Consider the following database. Employee has a unique eid, name, age and salary
ID: 3736023 • 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 deparments 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 department names with no manager assigned yet.
2. Select the employee names who work for more than 1 department.
3. Select the employee names and their manager names (keep in mind, one employee works for multiple departments so they will have multiple managers).
4. Select the name and salary of the employee with the second highest salary.
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
1. SELECT DNAME FROM DEPT WHERE MANAGERID IS NULL
The above query returns the department name where managerid has a null value in that row in the DEPT table
2. SELECT EID, count(*) as c FROM WORKS GROUP BY EID having c > 1
The above query groups the rows together that are having same EID. Thus for each EID count(*) in the quert returns the number of rows. By checking count for greater than 1, allows the query to filter only the employees who work for more than one department.
3. SELECT E.ENAME, M.ENAME D.MANAGERID FROM EMP AS E
LEFT JOIN WORKS AS W ON E.EID = W.EID
LEFT JOIN DEPT AS D W.DIP = D.DID
LEFT JOIN EMP AS M D.MANAGERID = M.EID
4. SELECT ENAME, SALARY FROM EMP AS E WHERE 2=(SELECT COUNT(DISTINCT SALARY) FROM EMP AS P WHERE E.SALARY <= P.SALARY)
The inner query returns the number of rows fetched whose salary is less than are equal to outer query fetched row.
If the inner query returns 2, then the outer query fetched row will have the second largest salary.
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