Solutions to Midterm Examination

Therefore, we need to create triggers for all the above events.
One trigger for INSERT and one for UPDATE on work are given below.
CREATE TRIGGER no_hard_working_1
BEFORE INSERT ON works
FOR EACH ROW
WHEN new.e_id NOT IN ( SELECT manager_id FROM dept )
DECLARE total INT;
BEGIN
SELECT sum(pct_time) INTO total
FROM works w
WHERE w.e_id = new.e_id;
IF (total > 100 - new.pct_time )
THEN raise_application_error(-2000, ’working too hard’);
ELSE IF
END
CREATE TRIGGER no_hard_working_2
BEFORE UPDATE ON works
FOR EACH ROW
WHEN new.e_id NOT IN ( SELECT manager_id FROM dept )
DECLARE total INT;
BEGIN
SELECT sum(pct_time) INTO total
FROM works w
WHERE w.e_id = new.e_id;
END
IF (total > 100 - new.pct_time + old.pct_time )
THEN raise_application_error(-2000, ’working too hard’);
ELSE IF
5. Consider a relation schema with attributes R = ABCGW XY Z and the set of dependencies
F = {XZ → ZY B, Y A → CG, C → W, B → G, XZ → G}.
(a) Find a minimal cover for F.
After right-reducing, we have
XZ → B
XZ → G
XZ → Y
Y A → C
Y A → G
C → W
B → G
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