1 Exercise 1 Given the following tables: employee (person_name, street, city) works (person_name, company_name, salary) company (company_name, city)

1

Exercise 1

Given the following tables:

employee (person_name, street, city)

works (person_name, company_name, salary)

company (company_name, city)

manages (person_name, manager_name)

2

Find the name of employees who earn more than $10,000 and live in Hong Kong.

select w.person_name from works as w where w.salary > 10000 and w.person_name in

(select e.person_name from employee as e where e.city = “Hong Kong”)

Alternative solutions

select w.person_name from works as w, employee as e where w.salary > 10000 and e.city = “Hong Kong”

and w.person_name = e.person_name

employee (person_name, street, city)works (person_name, company_name, alary)company (company_name, city)manages (person_name, manager_name)

3

Find the name of the employees who are not managers.

(select person_name

from employee)

except

(select manager_name

from manages)


4

Alternative solutions

select person_name

from employee

where not exists

(select *

from manages

where employee.person_name = manages.manager_name)

select person_name

from employee

where person_name not in

(select manager_name

from manages)


5

RA SQL Given the following tables:

employee (person_name, street, city)

works (person_name, company_name, salary)

company (company_name, city)

manages (person_name, manager_name)

Find the names of all persons who work for “First Bank Corporation” and live in the city where the company is located.

. _ . .

person_name company_name _ "First Bank Corporation"

(

( ( )))

employee person name employee city company city

company nameemployee works company

6

Find the names of all persons who work for “First Bank Corporation” and live in the city where the company is located.

select E.person_name

from employee as E, works as W, company as C

where E.person_name = W.person_name

and W.company_name = C.company_name

and C.company_name = “First Bank Corporation”

and E.city = C.city


7

Find the names, cities of employees who work for exactly ONE company

select e.person_name, e.city

from employee as e

where unique

(select *

from works as w

where w.person_name = e.person_name)

employee (person_name, street, city)works (person_name, company_name, salary)company (company_name, city)manages (person_naame, manager_name)

8

Find the names of all employees who earn more than SOME employee of Small Bank Corporation.

select w1.person_name

from works as w1

where w1.salary > some

(select w2.salary

from works as w2

where w2.company_name = “Small Bank Corporation”)

Alternative solution

select w1.person_name

from works as w1

where exists

(select *

from works as w2

where w2.company_name = “Small Bank Corporation” and w1.salary > w2.salary)


9

Find the company located in Hong Kong that has the largest number of employees

select temp.company_name

from (select w.company_name, count(distinct w.person_name) as cnt

from works as w, company as c

where w.company_name = c.company_name

and c.city = “Hong Kong”

group by w.company_name) as temp

where temp.cnt in (select max (cnt)

from temp)


Can also be ‘=’

10

Find all companies located in Hong Kong and have total payroll less than 100,000

select company.company_namefrom works, companywhere works.company_name = company.company_name

and company.city = “Hong Kong”group by company.company_namehaving sum(works.salary) < 100,000


11

Question: Consider the following schemas.CUST (cust-id, name), andWITHDRAW (w-id, cust-id, acc-id, date, amount)

Write an SQL query to retrieve all the names of the customers who have withdrawn more than 1k dollars in a single withdrawal. If a customer made several such withdrawals, her/his name should be reported only once.

12

Answer: Consider the following schemas.CUST (cust-id, name), andWITHDRAW (w-id, cust-id, acc-id, date, amount)

Write an SQL query to retrieve all the names of the customers who have withdrawn more than 1k dollars in a single withdrawal. If a customer made several such withdrawals, her/his name should be reported only once.

select distinct namefrom CUST as T1, WITHDRAW as T2where T1.cust-id = T2.cust-id and T2.amount > 1k

13


Sometimes there may be a “shared” account, namely, an account with multiple owners.

Write an SQL query to return the acc-id of all the shared accounts.You may assume that all the owners of a shared account have made withdrawals from the account.

14


Sometimes there may be a “shared” account, namely, an account (acc-id) with multiple owners (cust-id).

Write an SQL query to return the acc-id of all the shared accounts.You may assume that all the owners of a shared account have made withdrawals from the account.

select T1.acc-idfrom WITHDRAW as T1, WITHDRAW as T2where T1.cust-id <> T2.cust-id and T1.acc-id = T2.acc-id

15


Let us use the name “interesting account” to refer to the account from which the withdrawal with smallest amount was made.

Retrieve the acc-id of accounts from which withdrawals have been made, except the interesting account.

16


Let us use the name “interesting account” to refer to the account from which the withdrawal with smallest amount was made.

Retrieve the acc-id of accounts from which withdrawals have been made, except the interesting account.

select distinct acc-id from WITHDRAW where acc-id not in

(select acc-id from WITHDRAW where amount = (select min (amount) from WITHDRAW))

17

Consider table: deposit (dep-id, acc-id, cust-id, amount). We want to retrieve the cust-id of the customers who deposited into two accounts with acc-id ‘A1’ and ‘A2’, respectively.

Write an SQL query with intersect.

(select distinct cust-id from deposit where acc-id = ‘A1’)intersect(select distinct cust-id from deposit where acc-id = ‘A2’)

Write a nested SQL query without intersect.

select distinct cust-id from deposit where acc-id = ‘A1’ and cust-id in (select cust-id from deposit where acc-id = ‘A2’)

dep-id acc-id cust-id amount

070940 A1 1 2K

070941 A1 1 1K

070943 A2 1 1K

070945 A2 2 3K

070959 A3 3 2K

080341 A3 2 5K

18

Again consider table: deposit (dep-id, acc-id, cust-id, amount). We want to retrieve the cust-id of the customers who deposited into two accounts with acc-id ‘A1’ and ‘A2’, respectively.

Write an SQL query that contains only one select .

select distinct T1.cust-idfrom deposit as T1, deposit as T2where T1.cust-id = T2.cust-id and T1.acc-id = ‘A1’ and T2.acc-id = ‘A2’


070940 A1 1 2K

070941 A1 1 1K

070943 A2 1 1K

070945 A2 2 3K

070959 A3 3 2K

080341 A3 2 5K

19

Find the ids of the accounts which have been deposited into by more than one customer.

Write a SQL query without group by:

select D1.acc-idfrom deposit as D1, deposit as D2where D1.cust-id <> D2.cust-id and D1.acc-id = D2.acc-id


070940

A1 1 2K

070941

A1 1 1K

070943

A2 1 1K

070945

A2 2 3K

070959

A3 3 2K

080341

A3 2 5K

20

Find the ids of the accounts which have been deposited into by more than one customer.

Write a SQL query with group by:

select acc-idfrom deposit group by acc-idhaving count (distinct cust-id) >= 2

If there is no distinct here, A1 will also be displayed.


070940

A1 1 2K

070941

A1 1 1K

070943

A2 1 1K

070945

A2 2 3K

070959

A3 3 2K

080341

A3 2 5K

21

Again consider table: deposit (dep-id, acc-id, cust-id, amount). We want to retrieve the cust-id of the customers who deposited into the account with acc-id = ‘A1’ or ‘A2’ but not both.

Write an SQL query that contains only one SELECT.

select cust-id from deposit where acc-id = ‘A1’ or acc-id = ‘A2’group by cust-idhaving count (distinct acc-id) = 1

Indispensable!

dep_id acc-id cust-id amount

070940 A1 1 2K

070941 A1 1 1K

070943 A2 1 1K

070945 A2 2 3K

070959 A3 3 2K

080341 A3 2 5K

22

deposit (dep-id, acc-id, cust-id, amount). Retrieve the cust-id of the customer who deposited the largest number of times.

select cust-id from deposit group by cust-idhaving count (*) >= all (select count (*) from deposit

group by cust-id)


070940 A1 1 2K

070941 A1 1 1K

070943 A2 1 1K

070945 A2 2 3K

070959 A3 3 2K

080341 A3 2 5K

23

Question: As with natural join, outer joins are not compulsory operators. That is, we can implement an outer join using “conventional” SQL.

Let us verify this for left outer join.

CS-PROF (prof-id, name)

SUPERVISION (prof-id, stu-id)

Write an alternative query that returns the same information as

select prof-id, name, stu-idfrom CS-PROF left outer join SUPERVISION on CS-PROF.prof-id = SUPERVISION.prof-id

24

Answer: CS-PROF (prof-id, name) SUPERVISION (prof-id, stu-id)

select prof-id, name, stu-idfrom CS-PROF left outer join SUPERVISION on CS-PROF.prof-id = SUPERVISION.prof-id

(select T1.prof-id, name, stu-id from CS-PROF as T1, SUPERVISION as T2 where T1.prof-id = T2.prof-id)union(select prof-id, name, NULL from CS-PROF as T1 where not exists

(select * from SUPERVISION as T2 where T1.prof-id = T2.prof-id))

We can see that left outer join simplifies the query significantly.

25

Question: Consider MARKS(stu-id, course-id, score)

Write a query to retrieve the stu-id of every student who scored at least 80 in all the courses s/he took, but scored less than 90 in at least one course.

Your query should not contain more than 2 select.

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Answer: Consider MARKS(stu-id, course-id, score).


(select stu-id from MARKS where score < 90) except(select stu-id from MARKS where score < 80)

27

Answer: Consider MARKS (stu-id, course-id, score).


select stu-id from MARKSgroup by stu-idhaving min (score) >= 80 and min (score) < 90

28

What will happen if some of values of score are NULL in table ?

All aggregate operations except count(*) ignore tuples with null values on the aggregated attributes.

Documents

1 Exercise 1 Given the following tables: employee (person_name, street, city) works (person_name, company_name, salary) company (company_name, city)