SELECT pay.empid
FROM payroll pay
WHERE pay.salary != 199170);
---------------------------------------------
---------------------------------------------
|* 1 | HASH JOIN RIGHT ANTI NA| |
|* 2 | TABLE ACCESS FULL | PAYROLL |
---------------------------------------------
1 - access("PER"."EMPID"="PAY"."EMPID")
2 - filter("PAY"."SALARY"<>199170)

COMPARISON (11G RELEASE 2)
METHOD Plan Hash Elapsed Buffer Gets
(us)
--------------------------------------- ---------- ------- -----------
Uncorrelated subquery 3342999746 129 17
Correlated subquery 864898783 405 16
Relational algebra method 3901981856 426 16
Scalar subquery in the SELECT clause 750911849 701 16
Uncorrelated subquery (double negative) 2202369223 7702 241
Correlated subquery (double negative) 103534934 14499 241
Scalar subquery in the WHERE clause 3607962630 195999 10549
Aggregate function to check existence 3561519015 310690 10554

#8 STATISTICS ARE A DOUBLE-EDGED SWORD

“Oh, and by the way, could you please stop
gathering statistics constantly? I don’t
know much about databases, but I do
think I know the following: small tables
tend to stay small, large tables tend to stay
large, unique indexes have a tendency to
stay unique, and non-unique indexes often
stay non-unique.”
—Dave Ensor
(as remembered by Mogens Norgaard)
Statistics: How and When?
November 2010 issue of the NoCOUG
Journal

“Monitor the changes in execution plans
and/or performance for the individual SQL
statements … and perhaps as a
consequence re-gather stats. That way,
you’d leave stuff alone that works very
well, thank you, and you’d put your efforts
into exactly the things that have become
worse.”
—Mogens Norgaard
Journal

“There are some statistics about your data
that can be left unchanged for a long time,
possibly forever; there are some statistics
that need to be changed periodically; and
there are some statistics that need to be
changed constantly. … The biggest
problem is that you need to understand
the data.”
—Jonathan Lewis
Journal

“It astonishes me how many shops prohibit
any un-approved production changes and
yet re-analyze schema stats weekly.
Evidently, they do not understand that the
[possible consequence] of schema re-
analysis is to change their production SQL
execution plans, and they act surprised
when performance changes!”
—Don Burleson
Journal

#9 PHYSICAL DATABASE DESIGN MATTERS

“If a schema has no IOTs or clusters, that is a good
indication that no thought has been given to the matter of
optimizing data access.”
—Tom Kyte (quoting Steve Adams)
Effective Oracle by Design (page 379)

THE NOSQL COMPLAINT
“Using tables to store objects is like driving your car home and
then disassembling it to put it in the garage. It can be assembled
again in the morning, but one eventually asks whether this is the
most efficient way to park a car.”
—incorrectly attributed to Esther Dyson
Editor of Release 1.0
“You can keep a car in a file cabinet because you can file the
engine components in files in one drawer, and the axles and
things in another, and keep a list of how everything fits together.
You can, but you wouldn’t want to.”
—Esther Dyson
September 1989 issue of Release 1.0

DEMONSTRATION
Refer to the blog post The Twelve Days of NoSQL: Day
Six: The False Premise of NoSQL

#10 SOMETIMES THE OPTIMIZER NEEDS A HINT

WHO SAID THAT?
“No optimizer is perfect and directives such as Oracle’s
hints provide the simplest workaround [in] situations in
which the optimizer has chosen a suboptimal plan.
Hints are useful tools not just to remedy an occasional
suboptimal plan, but also for users who want to
experiment with access paths, or simply have full
control over the execution of a query.”
1. Homer Simpson
2. Dr. Edgar Codd (the inventor of relational database
theory)
3. Oracle white paper by Maria Colgan and Fred Lumpkin

SUGGESTIONS FROM DAN TOW
» Oracle’s Cost-based Optimizer (CBO) does a perfectly good job on
most SQL, requiring no manual tuning for most SQL.
» The CBO must parse quickly, use the data and indexes that it has,
make assumptions about what it does not know, and deliver exactly
the result that the SQL calls for.
» On a small fraction of the SQL, the constraints on the CBO result in
a performance problem.
» Find SQL worth tuning, ignoring the great majority that already
performs just fine.
» Find the true optimum execution plan (or at least one you verify is
fast enough), manually, without the CBO’s constraints or
assumptions.
» Compare your manually chosen execution plan, and its resulting
performance, with the CBO’s plan and consider why the CBO did
not select your plan, if it did not.
» Choose a solution that solves the problem.

#11 AWR AND STATSPACK ARE A GOLDMINE OF
HISTORICAL PERFORMANCE DATA

#12 READERS DO NOT BLOCK WRITERS;
WRITERS DO NOT BLOCK READERS

SERIALIZABILITY
“To describe consistent transaction behavior when
transactions run at the same time, database researchers
have defined a transaction isolation model called
serializability. The serializable mode of transaction behavior
tries to ensure that transactions run in such a way that they
appear to be executed one at a time, or serially, rather
than concurrently.”
—Oracle documentation up to 11g Release 1

DEMONSTRATION
create table parent (
parent_name varchar(8)
);
create table child (
child_name varchar(8),
parent_name varchar(8)
);
insert into parent values('Warbucks');
commit;

18:25:07 TRANSACTION A> alter session set
isolation_level=serializable;
Session altered.
18:25:07 TRANSACTION A> select * from parent where
parent_name='Warbucks';
1 row selected.
18:25:16 TRANSACTION B> alter session set
isolation_level=serializable;
Session altered.
18:25:16 TRANSACTION B> select * from child where
no rows selected

18:25:19 TRANSACTION A> insert into child values
('Annie','Warbucks');
1 row created.
18:25:21 TRANSACTION B> delete from parent where
1 row deleted.
18:25:23 TRANSACTION A> commit;
Commit complete.
18:25:25 TRANSACTION B> commit;
Commit complete.

Based on the blog post Day 12: The Twelve Days of SQL:
Readers do not block writers; writers do not block readers

#7 EXPLAIN PLAN LIES (SO DOES
AUTOTRACE)

WHAT ABOUT …
» Bind variables
» Adaptive cursor sharing
» Cardinality feedback
» Adaptive query optimization

#5 THE QUERY COST IS ONLY AN ESTIMATE

#6 THE EXECUTION PLAN IS A TREE

HOW TO READ EXPLAIN PLAN OUTPUT
(NOT!)
“The execution order in EXPLAIN PLAN output begins with
the line that is the furthest indented to the right. The next
step is the parent of that line. If two lines are indented
equally, then the top line is normally executed first.”
—Oracle documentation

THE REAL SCOOP
An Oracle EXPLAIN PLAN is a “tree” structure
corresponding to a relational algebra expression. It is
printed in “pre-order” sequence (visit the root of the tree,
then traverse each subtree—if any—in pre-order
sequence) but should be read in “post-order” sequence
(first traverse each subtree—if any—in post-order
sequence, then only visit the root of the tree).

----------------------------------------------------------------
----------------------------------------------------------------
| 1 | MINUS | |
| 2 | SORT UNIQUE NOSORT | |
| 3 | INDEX FULL SCAN | EMP_EMP_ID_PK |
| 5 | VIEW | |
| 6 | MINUS | |
| 8 | MERGE JOIN CARTESIAN | |
| 9 | INDEX RANGE SCAN | JOB_ID_PK |
| 10 | BUFFER SORT | |
| 11 | INDEX FAST FULL SCAN | EMP_EMP_ID_PK |
| 13 | UNION-ALL | |
| 14 | VIEW | index$_join$_006 |
| 15 | HASH JOIN | |
| 16 | INDEX FAST FULL SCAN| JHIST_EMP_ID_ST_DATE_PK |
| 17 | INDEX FAST FULL SCAN| JHIST_JOB_IX |
| 18 | VIEW | index$_join$_007 |
| 19 | HASH JOIN | |
| 20 | INDEX FAST FULL SCAN| EMP_EMP_ID_PK |
| 21 | INDEX FAST FULL SCAN| EMP_JOB_IX |
----------------------------------------------------------------

SELECT
e.first_name, e.last_name, e.salary,
j.job_title,
d.department_name,
l.city, l.state_province,
c.country_name,
r.region_name
FROM employees e, jobs j, departments d, locations l, countries c, regions r
WHERE
e.department_id = 90
AND j.job_id = e.job_id
AND d.department_id = e.department_id
AND l.location_id = d.location_id
AND c.country_id = l.country_id
AND r.region_id = c.region_id;

HINTS FOR DEEP LEFT TREE
LEADING (e j d l c r)
USE_NL(j)
USE_NL(d)
USE_NL(l)
USE_NL(c)
USE_NL(r)

HINTS FOR DEEP RIGHT TREE
USE_HASH(j) SWAP_JOIN_INPUTS(j)
USE_HASH(d) SWAP_JOIN_INPUTS(d)
USE_HASH(l) SWAP_JOIN_INPUTS(l)
USE_HASH(c) SWAP_JOIN_INPUTS(c)
USE_HASH(r) SWAP_JOIN_INPUTS(r)

HINTS FOR ZIG-ZAG TREE
USE_HASH(j) SWAP_JOIN_INPUTS(j)
USE_HASH(d) NO_SWAP_JOIN_INPUTS(d)
USE_HASH(l) SWAP_JOIN_INPUTS(l)
USE_HASH(c) NO_SWAP_JOIN_INPUTS(c)
USE_HASH(r) SWAP_JOIN_INPUTS(r)

BUSHY TREE (INLINE VIEW)SELECT /*+ LEADING(e j d) USE_NL(j) USE_NL(d) */
e.first_name, e.last_name, e.salary,
j.job_title,
d.department_name, d.city, d.state_province, d.country_name, d.region_name
FROM
employees e,
jobs j,
(
SELECT /*+ NO_MERGE */
d.department_id, d.department_name,
l.city, l.state_province,
c.country_name,
r.region_name
FROM departments d, locations l, countries c, regions r
WHERE l.location_id = d.location_id
AND c.country_id = l.country_id
AND r.region_id = c.region_id
) d
WHERE e.department_id = 90
AND j.job_id = e.job_id
AND d.department_id = e.department_id;

BUSHY TREES
» (A B) C) D
» (A (B C)) D
» (A B) (C D)
» A ((B C) D)
» A (B (C D))

TOTAL NUMBER OF TREES
N FACTORIAL(N) CATALAN(N-1) Total trees
2 2 1 2
3 6 2 12
4 24 5 120
5 120 14 1,680
6 720 42 30,240
7 5,040 132 665,280
8 40,320 429 17,297,280
9 362,880 1,430 518,918,400
10 3,628,800 4,862 17,643,225,600

THANK YOU FOR LISTENING!
iggy_fernandez@hotmail.com
The Hitchhiker’s Guide to the
EXPLAIN PLAN
NoCOUG Journal Archive

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