1. Query Examples in Relational Algebra and SQL
Consider the relation schemas as follows.
works(person name, company name, salary);
lives(person name, street, city);
located in(company name, city);
managers(person name, manager name);
where manager name refers to person name.
a Find the names of the persons who work for company ’FBC’ (company name=’FBC’).
Relational algebra:
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SQL:
Select person_name
From works
Where company_name = ’FBC’
b List the names of the persons who work for company ’FBC’ along with the cities they live in.
Relational algebra:
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cXd ¡Uef§g¥h¢¡pirqs I¢P¥Q ¡¢£SR¢TU VRXW¢Yut$vs¦¢g¥h¢¡¢£
¢¡¢£¥¤§¦©¨wa!$# #%$x 01y$€@2S' cXd ¡Uef§g¥h¢¡pirqXH
SQL:
Select lives.person_name, city
From works, lives
Where company_name = ’FBC’ and
works.person_name = lives.person_name
c Find the persons who work for company ’FBC’ with a salary of more than 10000. List the names of these
persons along with the streets and cities where they live.
Relational algebra:
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¢¡¢£¥¤§¦©¨w I¢P¥Q ¡©ˆ‰g¥¨ d W‰g¥¢‘ P ¦—t$vs¦¢g¥h¢¡¢£
SQL:
Select lives.person_name, stree, city
From lives, works
Where lives.person_name = works.person_name and salary 10000
and works.company_name = ’FBC’
1

2. d Find the names of the persons who live and work in the same city.
Relational algebra:
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SQL:
Select person_name
From works, lives, locates_in
Where works.person_name = lives.person_name and
works.company_name = located_in.company_name and
located_in.city = lives.city
e Find the names of the persons who live in the same city and on the same street as their managers.
Relational algebra:
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¢¡¢£¥¤§¦©¨wa¡'!¦ !$# #%$¥'$‘ P¥Q ¡V‘U¨V ¢¡V¡U¨Y§g¥¨Ve©H
SQL:
Select e.person_name
From lives e, lives m, managers
Where e.person_name = managers.person_name and
m.person_name = managers.manager_name and
e.street = m.street and e.city = m.city
f Find the names of the persons who do not work for company ’FBC’.
Relational algebra:
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ˆpTU V‡  ¡U §£UT©q‚£’a!$# #%$S'†…pTU V‡§£SH
¢¡¢£¥¤§¦©¨wˆpTU V‡  ¡U §£UT©q‚£$2  ¡U §£UT©qR¢TU VRXW¢Y
SQL:
Select person_name
From works
Where person_name not in (Select person_name
From works
Where company_name = ’FBC’)
g Find the persons whose salaries are more than the salary of everybody who work for company ’SBC’.
Relational algebra:
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2

3. SQL:
Select person_name
From works
Where salary all (Select salary
From works
Where company_name = ’SBC’)
h Find the names of the companies that is located in every city where company ’SBC’ is located in.
Relational algebra:
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i Q‚P ‚grq ¦V¦UYT Q¡ pP qVe ¨ ¦X¦X‘SW¢YVfVT¡ P ¨‚gST©q§£ a‚0!$1%$#32 #%$S'£¢)C¥¤§¦©¨ H ¨ ¦V¦X‘©W¢YVY§g¥¨‚g©¡¢£
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Relational algebra (another solution):
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¢¡¢£¥¤§¦©¨w ¦XT¡ P ¨¢¡£¢ grq ¦V¦X‘©W¢YVY§g¥¨‚g©¡¢£
SQL:
Select company_name
From located_in t
Where (Select city
From located_in s
Where t.company_name = s.company_name)
contains (Select city
From located_in s1
Where s1.company_name = ’SBC’)
SQL (another solution):
Select company_name
From located_in t
Where not exists
(Select *
From located_in s
Where s.company_name = ’SBC’ and
s.city not in
(Select city
From located_in l
Where l.company_name = t.company_name))
3