Consider the following database. Employee has a unique eid, name, age and salary
ID: 3737137 • Letter: C
Question
Consider the following database. Employee has a unique eid, name, age and salary. Department has a unique did, name and an employee as a manager. A department may or may not have a manager assigned. An employee can work for multiple departments. The pct_time in Works table is the percentage of time an employee works for a department. The salary of the employees is fixed and is not based on pct_time or the departments they work in.
Emp (eid: INT(11), ename: VARCHAR(50), age: INT(11), salary: FLOAT)
Works (eid: INT(11), did: INT(11), pct_time: INT(11))
Dept (did: INT(11), dname: VARCHAR(50, managerid: INT(11))
Constraints
-eid is primary key for Emp table and cannot be NULL
-did is primary key for Dept table and cannot be NULL
-eid and did are foreign keys referencing to Emp and Dept tables respectively in Works table. combination of eid and did is the primary key or works table.
-managerid is foreign key referencing to Emp table and can have NULL values if the department doesn’t have a manager assigned.
Sample values.
Emp
eid
ename
age
salary
1001
Georgi Facello
34
45000.00
1002
Parto Bamford
28
34000.00
1003
Christian Koblick
42
55000.00
1004
Saniya Peac
24
28000.00
Dept
did
dname
managerid
d001
Marketing
1003
d002
Finance
1002
d003
Information Security
NULL
Works
eid
did
pct_time
1001
d001
60
1002
d002
100
1003
d001
100
1001
d003
40
1004
d003
50
1004
d002
50
1.Select the employee name and no of departments each employee works for.
2. Select the employee names with salary greater than average salary of all employees,
eid
ename
age
salary
1001
Georgi Facello
34
45000.00
1002
Parto Bamford
28
34000.00
1003
Christian Koblick
42
55000.00
1004
Saniya Peac
24
28000.00
Explanation / Answer
A) select ename , count(*)
from Emp,Works
where Emp.eid=Works.eid
group by ename
B) select ename
from Emp
where salary > ( select avg(salary) from Emp)