More on SQL

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More on SQL

• Lecture Week 9
• INFS 614, Fall 2008

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Example Instances
R1
S1
S2
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SQL Seen so far

• Syntax of Basic SQL query
• Relation-list (FROM)
• Target-list (SELECT)
• Qualification (WHERE)
• Semantic Conceptual Evaluation Strategy
• String operators LIKE, NOT LIKE
• Set operators Union (all), Intersect (all),
  Except (all)
• Qualification involving sets IN, EXISTS,
  UNIQUE, ANY, ALL (and their negated version)
• Nested queries.

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Post Processing

• Processing on the result of an SQL query
• Sorting can sort the tuples in the output by any
  column (even the ones not appearing in the SELECT
  clause)
• Duplicate removal
• Example
• Aggregation operators

SELECT Distinct S.sname FROM Sailors S,
Reserves R WHERE S.sidR.sid and R.bid103 Order
by S.sid asc
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Order By Example
SELECT S.sname, S.age FROM Sailors S WHERE
S.sname LIKE U ORDER BY S.age,S.sname

• For each attribute in the sort list, we can
  specify its order
• ASC Ascending Order
• DESC Descending Order
• Default Order is Ascending

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Aggregate Operators

• Significant extension of relational algebra.

COUNT () COUNT ( DISTINCT A) SUM ( DISTINCT
A) AVG ( DISTINCT A) MAX (A) MIN (A)
single column
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COUNT () COUNT ( DISTINCT A) SUM ( DISTINCT
A) AVG ( DISTINCT A) MAX (A) MIN (A)
Aggregate Operators
SELECT COUNT () FROM Sailors S
SELECT AVG (S.age) FROM Sailors S WHERE
S.rating10
single column
SELECT COUNT (DISTINCT S.rating) FROM Sailors
S WHERE S.snameBob
SELECT AVG ( DISTINCT S.age) FROM Sailors
S WHERE S.rating10
SELECT S.sname FROM Sailors S WHERE S.rating
(SELECT MAX(S2.rating)
FROM Sailors S2)
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Find name and age of the oldest sailor(s)
SELECT S.sname, MAX (S.age) FROM Sailors S

• The first query is illegal! (Well 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, 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
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GROUP BY and HAVING

• So far, weve 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 dont 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
  (!)

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Queries With GROUP BY and HAVING
SELECT DISTINCT target-list FROM
relation-list WHERE qualification GROUP
BY grouping-list HAVING group-qualification

• 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.)

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Conceptual Evaluation

• The cross-product of relation-list is computed,
  tuples that fail qualification are discarded,
  unnecessary fields are deleted, and the
  remaining tuples are partitioned into groups by
  the value of attributes in grouping-list.
• The group-qualification is then applied to
  eliminate some groups. Expressions in
  group-qualification must have a single value per
  group!
• In effect, an attribute in group-qualification
  that is not an argument of an aggregate op also
  appears in grouping-list. (SQL does not exploit
  primary key semantics here!)
• One answer tuple is generated per qualifying
  group.

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Query with GROUP BY and HAVING

• Query Find the age of the youngest sailor for
  each rating level.

SELECT S.rating, MIN (S.age) FROM Sailors
S GROUP BY S.rating
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Find the age of the youngest sailor older than
18, for each rating with at least 2 such sailors
older than 18
SELECT S.rating, MIN (S.age) FROM Sailors
S WHERE S.age gt 18 GROUP BY S.rating HAVING
COUNT () gt 1

• Only S.rating and S.age are mentioned in the
  SELECT, GROUP BY or HAVING clauses other
  attributes unnecessary.
• 2nd column of result is unnamed. (Use AS to name
  it.)

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For each red boat, find the number of
reservations for this boat
SELECT B.bid, COUNT () AS rescount FROM Boats
B, Reserves R WHERE R.bidB.bid AND
B.colorred GROUP BY B.bid

• Grouping over a join of two relations.
• What do we get if we remove B.colorred from
  the WHERE clause and add a HAVING clause with
  this condition?

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Find the age of the youngest sailor with age gt
18, for each rating with at least 2 sailors (of
any age)
SELECT S.rating, MIN (S.age) FROM Sailors
S WHERE S.age gt 18 GROUP BY S.rating HAVING 1
lt (SELECT COUNT ()
FROM Sailors S2 WHERE
S.ratingS2.rating)

• HAVING clause can also contain a correlated
  subquery.
• Compare this with the query where we considered
  only ratings with 2 sailors over 18!
• What if HAVING clause is replaced by
• HAVING COUNT() gt1

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For each sailor with more than three
reservations, find the number of his reservations
SELECT R.sid, COUNT () AS rescount FROM
Reserves R GROUP BY R.sid HAVING COUNT() gt 3
Find the age of the youngest sailor, for each
rating level gt 5 with an average age gt 25
SELECT S.rating, MIN(S.age) AS minAge FROM
Sailors S WHERE S.rating gt 5 GROUP BY
S.rating HAVING AVG(S.age) gt 25
SELECT S.rating, MIN(S.age) AS minAge FROM
Sailors S GROUP BY S.rating HAVING S.rating gt 5
AND AVG(S.age) gt 25
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Find those ratings for which the average age of
sailors is the minimum over all ratings

• Aggregate operations cannot be nested! WRONG

SELECT S.rating FROM Sailors S WHERE
AVG(S.age) (SELECT MIN (AVG (S2.age))
FROM Sailors S2 GROUP BY S2.rating)
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Continue from previous slide

• Correct solution (in SQL/92)

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)
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Note on Oracle SQL in the labs
Do not use the keyword AS to name a temporary
table within the FROM clause.
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Note on Oracle SQL in the labs
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)
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Conclusions (so far)

• Post processing on the result of queries is
  supported.
• Aggregation is the most complex post processing
• Group by clause partitions the results into
  groups
• Having clause puts condition on groups (just
  like Where clause on tuples).

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Null Values

• Field values in a tuple are sometimes unknown
  (e.g., a rating has not been assigned) or
  inapplicable (e.g., no spouses name).
• SQL provides a special value null for such
  situations.

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Dealing with the null value

• Special operators are needed to check if value
  is/is not null.
• Attribute_name IS NULL TRUE if the attributes
  value is null. FALSE otherwise
• Attribute_name IS NOT NULL TRUE if the
  attributes value is not null. FALSE otherwise
• Is ratinggt8 true or false when rating is equal to
  null?
• Actually, its unknown.
• How about rating gt 8 AND age lt 40 ?
• Three-valued logic.

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Three Valued Logic

• Logical operators AND, OR and NOT using a
  3-valued logic (TRUE, FALSE and unknown).

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Null Values Some issues

• The presence of null complicates many issues
• WHERE clause eliminates tuples for which the
  qualification does not evaluate to TRUE.
• Important in nested queries involving EXISTS or
  UNIQUE
• Two tuples in SQL are duplicates in a relation if
  corresponding attributes have the same value or
  both contains null.

SELECT S1.sname FROM SAILORS S1 WHERE EXISTS
( SELECT FROM SAILORS S2 WHERE S1.sname
S2.sname AND AND S1.SID lt S2.SID AND
S1.RATING ltgt S2.RATING)
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Issues with the null value Summary

• WHERE and HAVING clause eliminates rows (groups)
  for which the qualification does not evaluate to
  true (i.e., evaluate to false or unknown).
• Aggregate functions ignore null values (except
  count()).
• Distinct treats all null values as the same.
• The arithmetic operations , -, , / return null
  if one of the arguments is null.

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Outer joins
(left outer-join)
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In Oracle

• Select
• From Sailor S, Reserves R
• Where S.sid R.sid ()

How about Select S.sid, count(R.bid) From Sailor
S, Reserves R Where S.sid R.sid () Group by
S.sid
OR Select S.sid, count() From Sailor S, Reserves
R Where S.sid R.sid () Group by S.sid
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More outer joins

• Left outer join
• sign on the right in Oracle
• Select from R, S where R.idS.id()
• Right outer join
• sign on the left in Oracle
• Select from R, S where R.id()S.id
• Full outer join
• not implemented in Oracle

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More on value functions

• Values can be transformed before aggregated
• e.g. SELECT sum(S.A/2) FROM S
• An interesting decode function (Oracle specific)
• decode(value, if1, then1, if2, then2, , else)
• SELECT sum(decode(Major, INFS, 1, 0)) as
  No_IS_Stu, sum(decode(Major, INFS, 0,
  1)) as No_NonIS_Stu
• FROM Student
• Calculating GPA from letter grades?

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Examples

• Department (D-code, D-Name, Chair-SSn)Course
  (D-code, C-no, Title, Units)Prereq (D-code,
  C-no, P-code, P-no)Class (Class-no, D-code,
  C-no, Instructor-SSn)Faculty (Ssn, F-Name,
  D-Code, Rank)Student (Ssn, S-Name, Major,
  Status)Enrollment (Class-no, Student-Ssn)Transcr
  ipt (Student-Ssn, D-Code, C-no, Grade)

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Query 1

• Department (D-code, D-Name,
  Chair-SSn)Course (D-code, C-no, Title,
  Units)Prereq (D-code, C-no, P-code, P-no)Class
  (Class-no, D-code, C-no, Instructor-SSn)Faculty
  (Ssn, F-Name, D-Code, Rank)Student (Ssn, S-Name,
  Major, Status)Enrollment (Class-no,
  Student-Ssn)Transcript (Student-Ssn, D-Code,
  C-no, Grade)

List the classes (Class-no) currently taken by
students whose names start with 'T'.
SELECT distinct e.Class-no FROM Enrollment e,
Student s WHERE e.Student-Ssn s.Ssn AND
s.S-Name LIKE 'T'
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Query 2

• Department (D-code, D-Name,
  Chair-SSn)Course (D-code, C-no, Title,
  Units)Prereq (D-code, C-no, P-code, P-no)Class
  (Class-no, D-code, C-no, Instructor-SSn)Faculty
  (Ssn, F-Name, D-Code, Rank)Student (Ssn, S-Name,
  Major, Status)Enrollment (Class-no,
  Student-Ssn)Transcript (Student-Ssn, D-Code,
  C-no, Grade)

List the students (SSN) who are currently taking
exactly one class.
SELECT e.Student-Ssn FROM Enrollment e GROUP BY
e.Student-Ssn HAVING 1count()
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Query 3

• Department (D-code, D-Name,
  Chair-SSn)Course (D-code, C-no, Title,
  Units)Prereq (D-code, C-no, P-code, P-no)Class
  (Class-no, D-code, C-no, Instructor-SSn)Faculty
  (Ssn, F-Name, D-Code, Rank)Student (Ssn, S-Name,
  Major, Status)Enrollment (Class-no,
  Student-Ssn)Transcript (Student-Ssn, D-Code,
  C-no, Grade)

Give the percentage of the students (among all
students) who are currently taking courses
offered by ISE (D-code'ISE').
SELECT count(distinct e.Student-Ssn)/count(distinc
t s.Ssn) as Percent FROM Enrollment e, Class
c, Student s WHERE e.Class-noc.Class-no and
c.D-code'ISE'
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Query 4

• Department (D-code, D-Name, Chair-SSn)Course
  (D-code, C-no, Title, Units)Prereq (D-code,
  C-no, P-code, P-no)Class (Class-no, D-code,
  C-no, Instructor-SSn)Faculty (Ssn, F-Name,
  D-Code, Rank)Student (Ssn, S-Name, Major,
  Status)Enrollment (Class-no, Student-Ssn)Transcr
  ipt (Student-Ssn, D-Code, C-no, Grade)

List the faculty members (F-Name) who teach 2 or
more classes. List these faculty members by the
number of classes they teach (ascending order).
SELECT f.F-Name FROM Faculty f, Class c WHERE
f.Ssnc.Instructor-SSn GROUP BY f.Ssn,
f.F-Name HAVING count(distinct c.Class-no)gt2 ORDE
R BY count(distinct c.Class-no), F-Name
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Query 5

• Department (D-code, D-Name,
  Chair-SSn)Course (D-code, C-no, Title,
  Units)Prereq (D-code, C-no, P-code, P-no)Class
  (Class-no, D-code, C-no, Instructor-SSn)Faculty
  (Ssn, F-Name, D-Code, Rank)Student (Ssn, S-Name,
  Major, Status)Enrollment (Class-no,
  Student-Ssn)Transcript (Student-Ssn, D-Code,
  C-no, Grade)

List the students (SSN and Name) along with the
number of classes they are taking. If a student
is not taking any class, the student should also
be listed (with 0 as the number of classes he/she
is taking). The list should be ordered by the
number of classes (in an ascending order), and in
case of a tie, by the SSN of the students.
SELECT s.Ssn, s.S-Name, count(distinct
Class-no) FROM Student s, Enrollment e WHERE
s.Ssn e.Student-Ssn () GROUP BY s.Ssn,
s.S-Name ORDER BY count(distinct Class-no), Ssn
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Query 6

• Department (D-code, D-Name,
  Chair-SSn)Course (D-code, C-no, Title,
  Units)Prereq (D-code, C-no, P-code, P-no)Class
  (Class-no, D-code, C-no, Instructor-SSn)Faculty
  (Ssn, F-Name, D-Code, Rank)Student (Ssn, S-Name,
  Major, Status)Enrollment (Class-no,
  Student-Ssn)Transcript (Student-Ssn, D-Code,
  C-no, Grade)

List the faculty members (F-Name) who teach more
than twice as many classes as Professor Smith
(F-Name'Smith') is teaching. (Note that if
Professor Smith is not teaching anything, then
any professor who teaches at least one class will
satisfy the above query.)
SELECT f.F-Name FROM Faculty f, Class c WHERE
f.Ssnc.Instructor-SSn GROUP BY f.Ssn,
f.F-Name HAVING count(distinct c.Class-no) gt
(SELECT 2count(distinct Class-no)
FROM Faculty f, Class c WHERE
f.F-Name'Smith' and
f.Ssnc.Instructor-SSn)
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Query 7

• Department (D-code, D-Name,
  Chair-SSn)Course (D-code, C-no, Title,
  Units)Prereq (D-code, C-no, P-code, P-no)Class
  (Class-no, D-code, C-no, Instructor-SSn)Faculty
  (Ssn, F-Name, D-Code, Rank)Student (Ssn, S-Name,
  Major, Status)Enrollment (Class-no,
  Student-Ssn)Transcript (Student-Ssn, D-Code,
  C-no, Grade)

Find the number of departments which do not have
a chairman (Chair_ssn is 'NULL').
SELECT count(d.D-code) FROM Department d WHERE
d.Chair-SSn is NULL
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Query 8

• Department (D-code, D-Name,
  Chair-SSn)Course (D-code, C-no, Title,
  Units)Prereq (D-code, C-no, P-code, P-no)Class
  (Class-no, D-code, C-no, Instructor-SSn)Faculty
  (Ssn, F-Name, D-Code, Rank)Student (Ssn, S-Name,
  Major, Status)Enrollment (Class-no,
  Student-Ssn)Transcript (Student-Ssn, D-Code,
  C-no, Grade)

For each department (i.e., students Major), give
the number of graduate students (status'Grad')
and the number of other students (status ltgt
'Grad'). The two numbers must be shown in the
same row as the department code. Hint use
decode.
SELECT Major, sum(decode(Status, 'Grad', 1,0)) as
Grad, sum(decode(Status, 'Grad',
0,1)) as NoGrad FROM Student GROUP BY Major
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Query 9
For each department (D-code), give the highest
rank of the professors in the department along
with the number of the faculty with that highest
rank. The output contains one row for each
department. Assume FullgtAssociategtAssistant,
i.e., lexicographic order is fine. Note that some
department may not have professors in some ranks
(e.g., a department may not have full, or
associate or assistant professors).
SELECT f.D-Code as dept, f.maxrank, count(e.Ssn)
as num FROM (SELECT D-code, max(Rank) as
maxrank FROM Faculty GROUP BY
D-code) AS f, Faculty e WHERE
f.D-codee.D-code AND e.Rankf.maxrank GROUP BY
f.D-code, f.maxrank ORDER BY f.D-code