Formulate the following queries using SQL:

Question

Consider the following relational schema:
Staff (staffNo, name, dept, skillCode)
Skill (skillCode, description, chargeOutRate)
Project (projectNo, startDate, endDate, budget, projectManagerStaffNo)
Booking (staffNo, projectNo, dateWorkedOn, timeWorkedOn)
where: Staff contains staff details and staffNo is the key.
Skil contains descriptions of skill codes (e.g. Programmer, Analyst,
Manager, etc.) and the charge out rate per hour for that skill; the
key is skillCode.
Project contains project details and projectNo is the key.
Booking contains details of the date and the number of hours that a member of
staff worked on a project and the key is staffNo/projectNo.

(a) (1) List all skills with a charge out rate greater than ?60 per hour, in alphabetical
order of description.
(2) List all staff with the skill description ‘Programmer’ who work in the ‘Special
Projects’ department.
(3) For all projects that were active in July 1995, list the staff name, project
number and the date and number of hours worked on the project, ordered by
staff name, within staff name by the project number and within project number
by date.
(4) How many staff have the skill ‘Programmer’?
(5) List all projects that have at least two staff booking to it.
(6) List the average charge out rate.
(7) List all staff with a charge out rate greater than the average charge out rate.

(b) Create a view of staff details giving the staff number, staff name, skill description, and department, but excluding the skill number and charge out rate.

Answer 1

(a) (1) SELECT *
FROM Skill
WHERE chargeOutRate > 60
ORDER BY description;
(2) SELECT *
FROM Staff s, Skill k
WHERE s.skillCode = k.skillCode AND
description = ‘Programmer’ AND dept = ‘Special Projects’;
(3) SELECT name, p.projectNo, dateWorkedOn, timeWorkedOn
FROM Staff s, Project p, Booking b
WHERE s.staffNo = b.staffNo AND
b.projectNo = p.projectNo AND
endDate >= DATE ‘1995-06-01’
ORDER BY name, p.projectNo, dateWorkedOn;
(4) SELECT COUNT(*)
FROM Staff s, Skill k
WHERE s.skillCode = k.skillCode AND
description = ‘Programmer’;
(5) SELECT projectNo, COUNT(*)
FROM Booking
GROUP BY projectNo
HAVING COUNT(*) >= 2;
(6) SELECT AVG(chargeOutRate)
FROM Skill;
(7) SELECT s.* FROM Staff s, Skill k
WHERE s.skillCode = k.skillCode AND
chargeOutRate > (SELECT AVG(chargeOutRate)
FROM Skill);
(b) CREATE VIEW SD (staffNo, name, dept, description)
AS SELECT staffNo, name, dept, description
FROM Staff s, Skill k
WHERE s.skillCode = k.skillCode;