Final Part I: Response Sheet
Question #:
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FINAL PART I
1. The information stored in the ____ is used by the DBMS for a recovery requirement triggered by a ROLLBACK statement, a program’s abnormal termination, or a system failure such as a network discrepancy or a disk crash A. data dictionaryB. transaction logC. metadataD. rollback manager
2. Which of the following is an example of lost update? A. The system assigns the last aisle seat on a flight to John whereas Peter receives a window seat, although both indicated preference for an aisle seatB. John and Peter receive each a window seat as they both requestedC. John is initially assigned the last window seat, but this seat is later on assigned to PeterD. None of the above
3. Julia has a lock on the savings account and Bill is trying to perform a conflicting action on the data: A. Bill’s transaction is granted permission to perform the desired actionB. Bill’s transaction must wait until Julia has read the savings account valueC. Bill’s transaction must wait until Julia’s lock is released D. None of the above
4. Julia has a shared lock on the savings account and Bill is trying to perform an action requesting a shared lock on the same data: A. Bill’s transaction must wait until Julia has read the savings account valueB. Bill’s transaction must wait until Julia’s lock is releasedC. Bill’s transaction is granted permission to perform the desired action D. None of the above
5. Julia has an exclusive lock on the savings account and Bill is trying to perform an action requesting a shared lock on the same data: A. Bill’s transaction must wait until Julia’s lock is released B. Bill’s transaction is granted permission to perform the desired actionC. Bill’s transaction must wait until Julia has read the savings account valueD. None of the above
6. Which of the following will cause a deadlock? A. Transaction T1 requests an exclusive lock on data A at the same time as transaction T2B. Transaction T1 places an exclusive lock on data A, then transaction T2 requests a shared lock on data AC. Transaction T1 and T2 modify data A without any lockD. Transaction T1 places an exclusive lock on data A, then requests one on data B. Simultaneously, transaction T2 places an exclusive lock on data B, then requests one on data A
7. Consider a transaction that updates each of the 100 employees’ salaries in a table (there is one row per employee) under the immediate update mode. The transaction terminates abnormally after 10 rows have been updated in the database (and the normal operations have taken place on the log). What should the recovery procedure perform to make sure the transaction completes as desired? A. RedoB. Redo followed by execution of the transaction againC. UndoD. Undo followed by execution of the transaction again
8. The _______________ query will output the table contents when the value of the character field P_CODE is alphabetically less than 1558-QW1.
A. SELECT P_CODE, P_DESCRIPT, P_QOH, P_MIN, P_PRICE
FROM PRODUCT
WHERE P_CODE <‘1558-QW1’;
B. SELECT P_CODE, P_DESCRIPT, P_QOH, P_MIN, P_PRICE
FROM PRODUCT
WHERE P_CODE = [1558-QW1]
C. SELECT P_CODE, P_DESCRIPT, P_QOH, P_MIN, P_PRICE
FROM PRODUCT
WHERE P_CODE = (1558-QW1)
D. SELECT P_CODE, P_DESCRIPT, P_QOH, P_MIN, P_PRICE
FROM PRODUCT
WHERE P_CODE = {1558-QW1}
9. The primary objective of database design is __________________.
A. To create the most efficient database possible
B. To create the fastest queries
C. To create complete, normalized, nonredundant, and fully integrated conceptual, logical, and physical database models
D. To be able to add data quickly
10. The command to join the P_DESCRIPT and P_PRICE fields from the PRODUCT table and the V_NAME, V_AREACODE, V_PHONE, and V_CONTACT fields from the VENDOR table where the value of V_CODE match is ________________.
A. SELECT P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONE
FROM PRODUCT, VENDOR
WHERE PRODUCT.V_CODE <> VENDOR.V_CODE;
B. SELECT P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONE
FROM PRODUCT, VENDOR
WHERE PRODUCT.V_CODE = VENDOR.V_CODE;
C. SELECT P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONE
FROM PRODUCT, VENDOR
WHERE PRODUCT.V_CODE <= VENDOR.V_CODE;
D. SELECT P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONE
FROM PRODUCT, VENDOR
WHERE PRODUCT.V_CODE => VENDOR.V_CODE;
11. User requirements, existing system evaluation, and logical system design are part of the __________ phase of the SDLC.
A. planning
B. analysis
C. detailed systems design
D. Implementation
12. The conceptual design steps determines end-user views, outputs, and transaction-processing requirements are _______________________.
A. Database analysis and requirements
B. Entity relationship modeling and normalization
C. Data model verification
D. Distributed database design
13. ___________________ is a disadvantage of a DDBMS.
A. Data is located near the “greatest demand” site
B. Growth facilitation
C. Danger of a single-point failure
D. Lack of standards
14. A consistent database is _____________________.
A. One in which all tables have foreign keys
B. One in which all data integrity constraints are satisfied
C. One in which all tables are normalized
D. One in which all SQL statements only update one table at a time
15. The ____ contains business data extracted from the operational database and from external data sources.
A. Data store
B. Data visualization tool
C. Data dictionary
D. ETL tool
16. The ________________ command is used to list a unique value for V_CODE, where the list will produce only a list of those values that are different from one another.
A. SELECT ONLY V_CODE
FROM PRODUCT;
B. SELECT UNIQUE V_CODE
FROM PRODUCT;
C. SELECT DIFFERENT V_CODE
FROM PRODUCT;
D. SELECT DISTINCT V_CODE
FROM PRODUCT;
17. All transactions must display _____________________.
A. atomicity, serializability, and durability
B. durability and isolation
C. serializability, durability, and isolation
D. atomicity, durability, serializability, and isolation
18. The ANSI has defined standards that govern SQL database transactions. Transaction support is provided by two SQL statements: ____________________ and ROLLBACK.
A. RETRIEVE
B. ASSIGN
C. UPDATE
D. COMMIT
19. ____________ query optimization takes place at compilation time.
A. Static
B. Dynamic
C. Automatic
D. Manual
Use the following tables for a car rental database to answer questions 20-25:
CUSTOMER In table CUSTOMER, CID is the primary key (Customer ID). RENTALS In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer’s unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned. RENTCOST RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below. RENTLENGTH RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.
20. SELECT DISTINCT CID, CNAME FROM CUSTOMERWHERE CID IN(SELECT CID FROM RENTALS WHERE MAKE IN (‘FORD’, ‘TOYOTA’))The CNAMEs shown by the execution of this query are: A. BLACKB. BLACK, JONESC. BLACK, JONES, MARTIND. BLACK, JONES, MARTIN, VERNON
21. SELECT DISTINCT CUSTOMER.CID, CNAME FROM CUSTOMER, RENTALS, RENTCOSTWHERE CUSTOMER.CID = RENTALS.CIDAND RENTALS.MAKE = RENTCOST.MAKE AND NOT EXISTS(SELECT * FROM RENTALS R, RENTCOST CWHERE R.MAKE = C.MAKEAND RENTALS.CID = R.CIDAND RENTCOST.COST <> C.COST)The meaning of this query is: A. List all customers with more than one car make rentedB. List all customers with one or more rentals for which the cost of each car make rented is the sameC. List all customers who have only rented one make D. None of the above
22. SELECT MAKE FROM RENTALS, CUSTOMERWHERE RENTALS.CID = CUSTOMER.CID AND RESID_CITY = ‘HEMET’GROUP BY MAKEHAVING COUNT (DISTINCT RENTALS.CID) =(SELECT COUNT(*) FROM CUSTOMERWHERE RESID_CITY = ‘HEMET’)The execution of this query produces the following number of rows: A. 0B. 1C. 2D. 3
23. SELECT MAKE FROM RENTALS, CUSTOMERWHERE RENTALS.CID = CUSTOMER.CID AND RESID_CITY = ‘HEMET’GROUP BY MAKEHAVING COUNT (DISTINCT RENTALS.CID) =(SELECT COUNT(*) FROM CUSTOMERWHERE RESID_CITY = ‘HEMET’)The meaning of this query is the following: A. List all makes of cars rented to customers residing in HemetB. List all makes of cars rented to at least one customer residing in HemetC. List all makes of cars rented to all customers residing in HemetD. None of the above
24. SELECT CID, CNAME FROM CUSTOMER WHERE 0 =(SELECT COUNT(*) FROM RENTALSWHERE CUSTOMER.CID = RENTALS.CID)What is the interpretation of this query? A. List the customers who do not have rentalsB. List the customers who have one rentalC. List the customers who have 0 or more rentalsD. List the customers who have 1 or more rentals
25. SELECT CNAME, DATE_OUT, RTN FROM CUSTOMER, RENTALSWHERE CUSTOMER.CID = RENTALS.CID AND BIRTHPLACE IN (‘ERIE’, ‘CARY’) AND EXISTS(SELECT * FROM RENTCOSTWHERE COST < 40 AND RENTALS.MAKE= RENTCOST.MAKE)The CNAMEs shown by the execution of this query are: A. GREEN B. GREEN, BLACK, SIMONC. SIMOND. GREEN, SIMON
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DBST 651
FINAL PART II
Name: Date:
Question 1:(20 Points)
Use the DBST651_final_Part2_q1.sql script attached to respond the following SQL problems:
Create SQL statements for the following scenarios. Your response should include SQL statement, output and any other assumptions you have made to arrive at the solution.
a. For Colorado customers compute the average amount of their orders and the number of orders placed. The result should include the customer number, customer last name, average order amount and the number of orders placed.
b. For Colorado customers compute the number of unique products ordered. If a product is purchased on multiple orders, it should be counted only one time. The result should include customer number, customer last name and the number of unique products ordered.
c. For each employee with a commission less than 0.04, compute the number of orders taken and the average number of products per order. The results should include the employee number, employee last name, number of orders taken and the average number of products per order.
d. For each Connex product compute the number of unique customers who ordered the product in Jan 2007. The results should include the product number, product name and the number of unique customers.
Question 2:
Please provide answers to problems 1-10 in Chapter 7 of Coronel book page 295-296. (10th edition page 282-284) (15 Points)
Question 3:
Please provide answers to problem 1a-1d in Chapter 10 of Coronel book page 483. (10th edition page 464) (15 Points)
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DROP TABLE ordline cascade constraints;Drop TABLE OrderTBL cascade constraints;DROP TABLE Customer cascade constraints;dROP TABLE Employee cascade constraints;DROP TABLE Product cascade constraints;
CREATE TABLE Product( ProdNo CHAR(8), ProdName VARCHAR2(50) CONSTRAINT ProdNameRequired NOT NULL, ProdMfg VARCHAR2(20) CONSTRAINT ProdMfgRequired NOT NULL, ProdQOH INTEGER, ProdPrice DECIMAL(12,2), ProdNextShipDate DATE, CONSTRAINT PKProduct PRIMARY KEY (ProdNo) );
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P0036566′,’17 inch Color Monitor’,’ColorMeg, Inc.’,12,’20-Feb-2007′,169.00);
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P0036577′,’19 inch Color Monitor’,’ColorMeg, Inc.’,10,’20-Feb-2007′,319.00);
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P1114590′,’R3000 Color Laser Printer’,’Connex’,5,’22-Jan-2007′,699.00);
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P1412138′,’10 Foot Printer Cable’,’Ethlite’,100,”,12.00);
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P1445671′,’8-Outlet Surge Protector’,’Intersafe’,33,”,14.99);
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P1556678′,’CVP Ink Jet Color Printer’,’Connex’,8, ’22-Jan-2007′,99.00);
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P3455443′,’Color Ink Jet Cartridge’,’Connex’,24,’22-Jan-2007′,38.00);
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P4200344′,’36-Bit Color Scanner’,’UV Components’,16,’29-Jan-2007′,199.99);
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P6677900′,’Black Ink Jet Cartridge’,’Connex’,44,”,25.69);
INSERT INTO product (ProdNo, ProdName, ProdMfg, ProdQOH, ProdNextShipDate, ProdPrice) VALUES (‘P9995676′,’Battery Back-up System’,’Cybercx’,12,’1-Feb-2007′,89.00);
CREATE TABLE Employee( EmpNo CHAR(8), EmpFirstName VARCHAR2(20) CONSTRAINT EmpFirstNameRequired NOT NULL, EmpLastName VARCHAR2(30) CONSTRAINT EmpLastNameRequired NOT NULL, EmpPhone CHAR(15), EmpEMail VARCHAR(50) CONSTRAINT EmpEMailRequired NOT NULL, SupEmpNo CHAR(8), EmpCommRate DECIMAL(3,3),CONSTRAINT PKEmployee PRIMARY KEY (EmpNo),CONSTRAINT UniqueEMail UNIQUE(EmpEMail),CONSTRAINT FKSupEmpNo FOREIGN KEY (SupEmpNo) REFERENCES Employee );
INSERT INTO employee (EmpNo, EmpFirstName, EmpLastName, EmpPhone, EmpEMail, SupEmpNo, EmpCommRate) VALUES (‘E9884325′,’Thomas’,’Johnson’,'(303) 556-9987′,'[email protected]’,”,0.05);
INSERT INTO employee (EmpNo, EmpFirstName, EmpLastName, EmpPhone, EmpEMail, SupEmpNo, EmpCommRate) VALUES (‘E8843211′,’Amy’,’Tang’,'(303) 556-4321′,'[email protected]’,’E9884325′,0.04);
INSERT INTO employee (EmpNo, EmpFirstName, EmpLastName, EmpPhone, EmpEMail, SupEmpNo, EmpCommRate) VALUES (‘E9345771′,’Colin’,’White’,'(303) 221-4453′,'[email protected]’,’E9884325′,0.04);
INSERT INTO employee (EmpNo, EmpFirstName, EmpLastName, EmpPhone, EmpEMail, SupEmpNo, EmpCommRate) VALUES (‘E1329594′,’Landi’,’Santos’,'(303) 789-1234′,'[email protected]’,’E8843211′,0.02);
INSERT INTO employee (EmpNo, EmpFirstName, EmpLastName, EmpPhone, EmpEMail, SupEmpNo, EmpCommRate) VALUES (‘E8544399′,’Joe’,’Jenkins’,'(303) 221-9875′,'[email protected]’,’E8843211′,0.02);
INSERT INTO employee (EmpNo, EmpFirstName, EmpLastName, EmpPhone, EmpEMail, SupEmpNo, EmpCommRate) VALUES (‘E9954302′,’Mary’,’Hill’,'(303) 556-9871′,'[email protected]’,’E8843211′,0.02);
INSERT INTO employee (EmpNo, EmpFirstName, EmpLastName, EmpPhone, EmpEMail, SupEmpNo) VALUES (‘E9973110′,’Theresa’,’Beck’,'(720) 320-2234′,'[email protected]’,’E9884325′);
CREATE TABLE Customer( CustNo CHAR(8), CustFirstName VARCHAR2(20) CONSTRAINT CustFirstNameRequired NOT NULL, CustLastName VARCHAR2(30) CONSTRAINT CustLastNameRequired NOT NULL, CustStreet VARCHAR2(50), CustCity VARCHAR2(30), CustState CHAR(2), CustZip CHAR(10), CustBal DECIMAL(12,2) DEFAULT 0, CONSTRAINT PKCustomer PRIMARY KEY (CustNo) );
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C0954327′,’Sheri’,’Gordon’,’336 Hill St.’,’Littleton’,’CO’,’80129-5543′,230.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C1010398′,’Jim’,’Glussman’,’1432 E. Ravenna’,’Denver’,’CO’,’80111-0033′,200.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C2388597′,’Beth’,’Taylor’,’2396 Rafter Rd’,’Seattle’,’WA’,’98103-1121′,500.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C3340959′,’Betty’,’Wise’,’4334 153rd NW’,’Seattle’,’WA’,’98178-3311′,200.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C3499503′,’Bob’,’Mann’,’1190 Lorraine Cir.’,’Monroe’,’WA’,’98013-1095′,0.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C8543321′,’Ron’,’Thompson’,’789 122nd St.’,’Renton’,’WA’,’98666-1289′,85.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C8574932′,’Wally’,’Jones’,’411 Webber Ave.’,’Seattle’,’WA’,’98105-1093′,1500.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C8654390′,’Candy’,’Kendall’,’456 Pine St.’,’Seattle’,’WA’,’98105-3345′,50.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C9128574′,’Jerry’,’Wyatt’,’16212 123rd Ct.’,’Denver’,’CO’,’80222-0022′,100.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C9403348′,’Mike’,’Boren’,’642 Crest Ave.’,’Englewood’,’CO’,’80113-5431′,0.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C9432910′,’Larry’,’Styles’,’9825 S. Crest Lane’,’Bellevue’,’WA’,’98104-2211′,250.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C9543029′,’Sharon’,’Johnson’,’1223 Meyer Way’,’Fife’,’WA’,’98222-1123′,856.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C9549302′,’Todd’,’Hayes’,’1400 NW 88th’,’Lynnwood’,’WA’,’98036-2244′,0.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C9857432′,’Homer’,’Wells’,’123 Main St.’,’Seattle’,’WA’,’98105-4322′,500.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C9865874′,’Mary’,’Hill’,’206 McCaffrey’,’Littleton’,’CO’,’80129-5543′,150.00);
INSERT INTO customer (CustNo, CustFirstName, CustLastName, CustStreet, CustCity, CustState, CustZip, CustBal) VALUES(‘C9943201′,’Harry’,’Sanders’,’1280 S. Hill Rd.’,’Fife’,’WA’,’98222-2258′,1000.00);
CREATE TABLE OrderTbl( OrdNo CHAR(8), OrdDate DATE CONSTRAINT OrdDateRequired NOT NULL, CustNo CHAR(8) CONSTRAINT CustNoRequired NOT NULL, EmpNo CHAR(8), OrdName VARCHAR2(50), OrdStreet VARCHAR2(50), OrdCity VARCHAR2(30), OrdState CHAR(2), OrdZip CHAR(10),CONSTRAINT PKOrderTbl PRIMARY KEY (OrdNo),CONSTRAINT FKCustNo FOREIGN KEY (CustNo) REFERENCES Customer,CONSTRAINT FKEmpNo FOREIGN KEY (EmpNo) REFERENCES Employee );
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O1116324′,’23-Jan-2007′,’C0954327′,’E8544399′,’Sheri Gordon’,’336 Hill St.’,’Littleton’,’CO’,’80129-5543′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O1231231′,’23-Jan-2007′,’C9432910′,’E9954302′,’Larry Styles’,’9825 S. Crest Lane’,’Bellevue’,’WA’,’98104-2211′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O1241518′,’10-Feb-2007′,’C9549302′,”,’Todd Hayes’,’1400 NW 88th’,’Lynnwood’,’WA’,’98036-2244′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O1455122′,’9-Jan-2007′,’C8574932′,’E9345771′,’Wally Jones’,’411 Webber Ave.’,’Seattle’,’WA’,’98105-1093′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O1579999′,’5-Jan-2007′,’C9543029′,’E8544399′,’Tom Johnson’,’1632 Ocean Dr.’,’Des Moines’,’WA’,’98222-1123′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O1615141′,’23-Jan-2007′,’C8654390′,’E8544399′,’Candy Kendall’,’456 Pine St.’,’Seattle’,’WA’,’98105-3345′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O1656777′,’11-Feb-2007′,’C8543321′,”,’Ron Thompson’,’789 122nd St.’,’Renton’,’WA’,’98666-1289′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O2233457′,’12-Jan-2007′,’C2388597′,’E9884325′,’Beth Taylor’,’2396 Rafter Rd’,’Seattle’,’WA’,’98103-1121′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O2334661′,’14-Jan-2007′,’C0954327′,’E1329594′,’Mrs. Ruth Gordon’,’233 S. 166th’,’Seattle’,’WA’,’98011′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O3252629′,’23-Jan-2007′,’C9403348′,’E9954302′,’Mike Boren’,’642 Crest Ave.’,’Englewood’,’CO’,’80113-5431′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O3331222′,’13-Jan-2007′,’C1010398′,”,’Jim Glussman’,’1432 E. Ravenna’,’Denver’,’CO’,’80111-0033′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O3377543′,’15-Jan-2007′,’C9128574′,’E8843211′,’Jerry Wyatt’,’16212 123rd Ct.’,’Denver’,’CO’,’80222-0022′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O4714645′,’11-Jan-2007′,’C2388597′,’E1329594′,’Beth Taylor’,’2396 Rafter Rd’,’Seattle’,’WA’,’98103-1121′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O5511365′,’22-Jan-2007′,’C3340959′,’E9884325′,’Betty White’,’4334 153rd NW’,’Seattle’,’WA’,’98178-3311′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O6565656′,’20-Jan-2007′,’C9865874′,’E8843211′,’Mr. Jack Sibley’,’166 E. 344th’,’Renton’,’WA’,’98006-5543′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O7847172′,’23-Jan-2007′,’C9943201′,”,’Harry Sanders’,’1280 S. Hill Rd.’,’Fife’,’WA’,’98222-2258′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O7959898′,’19-Feb-2007′,’C8543321′,’E8544399′,’Ron Thompson’,’789 122nd St.’,’Renton’,’WA’,’98666-1289′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O7989497′,’16-Jan-2007′,’C3499503′,’E9345771′,’Bob Mann’,’1190 Lorraine Cir.’,’Monroe’,’WA’,’98013-1095′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O8979495′,’23-Jan-2007′,’C9865874′,”,’HelenSibley’,’206 McCaffrey’,’Renton’,’WA’,’98006-5543′);
INSERT INTO ordertbl (OrdNo, OrdDate, CustNo, EmpNo, OrdName, OrdStreet, OrdCity, OrdState, OrdZip) VALUES (‘O9919699′,’11-Feb-2007′,’C9857432′,’E9954302′,’Homer Wells’,’123 Main St.’,’Seattle’,’WA’,’98105-4322′);
CREATE TABLE OrdLine( OrdNo CHAR(8), ProdNo CHAR(8), Qty INTEGER DEFAULT 1,CONSTRAINT PKOrdLine PRIMARY KEY (OrdNo, ProdNo), CONSTRAINT FKOrdNo FOREIGN KEY (OrdNo) REFERENCES OrderTbl ON DELETE CASCADE, CONSTRAINT FKProdNo FOREIGN KEY (ProdNo) REFERENCES Product );
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1116324′,’P1445671’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1231231′,’P0036566’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1231231′,’P1445671’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1241518′,’P0036577’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1455122′,’P4200344’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1579999′,’P1556678’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1579999′,’P6677900’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1579999′,’P9995676’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1615141′,’P0036566’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1615141′,’P1445671’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1615141′,’P4200344’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1656777′,’P1445671’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O1656777′,’P1556678’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O2233457′,’P0036577’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O2233457′,’P1445671’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O2334661′,’P0036566’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O2334661′,’P1412138’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O2334661′,’P1556678’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O3252629′,’P4200344’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O3252629′,’P9995676’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O3331222′,’P1412138’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O3331222′,’P1556678’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O3331222′,’P3455443’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O3377543′,’P1445671’,1);
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INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O4714645′,’P0036566’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O4714645′,’P9995676’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O5511365′,’P1412138’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O5511365′,’P1445671’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O5511365′,’P1556678’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O5511365′,’P3455443’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O5511365′,’P6677900’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O6565656′,’P0036566’,10);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O7847172′,’P1556678’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O7847172′,’P6677900’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O7959898′,’P1412138’,5);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O7959898′,’P1556678’,5);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O7959898′,’P3455443’,5);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O7959898′,’P6677900’,5);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O7989497′,’P1114590’,2);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O7989497′,’P1412138’,2);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O7989497′,’P1445671’,3);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O8979495′,’P1114590’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O8979495′,’P1412138’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O8979495′,’P1445671’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O9919699′,’P0036577’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O9919699′,’P1114590’,1);
INSERT INTO ordline (OrdNo, ProdNo, Qty) VALUES(‘O9919699′,’P4200344’,1);
commit;
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