程序代写代做代考 database SQL module8
module8
Database Management – GMU. Prof. Alex Brodsky. Module 8 1
Still More on SQL
Module 8
Prof. Alex Brodsky
Database Systems
Database Management – GMU. Prof. Alex Brodsky. Module 8 2
Example Instances
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?
Database Management – GMU. Prof. Alex Brodsky. Module 8 3
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
Database Management – GMU. Prof. Alex Brodsky. Module 8 4
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.
Database Management – GMU. Prof. Alex Brodsky. Module 8 5
For each red boat, find the number of
reservations for this boat
❖ Grouping over a join of three relations.
❖ What do we get if we remove B.color=‘red’ from the
WHERE clause and add a HAVING clause with this
condition?
❖ What if we drop Sailors and the condition involving
S.sid?
SELECT B.bid, COUNT (*) AS scount
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’
GROUP BY B.bid
Database Management – GMU. Prof. Alex Brodsky. Module 8 6
Find those ratings for which the average
age is the minimum over all ratings
❖ Aggregate operations cannot be nested! WRONG:
SELECT S.rating
FROM Sailors S
WHERE S.age = (SELECT MIN (AVG (S2.age)) FROM Sailors S2)
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):
Database Management – GMU. Prof. Alex Brodsky. Module 8 7
Continue from previous
However, this should work on Oracle 8:
SELECT S.rating
FROM Sailors S
Group by S.rating
Having AVG(S.age) = (SELECT MIN (AVG (S2.age))
FROM Sailors S2
Group by rating);
Database Management – GMU. Prof. Alex Brodsky. Module 8 8
Conclusion so far
❖ Post processing on the result of queries is
supported.
❖ Aggregation is the most complex “post
processing”
– “Group by” clause partition the results into
groups
– “Having” clause puts condition on groups (just
like Where clause on tuples).
Database Management – GMU. Prof. Alex Brodsky. Module 8 9
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.
Database Management – GMU. Prof. Alex Brodsky. Module 8 10
Deal with the null value
❖ Special operators needed to check if value
is/is not null.
– “is null” always true or false (never unknown)
– “is not null”
❖ Is rating>8 true or false when rating is equal to
null?
– Actually, it’s unknown.
– Three-valued logic
Database Management – GMU. Prof. Alex Brodsky. Module 8 11
Three valued logic
AND False True Unknown
False False False False
True False True Unknown
Unknown False Unknown Unknown
OR False True Unknown
False False True Unknown
True True True True
Unknown Unknown True Unknown
NOT
False True
True False
Unknown Unknown
Database Management – GMU. Prof. Alex Brodsky. Module 8 12
Other issues with the null value
❖ WHERE and Having clause eliminates rows that
don’t evaluate to true (i.e., rows evaluate to
false or unknown.
❖ Aggregate functions ignore nulls (except
count(*))
❖ Distinct treats all nulls as the same
Database Management – GMU. Prof. Alex Brodsky. Module 8 13
Outer joins
(left outer-join)
= sid sname rating age bid day
22 dustin 7 45.0 101 10/10/96
31 lubber 8 55.5 Null Null
58 rusty 10 35.0 103 11/12/96
Database Management – GMU. Prof. Alex Brodsky. Module 8 14
In Oracle Select *
From Sailor S, Reserve R
Where S.sid = R.sid (+);
How about:
Select S.sid, count(R.bid)
From Sailor S, Reserve R
Where S.sid = R.sid (+)
Group by S.sid;
OR
Select S.sid, count(*)
From Sailor S, Reserve R
Where S.sid = R.sid (+)
Group by S.sid;
Database Management – GMU. Prof. Alex Brodsky. Module 8 15
More outer joins
❖ Left outer join
+ sign on the right in Oracle:
Select * from R, S where R.id=S.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
Database Management – GMU. Prof. Alex Brodsky. Module 8 16
More on value functions
❖ Values can be transformed before aggregated:
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 Non_NonIS_Stu
From student ;
❖ Calculating GPA from letter grades (HW4)?
Database Management – GMU. Prof. Alex Brodsky. Module 8 17
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)
Transcript (Student-Ssn, D-Code, C-no, Grade)
Database Management – GMU. Prof. Alex Brodsky. Module 8 18
Query 1
List the classes (class_no) taken by students
whose names start with ‘T’.
select distinct class_no
from enrollment e, student s
where e.student_ssn = s.ssn and s.s_name like ‘T%’;
Database Management – GMU. Prof. Alex Brodsky. Module 8 19
Query 2
List the students (SSN) who are currently
taking exactly one class.
select distinct student_ssn
from enrollment
group by student_ssn
having 1=count(*);
Database Management – GMU. Prof. Alex Brodsky. Module 8 20
Query 3
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(distinct s.ssn) as Percent
from enrollment e, class c, student s
where e.class_no=c.class_no and d_code=’ISE’;
Database Management – GMU. Prof. Alex Brodsky. Module 8 21
Query 4
List the faculty members (F_Name) who teach 2
or more classes. List these faculty members by
the number of classes they teach.
select f.f_name
from faculty f, class c
where f.ssn=c.instructor_ssn
group by f.ssn, f_name
having count(distinct c.class_no)>=2
order by count(distinct c.class_no), f_name;
Database Management – GMU. Prof. Alex Brodsky. Module 8 22
Query 5
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 ssn, s_name, count(distinct class_no)
from student s, enrollment e
where s.ssn = e.student_ssn (+)
group by ssn, s_name
order by count(distinct class_no), ssn;
Database Management – GMU. Prof. Alex Brodsky. Module 8 23
Query 6
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_name
from faculty f, class c
where f.ssn=c.instructor_ssn
group by f.ssn, f_name
having count(distinct c.class_no) > (select 2*count(class_no)
from faculty f, class c
where f.f_name=’Smith’
and f.ssn=c.instructor_ssn);
Database Management – GMU. Prof. Alex Brodsky. Module 8 24
Query 7
Find the number of departments which do
not have a chairman (Chair_ssn is ‘NULL’).
select count(d_code)
from department
where chair_ssn is NULL;
Database Management – GMU. Prof. Alex Brodsky. Module 8 25
Query 8
For each department (d_code), give the number of graduate
students (status=’Grad’) and the number of other students
(status <> ‘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;
Database Management – GMU. Prof. Alex Brodsky. Module 8 26
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 Full>Associate>Assistant, 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) f,
faculty e
where f.d_code=e.d_code and e.rank=f.maxrank
group by f.d_code, f.maxrank
order by f.d_code
Database Management – GMU. Prof. Alex Brodsky. Module 8 27
Conclusion
❖ Done with most SQL stuff
❖ More on programming with SQL
– In lecture week 14 (advanced topics)
❖ More practice!
