1. More SQL Statements To Modify Tables

2. Library Database Books ISBN bkTitle bkPrice pubID Publishers pubID pubName pubPhone Authors auID auName auPone Books/Authors ISBN auID

3. INNER JOIN Simplified in SELECT Query SELECT bkTitle, pubName FROM Books INNER JOIN Publishers ON Books.pubID = Publishers.pubID WHERE Books.bkPrice <= 50; SELECT bkTitle, pubName FROM Books, Publishers WHERE Books.pubID = Publishers.pubID AND Books.bkPrice <= 50; is equivalent to

4. INNER JOIN Simplified in SELECT Query (2) SELECT bkTitle, auName FROM Books INNER JOIN BooksAuthors INNER JOIN Authors ON BooksAuthors.auID = Authors.auID ON Books.ISBN = BooksAuthors.ISBN WHERE Books.bkPrice <= 50; SELECT bkTitle, auName FROM Books, BooksAuthors, Authors WHERE Books.ISBN = BooksAuthors.ISBN AND (BooksAuthors.auID = Authors.auID AND (Books.bkPrice <= 50) ); is equivalent to

5. UPDATE More Than One Column Change the price of “Iliad” to $50 and pubID to 3. UPDATE Books SET bkPrice = 50, pubID = 3 WHERE title= ‘Iliad’; What will result from the following statement? UPDATE Books SET bkPrice = 50;

6. UPDATE Statement with Subquery Raise the price of all books from publisher “Big House” by 10 % Required Tables: Books (ISBN, bkTitle, bkPrice, pubID) Publishers (pubID, pubName, pubPhone)

7. UPDATE Statement with Subquery Raise the price of all books from publisher “Big House” by 10 % UPDATE Books SET bkPrice = bkPrice * 1.1 WHERE pubID in Subquery Note: subquery SELECT pubID FROM Publishers WHERE pubName = “Big House”

8. Raise the price of all books from publisher “Big House” by 10 % UPDATE Books SET bkPrice = bkPrice * 1.1 WHERE Books.pubID in (SELECT Publishers.pubID FROM Publishers WEHRE pubName = “Big House”)

9. Alternately… UPDATE Books INNER JOIN Publishers ON Books.pubID = Publishers.pubID SET Books.bkPrice = Books.bkPrice * 1.1 WHERE Publishers.pubName=”Big House";

10. UPDATE with Data from Another Table Update price column in Books table, with new prices from table NewPrices (ISBN, price). NewPrices ISBN price Books ISBN bkTitle bkPrice pubID

11. UPDATE with Data from Another Table Update price column in Books table, with new prices from table NewPrices (ISBN, price). UPDATE Books INNER JOIN NewPrices ON Books.ISBN = NewPrices.ISBN SET Books.bkPrice = NewPrices.price WHERE Books.bkPrice <> NewPrices.price

12. Previewing Update To check which records will be modified by the UPDATE statement… SELECT Books.* FROM Books INNER JOIN NewPrices ON Books.ISBN = NewPrices.ISBN WHERE BOOKS.bkPrice<>NewPrices.price;

13. Your Turn  Change author “Shakespeare’s” telephone number to “123-4568”.  Reduce the price of all books from publisher “Small House” by 10%.  Raise the price of all books by author “Shakespeare” by 10%.

14. Change author Shakespear's phone number to "123-4568“. UPDATE Authors SET auPhone = "123-4568“ WHERE auName = "Shakespeare";

15. Reduce the price of all books published by "Small House" by 10%. UPDATE Books SET Books.bkPrice = Books.bkPrice * 0.90 WHERE Books.pubID IN (SELECT Publishers.pubID FROM Publishers WHERE Publishers.pubName="Small House");

16. Difference Between “IN” and “=“ WHERE Books.pubID = (single value) But, SELECT Publishers.pubID FROM Publishers WHERE Publishers.pubName = “XXX” OR Publishers.pubName = “YYY” may return multiple values. WHERE Books.pubID IN (No1, No2) same as Books.pubID = No1 OR Books.pubID = No2

17. Raise by 10% the price of all books authored by Shakespeare. Books ISBN bkPrice BooksAuthos ISBN auID Authors auID auName SELECT BooksAuthors.ISBN FROM BooksAuthors WHERE BooksAuthors.auID = UPDATE Books SET bkPrice = bkPrice * 1.10 WHERE ISBN IN SELECT auID FROM Authors WHERE auName = “Shakespeare”

18. Raise by 10% the price of all books authored by "Shakespeare”. UPDATE Books SET Books.bkPrice = Books.bkPrice * 1.10 WHERE Books.ISBN IN (SELECT BooksAuthors.ISBN FROM BooksAuthors WHERE BooksAuthors.auID = (SELECT Authors.auID FROM Authors WHERE Authors.auName = "Shakespeare“ ) );

19. Raise the price by 5% of all books by Author Shakespeare. UPDATE Books SET bkPrice = bkPrice * 1.05 FROM Books INNER JOIN BooksAuthors INNER JOIN Authors ON BooksAuthors.auID = Authors.auID ON Books.ISBN = BooksAuthors.ISBN WHERE Authors.auName = “Shakespeare”;

20. Adding New Records INSERT INTO Statement INSERT INTO TableName VALUES (“value-1”, “value-2”, “value-3”) INSERT INTO TableName (field-2, field-3) VALUES (“value-2”, “value-3”)

21. Adding New Records INSERT INTO Statement INSERT INTO Books VALUES (“1-1111-1111-1”, “SQL Is Fun”, 1, 25.00) INSERT INTO Books (ISBN, Title) VALUES (“2-2222-2222-2”, “Born to Code”)

22. Your Turn  Add the following record to Books ISBN: 1-2345-6789-0 title: SQL for Dummies pubID: 3 price: $35.50  Add the following record to Authors auID: 14 auName: Brando

23. Inserting from Another Table Create a new table called NewAuthors with same fields as Authors. Add to NewAuthors a field called new (boolean) Add 3 new authors in NewAuthors.

24. Inserting from Another Table To copy data from NewAuthors to Authors INSERT into Authors (auName, auPhone) SELECT NewAuthors,auName, NewAuthors.auPhone FROM NewAuthors WHERE NewAuthors.new = true;

25. Removing Records DELETE Statement Syntax DELETE FROM TableName WHERE Criteria Deletes whole records (rows) Criteria chooses the rows to be deleted Can delete whole table data, but not structure

26. Delete Statement (cont) Delete the book whose ISBN is 0-12- 345678-9 DELETE FROM Books WHERE ISBN=“0-12-345678-9”

27. Delete Statement (cont) Delete all books published by “Small House” DELETE FROM Books WHERE Books.pubID = (Subquery to return pubID for “Small House” from Publishers table)

28. Delete Statement (cont) Delete all books published by “Small House” DELETE FROM Books WHERE Books.pubID = (SELECT Publishers.pubID FROM Publishers WHERE Publishers.pubName="Small House“ );

29. Your Turn 1. Delete a book titled “Born in Maui” 2. Delete all books priced $50 or above 3. Delete all books published by “Small House” 4. Delete all books written by Shakespeare

30. Delete all books priced $50 or above. DELETE FROM Books WHERE bkPrice >= 50.00;

31. Delete all books published by “Small House” DELETE FROM Books WHERE Books.pubID IN (SELECT Publishers.pubID FROM Publishres WHERE Publishers.pubName=“Small House” );

32. Delete all books written by Shakespeare DELETE FROM Books WHERE Books.ISBN IN (SELECT BooksAuthors.ISBN FROM BooksAuthors WHERE BooksAuthors.auID = (SELECT Authors.auID FROM Authors WHERE Authors.auName = "Shakespeare“ ) );

33. Used-Car Dealership DB (available from Resources/320/ on the server)

34. Your Turn (dealership DB) SELECT 1.List all cars that were made in 2000 or later. 2.Who was the salesperson that took care of Customer “John Wayne”? 3.Who was the salesperson that sold “Ford” “Edsel” in carlot at “Windward Cars”? Update 1.Reduce price of all Honda cars by 30%. 2.Reduce the price of all non-Honda cars by 50%. 3.Mark as sold all cars which are located at “Windward Cars”

35. Your Turn Delete 1.Remove from Salespeople table an employee named “Linda” “Gingle”. 2.Remove from Cars table all cars which are located at “Windward Cars”. Insert 1.Add a customer whose name is “Zeke” “Zorro” with a phone number of “808-123-4567”. 2.Add to Cars table the cars from the Newcars table.

36. Solutions SELECT 1.List all cars that were made in 2000 or later. SELECT * FROM cars WHERE year >= 2000; 2.Who was the salesperson that took care of Customer “John Wayne”? SELECT salespeople.firstName, salespeople.lastName FROM salespeople, transactions, customers WHERE salespeople.salesID = transactions.salesID AND transactions.custID = customers.custID AND customers.nameFirst = “John” AND customers.nameLast = “Wayne”;

37. Solutions SELECT 3.Who was the salesperson that sold “Ford” “Edsel” in carlot at “Windward Cars”? SELECT salespeople.firstName, salespeople.lastName FROM salespeople, transactions, cars WHERE salespeople.salesID = transactions.salesID AND transactions.carID = cars.carID AND cars.make = “Ford” AND cars.model = “Edsel”;

38. Solutions Update 1.Reduce price of all Honda cars by 30%. UPDATE cars SET carListPrice = carListPrice * 0.7 WHERE carMake = 'Honda'; 2.Reduce the price of all non-Honda cars by 50%. UPDATE Cars SET carListPrice = carListPrice * 0.5 WHERE carMake <> 'Honda';

39. Solutions Update 3.Mark as sold all cars which are located at “Windward Cars” UPDATE cars SET carSold = true WHERE cars.lotID IN (SELECT carlots.lotID FROM cars, carlots WHERE cars.lotID = carLots.lotID AND carlots.lotName = 'Windward Cars');

40. Solutions Delete 1.Remove from Salespeople table an employee named “Elmo” “Jones”. DELETE FROM salePeople WHERE salNameFirst = 'Elmo' AND salNameLast = 'Jones'; 2.Remove from Cars table all cars which are located at “Lemons Are Us”. DELETE FROM Cars WHERE Cars.lotID = (SELECT Carlots.lotID FROM Carlots WHERE Carlots.lotName = 'Lemons Are Us');

41. Solutions Insert 1.Add a customer whose name is “Zeke” “Zorro” with a phone number of “808-123-4567”. INSERT INTO Customers (nameFirst, nameLast, phoneNumber) VALUES (“Zeke”, “Zorro”, “808-123-457”); Add to Cars table the cars from the Newcars table. INSERT INTO Cars (carMake, carModel, carYear, carDoors, carListPrice,lotID) (SELECT make, model, year, doors, listPrice, lotID FROM newcars );

42. SQL Statements SELECT field1, field2 FROM Table WHERE someCondtion; SELECT Table1.field1, Table2.field1 FROM Table1, Table2 WHERE Table1.prKey = forKey AND someCondtion;

43. SQL Statements UPDATE Table SET field1 = value1, field2 = value2 WHERE someCondition UPDATE Table SET field1 = value1 WHERE (subquery)

44. SQL Statements INSERT INTO Table(field1, field2) VALUES (value1, value2);

45. SQL Statements DELETE FROM Table WHERE someCondition;

46. Functions Aggregate Functions—perform on set of data Value Functions—perform on individual values

47. Aggregate Functions COUNT – returns count of rows SELECT COUNT(auName) FROM Authors AVG – returns average value of column SELECT AVG(bkPrice) FROM BOOKS MAX MIN SUM

48. VALUE FUNCTIONS String Value Functions Numeric Value Functions Datetime Value Functions

49. String Value Functions SUBSTRING(sourceString FROM start [FOR length]) sourceString=“Ala Moana” SUBSTRING(sourceString FROM 5 FOR 3) -- “Moa” UPPER(sourceString) -- “ALA MOANA”

50. Numeric Value Function POSITION (subString IN sourceString) sourceString = “Ala Moana” POSITION (“Moana” in sourceString) -- 5 CHARACTER_LENGTH(sourceString) CHARACTER_LENGTH(“Ala Moana”) -- 9