Introduction to SQL brief notes for academic.ppt

SQL: Queries, Programming,
Triggers

Example Instances
sid sname rating age
22 dustin 7 45.0
31 lubber 8 55.5
58 rusty 10 35.0
28 yuppy 9 35.0
31 lubber 8 55.5
44 guppy 5 35.0
58 rusty 10 35.0
sid bid day
22 101 10/10/96
58 103 11/12/96
R1
S1
S2
 We will use these
instances of the
Sailors and
Reserves relations
in our examples.
 If the key for the
Reserves relation
contained only the
attributes sid and
bid, how would the
semantics differ?

Basic SQL Query
 relation-list A list of relation names (possibly with a
range-variable after each name).
 target-list A list of attributes of relations in relation-list
 qualification Comparisons (Attr op const or Attr1 op
Attr2, where op is one of <,>,=,<=,>==,like)
combined using AND, OR and NOT.
 DISTINCT is an optional keyword indicating that the
answer should not contain duplicates. Default is that
duplicates are not eliminated!
SELECT [DISTINCT] target-list
FROM relation-list
WHERE qualification

Conceptual Evaluation Strategy
 Semantics of an SQL query defined in terms of the
following conceptual evaluation strategy:
 Compute the cross-product of relation-list.
 Discard resulting tuples if they fail qualifications.
 Delete attributes that are not in target-list.
 If DISTINCT is specified, eliminate duplicate rows.
 This strategy is probably the least efficient way to
compute a query! An optimizer will find more
efficient strategies to compute the same answers.

Conceptual Evaluation Strategy
 Semantics of an SQL query based on R.A:
SELECT R.A,S.B
FROM R, S
WHERE R.C=S.C
==============>
R.A,S.B R.C=S.C(R x S)

Example of Conceptual Evaluation
SELECT S.sname
FROM Sailors S, Reserves R ---->range variable
WHERE S.sid=R.sid AND R.bid=103
(sid) sname rating age (sid) bid day
22 dustin 7 45.0 22 101 10/10/96
22 dustin 7 45.0 58 103 11/12/96
31 lubber 8 55.5 22 101 10/10/96
31 lubber 8 55.5 58 103 11/12/96
58 rusty 10 35.0 22 101 10/10/96
58 rusty 10 35.0 58 103 11/12/96

A Note on Range Variables
 Really needed only if the same relation
appears twice in the FROM clause. The
previous query can also be written as:
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND bid=103
SELECT sname
FROM Sailors, Reserves
WHERE Sailors.sid=Reserves.sid
AND bid=103
It is good style,
however, to use
range variables
always!
OR

Find sailors who’ve reserved at least one boat
 Would adding DISTINCT to this query make a
difference?
 What is the effect of replacing S.sid by S.sname in
the SELECT clause? Would adding DISTINCT to
this variant of the query make a difference?.
SELECT S.sid
FROM Sailors S, Reserves R
WHERE S.sid=R.sid

Expressions and Strings
 Illustrates use of arithmetic expressions and string
pattern matching: Find triples (of ages of sailors and two
fields defined by expressions) for sailors whose names
begin and end with B and contain at least three characters.
 AS and = are two ways to name fields in result.
 LIKE is used for string matching. `_’ stands for any
one character and `%’ stands for 0 or more arbitrary
characters.
SELECT S.age, age1=S.age-5, 2*S.age AS age2
FROM Sailors S
WHERE S.sname LIKE ‘B_%B’

Find sid’s of sailors who’ve reserved a red or a green boat
 UNION: Can be used to
compute the union of any
two union-compatible sets
of tuples (which are
themselves the result of
SQL queries).
 If we replace OR by AND in
the first version, what do
we get?
 Also available: EXCEPT
(What do we get if we
replace UNION by EXCEPT?)
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND (B.color=‘red’ OR B.color=‘green’)
SELECT S.sid
WHERE S.sid=R.sid AND
R.bid=B.bid
AND B.color=‘red’
UNION
SELECT S.sid
R.bid=B.bid

Find sid’s of sailors who’ve reserved a red and a green boat
 INTERSECT: Can be used to
compute the intersection
of any two union-
compatible sets of tuples.
 Included in the SQL/92
standard, but some
systems don’t support it.
SELECT S.sid
FROM Sailors S, Boats B1, Reserves R1,
Boats B2, Reserves R2
WHERE S.sid=R1.sid AND R1.bid=B1.bid
AND S.sid=R2.sid AND R2.bid=B2.bid
AND (B1.color=‘red’ AND B2.color=‘green’)
SELECT S.sid
R.bid=B.bid
AND B.color=‘red’
INTERSECT
SELECT S.sid
R.bid=B.bid
AND B.color=‘green’
Key field!

Nested Queries
 A very powerful feature of SQL: a WHERE clause can itself contain
an SQL query! (Actually, so can FROM and HAVING clauses, not
supported by all systems.)
 To find sailors who’ve not reserved #103, use NOT IN.
 To understand semantics of nested queries, think of a nested loops
evaluation: For each Sailors tuple, check the qualification by computing
the subquery.
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT R.sid
FROM Reserves R
WHERE R.bid=103)
Find names of sailors who’ve reserved boat #103:

Nested Queries with Correlation
 EXISTS is another set comparison operator, like IN.
 If UNIQUE is used, and * is replaced by R.bid, finds sailors with at
most one reservation for boat #103. (UNIQUE checks for duplicate
tuples; * denotes all attributes. Why do we have to replace * by
R.bid?)
 Illustrates why, in general, subquery must be re-computed for
each Sailors tuple.
SELECT S.sname
FROM Sailors S
WHERE EXISTS (SELECT *
FROM Reserves R
WHERE R.bid=103 AND S.sid=R.sid)
Find names of sailors who’ve reserved boat #103:

More on Set-Comparison Operators
 We’ve already seen IN, EXISTS and UNIQUE. Can also
use NOT IN, NOT EXISTS and NOT UNIQUE.
 Also available: op ANY, op ALL, op IN
 Find sailors whose rating is greater than that of some
sailor called Horatio:
  
, , , , ,
SELECT *
FROM Sailors S
WHERE S.rating > ANY (SELECT S2.rating
FROM Sailors S2
WHERE S2.sname=‘Horatio’)

Rewriting INTERSECT Queries Using IN
 Similarly, EXCEPT queries re-written using NOT IN.
 To find names (not sid’s) of Sailors who’ve reserved
both red and green boats, just replace S.sid by S.sname
in SELECT clause. (What about INTERSECT query?)
Find sid’s of sailors who’ve reserved both a red and a green boat:
SELECT S.sid
WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’
AND S.sid IN (SELECT S2.sid
FROM Sailors S2, Boats B2, Reserves R2
WHERE S2.sid=R2.sid AND R2.bid=B2.bid
AND B2.color=‘green’)

Division in SQL
 Let’s do it the hard
way, without EXCEPT:
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS
((SELECT B.bid
FROM Boats B)
EXCEPT
(SELECT R.bid
FROM Reserves R
WHERE R.sid=S.sid))
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS (SELECT B.bid
FROM Boats B
WHERE NOT EXISTS (SELECT R.bid
FROM Reserves R
WHERE R.bid=B.bid
AND R.sid=S.sid))
Sailors S such that ...
there is no boat B without ...
a Reserves tuple showing S reserved B
Find sailors who’ve reserved all boats.
(1)
(2)

Aggregate Operators
 Significant extension of
relational algebra.
COUNT (*)
COUNT ( [DISTINCT] A)
SUM ( [DISTINCT] A)
AVG ( [DISTINCT] A)
MAX (A)
MIN (A)
SELECT AVG (S.age)
FROM Sailors S
WHERE S.rating=10
SELECT COUNT (*)
FROM Sailors S
SELECT AVG ( DISTINCT S.age)
FROM Sailors S
WHERE S.rating=10
SELECT S.sname
FROM Sailors S
WHERE S.rating= (SELECT MAX(S2.rating)
FROM Sailors S2)
single column
SELECT COUNT (DISTINCT S.rating)
FROM Sailors S
WHERE S.sname=‘Bob’

Find name and age of the oldest sailor(s)
 The first query is illegal!
(We’ll look into the reason
a bit later, when we
discuss GROUP BY.)
 The third query is
equivalent to the second
query, and is allowed in
the SQL/92 standard, but
is not supported in some
systems.
SELECT S.sname, MAX (S.age)
FROM Sailors S
SELECT S.sname, S.age
FROM Sailors S
WHERE S.age =
(SELECT MAX (S2.age)
FROM Sailors S2)
SELECT S.sname, S.age
FROM Sailors S
WHERE (SELECT MAX (S2.age)
FROM Sailors S2)
= S.age

GROUP BY and HAVING
 So far, we’ve applied aggregate operators to all
(qualifying) tuples. Sometimes, we want to apply
them to each of several groups of tuples.
 Consider: Find the age of the youngest sailor for each
rating level.
 In general, we don’t know how many rating levels
exist, and what the rating values for these levels are!
 Suppose we know that rating values go from 1 to 10;
we can write 10 queries that look like this (!):
SELECT MIN (S.age)
FROM Sailors S
WHERE S.rating = i
For i = 1, 2, ... , 10:

Queries With GROUP BY and HAVING
 The target-list contains (i) attribute names (ii) terms
with aggregate operations (e.g., MIN (S.age)).
 The attribute list (i) must be a subset of grouping-list.
Intuitively, each answer tuple corresponds to a group, and
these attributes must have a single value per group. (A
group is a set of tuples that have the same value for all
attributes in grouping-list.)
SELECT [DISTINCT] target-list
FROM relation-list
WHERE qualification
GROUP BY grouping-list
HAVING group-qualification

Find the age of the youngest sailor with age 18,
for each rating with at least 2 such sailors
 Only S.rating and S.age are
mentioned in the SELECT,
GROUP BY or HAVING clauses;
 2nd column of result is
unnamed. (Use AS to name it.)
SELECT S.rating, MIN (S.age)
FROM Sailors S
WHERE S.age >= 18
GROUP BY S.rating
HAVING COUNT (*) > 1
22 dustin 7 45.0
31 lubber 8 55.5
71 zorba 10 16.0
64 horatio 7 35.0
29 brutus 1 33.0
58 rusty 10 35.0
rating age
1 33.0
7 45.0
7 35.0
8 55.5
10 35.0
rating
7 35.0
Answer relation


For each red boat, find the number of
reservations for this boat
 Grouping over a join of three relations.
SELECT B.bid, COUNT (*) AS scount
WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’
GROUP BY B.bid

Find the age of the youngest sailor with age > 18,
for each rating with at least 2 sailors (of any age)
 Shows HAVING clause can also contain a subquery.
 Compare this with the query where we considered
only ratings with 2 sailors over 18!
SELECT S.rating, MIN (S.age)
FROM Sailors S
WHERE S.age > 18
GROUP BY S.rating
HAVING 1 < (SELECT COUNT (*)
FROM Sailors S2
WHERE S.rating=S2.rating)

Find those ratings for which the average
age is the minimum over all ratings
 Aggregate operations cannot be nested!
SELECT Temp.rating, Temp.avgage
FROM (SELECT S.rating, AVG (S.age) AS avgage
FROM Sailors S
GROUP BY S.rating) AS Temp
WHERE Temp.avgage = (SELECT MIN (Temp.avgage)
FROM Temp)
 Correct solution (in SQL/92):

Null Values
 Field values in a tuple are sometimes unknown (e.g., a rating
has not been assigned) or inapplicable (e.g., no spouse’s
name).
 SQL provides a special value null for such situations.
 The presence of null complicates many issues. E.g.:
 Special operators needed to check if value is/is not null.
 Is rating>8 true or false when rating is equal to null? What about
AND, OR and NOT connectives?
 We need a 3-valued logic (true, false and unknown).
 Meaning of constructs must be defined carefully. (e.g., WHERE
clause eliminates rows that don’t evaluate to true.)
 New operators (in particular, outer joins) possible/needed.

Integrity Constraints (Review)
 An IC describes conditions that every legal instance
of a relation must satisfy.
 Inserts/deletes/updates that violate IC’s are disallowed.
 Can be used to ensure application semantics (e.g., sid is a
key), or prevent inconsistencies (e.g., sname has to be a
string, age must be < 200)
 Types of IC’s: Domain constraints, primary key
constraints, foreign key constraints, general
constraints.
 Domain constraints: Field values must be of right type.
Always enforced.

General Constraints
 Useful when
more general
ICs than keys
are involved.
 Can use queries
to express
constraint.
 Constraints can
be named.
CREATE TABLE Sailors
( sid INTEGER,
sname CHAR(10),
rating INTEGER,
age REAL,
PRIMARY KEY (sid),
CHECK ( rating >= 1
AND rating <= 10 )
CREATE TABLE Reserves
( sname CHAR(10),
bid INTEGER,
day DATE,
PRIMARY KEY (bid,day),
CONSTRAINT noInterlakeRes
CHECK (`Interlake’ <>
( SELECT B.bname
FROM Boats B
WHERE B.bid=bid)))

Constraints Over Multiple Relations
CREATE TABLE Sailors
( sid INTEGER,
sname CHAR(10),
rating INTEGER,
age REAL,
PRIMARY KEY (sid)
)
 Awkward and
wrong!
 If Sailors is
empty, the
number of Boats
tuples can be
anything!
 ASSERTION is the
right solution;
not associated
with either table.
CREATE ASSERTION smallClub
CHECK
( (SELECT COUNT (S.sid) FROM Sailors S)
+ (SELECT COUNT (B.bid) FROM Boats B) < 100 )
Number of boats
plus number of
sailors is < 100

Triggers
 Trigger: procedure that starts automatically if
specified changes occur to the DBMS
 Three parts (ECA rules):
 Event (activates the trigger)
 Condition (tests whether the triggers should run)
 Action (what happens if the trigger runs)

Triggers: Example (SQL:1999)
CREATE TRIGGER youngSailorUpdate
AFTER INSERT ON SAILORS
REFERENCING NEW TABLE NewSailors
FOR EACH STATEMENT
INSERT
INTO YoungSailors(sid, name, age, rating)
SELECT sid, name, age, rating
FROM NewSailors N
WHERE N.age <= 18

Summary
 SQL was an important factor in the early acceptance
of the relational model; more natural than earlier,
procedural query languages.
 Relationally complete; in fact, significantly more
expressive power than relational algebra.
 Even queries that can be expressed in RA can often
be expressed more naturally in SQL.
 Many alternative ways to write a query; optimizer
should look for most efficient evaluation plan.
 In practice, users need to be aware of how queries are
optimized and evaluated for best results.

Summary (Contd.)
 NULL for unknown field values brings many
complications
 SQL allows specification of rich integrity
constraints
 Triggers respond to changes in the database

Introduction to SQL brief notes for academic.ppt

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