SQL EXERCISE
Retrieve the required information using SQL language.
- How many copies of the book titled The Lost Tribe are owned by the library branch whose name is "Sharpstown"?
- How many copies of the book titled The Lost Tribe are owned by each library branch?
- Retrieve the names of all borrowers who do not have any books checked out .
- For each book that is loaned out from the "Sharpstown" branch and whose DueDate is today, retrieve the book title, the borrower's name, and the borrower's address.
- For each library branch, retrieve the branch name and the total number of books loaned out from that branch.
- Retrieve the names, addresses, and number of books checked out for all borrowers who have more than five books checked out.
- For each book authored (or co-authored) by "Stephen King", retrieve the title and the number of copies owned by the library branch whose name is "Central"
Part II Give a database schema of a company as the following picture.
- Retrieve the names of employees in department 5 who work more than 10 hours per week on the 'ProductX' project.
- For each project, list the project name and the total hours per week (by all employees) spent on that project.
- Retrieve the names of employees who work on every project.
- Retrieve the names of employees who do not work on any project.
- Find the names and addresses of employees who work on at least one project located in
Houston but whose department has no location inHouston - List the last names of department managers who have no dependents.
- Find details of those employees whose salary is > the average salary for all employees. Output salary in descending order.
- Find details of those employees whose salary is > the average salary for all employees in his/her department. Output salary in ascending order.
CREATE TABLE Manufacturers ( Code INTEGER PRIMARY KEY NOT NULL, Name TEXT NOT NULL );CREATE TABLE Products ( Code INTEGER PRIMARY KEY NOT NULL, Name TEXT NOT NULL , Price REAL NOT NULL , Manufacturer INTEGER NOT NULL CONSTRAINT fk_Manufacturers_Code REFERENCES MANUFACTURERS(Code));2. Select the name of each manufacturer along with the name and price of its most expensive product.
3. Select the name and price of all products with a price larger than or equal to $180, and sort first by price (in descending order), and then by name (in ascending order).
4. Select the names of manufacturer whose products have an average price larger than or equal to $150.
Answer:
- Should not refer to the answer before trying to write down your solutions
- The answer for each question is only one (or two) of the many other solutions
Solution 1:
SELECT bc.No_Of_Copies
FROM BOOK b, BOOK_COPIES bc, LIBRARY_BRANCH bl
WHERE b.BookId = bc.BookId AND
bc.BranchId = bl.BranchId AND
Title='The Lost Tribe' AND BranchName='Sharpstown';
Solution 2:
SELECT No_Of_Copies
FROM ((BOOK NATURAL JOIN BOOK_COPIES ) NATURAL JOIN LIBRARY_BRANCH )
WHERE Title='The Lost Tribe' AND BranchName='Sharpstown';
- SELECT BranchName, No_Of_Copies
FROM ((BOOK NATURAL JOIN BOOK_COPIES ) NATURAL JOIN LIBRARY_BRANCH )
WHERE Title='The Lost Tribe';
Solution 1:
SELECT Name
FROM BORROWER B
WHERE CardNo NOT IN (SELECT CardNo
FROM BOOK_LOANS );
Solution 2:
SELECT Name
FROM BORROWER B
WHERE NOT EXISTS (SELECT *
FROM BOOK_LOANS L
WHERE B.CardNo = L.CardNo );
- SELECT B.Title, R.Name, R.Address
FROM BOOK B, BORROWER R, BOOK_LOANS BL, LIBRARY_BRANCH LB
WHERE LB.BranchName='Sharpstown' AND LB.BranchId=BL.BranchId AND
BL.DueDate='today' AND BL.CardNo=R.CardNo AND BL.BookId=B.BookId
- SELECT L.BranchName, COUNT(*)
FROM LIBRARY_BRANCH L, BOOK_LOANS BL
WHERE BL.BranchId = L.BranchId
GROUP BY L.BranchName;
- SELECT B.Name, B.Address, COUNT(*)
FROM BORROWER B, BOOK_LOANS L
WHERE B.CardNo = L.CardNo
GROUP BY B.CardNo, B.Name, B.Address
HAVING COUNT(*) > 5;
Solution 1:
SELECT Title, No_Of_Copies
FROM (((BOOK_AUTHORS NATURAL JOIN BOOK) NATURAL JOIN BOOK_COPIES) NATURAL JOIN LIBRARY_BRANCH)WHERE Author_Name='Stephen King' AND BranchName='Central';
Solution 2: Student should write the another solution not using the natural join.
- SELECT LNAME, FNAME
FROM EMPLOYEE, WORKS_ON, PROJECT
WHERE DNO=5 AND SSN=ESSN AND PNO=PNUMBER AND PNAME='ProductX' AND HOURS>10;
- SELECT PNAME, SUM (HOURS)
FROM PROJECT, WORKS_ON
WHERE PNUMBER=PNO
GROUP BY PNAME;
Note: The Group By clause should be replaced as GROUP BY (PNUMBER, PNAME) since there may be some projects have the same name.
Solution 1:
SELECT E.LNAME, E.FNAME
FROM EMPLOYEE E
WHERE NOT EXISTS (SELECT PNUMBER
FROM PROJECT
WHERE PNUMBER NOT IN (SELECT PNO
FROM WORKS_ON
WHERE ESSN=E.SSN ) );
Solution 2:
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE NOT EXISTS (SELECT PNUMBER
FROM PROJECT
WHERE NOT EXISTS (SELECT *
FROM WORKS_ON
WHERE PNUMBER=PNO AND ESSN=SSN ) );
Solution 1:
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE SSN NOT IN ( SELECT ESSN
FROM WORKS_ON);
Solution 2:
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE NOT EXISTS ( SELECT *
FROM WORKS_ON
WHERE ESSN=SSN );
Solution 1:
SELECT LNAME, FNAME, ADDRESS
FROM EMPLOYEE
WHERE EXISTS ( SELECT *
FROM WORKS_ON W, PROJECT P, DEPT_LOCATIONS DL
WHERE W.PNO = P.PNUMBER AND
P.DNUM = DL.DNUM AND
DL.DLOCATION <> ‘Houston
Solution 2:
SELECT LNAME, FNAME, ADDRESS
FROM EMPLOYEE
WHERE EXISTS ( SELECT *
FROM WORKS_ON, PROJECT
WHERE SSN=ESSN AND PNO=PNUMBER AND PLOCATION='Houston
AND
NOT EXISTS ( SELECT *
FROM DEPT_LOCATIONS
WHERE DNO=DNUMBER AND DLOCATION='Houston
Solution 1:
SELECT E.LNAME, E.FNAME
FROM EMPLOYEE E, DEPARTMENT D
WHERE E.SSN = D.MRGSSN AND
NOT EXISTS ( SELECT DEPENDENT_NAME
FROM DEPENDENT
WHERE ESSN=E.SSN )
Solution 2:
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE EXISTS ( SELECT *
FROM DEPARTMENT
WHERE SSN=MGRSSN )
AND
NOT EXISTS ( SELECT *
FROM DEPENDENT
WHERE SSN=ESSN );
- SELECT *
FROM Employee
WHERE Salary > (SELECT AVG (Salary)
FROM Employee )
ORDER BY Salary DESC;
Solution 1:
SELECT E.*
FROM EMPLOYEE E, (SELECT DNO, AVERAGE(SALARY) AS LTB
FROM EMPLOYEE
GROUP BY DNO) AS A
WHERE E.DNO = A.DNO AND
E.SALARY > LTB;
Solution 2:
SELECT E.*
FROM EMPLOYEE E
WHERE E.SALARY > (SELECT AVERAGE(SALARY)
FROM EMPLOYEE
WHERE DNO = E.DNO);
Note: if we want to display the average of the salary we should employ the solution 1 and put the “field name” LT
B in the select clause.
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