Section I : Section II : Question 1. Question 2. Question 3.

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1 Computer Science, Midterm Examiner: Ritu Chaturvedi Date: Oct. 27 th, 2011 Student Name: Student Number: INSTRUCTIONS (Please Read Carefully) Examination Period is 1 hour and 15 minutes Answer all questions. Write your answers in the spaces provided in the question paper. This is closed book and closed notes test. Total Marks =60. Total number of sections = 2 Total number of pages =14 Please read questions carefully! Misinterpreting a question intentionally or unintentionally results in getting a ZERO for that question. Good Luck!!! Section I : Section II : Question 1. Question 2. Question 3. Page 1 of 16

2 SECTION I : Multiple-Choice Questions : Each question has 1 answer, unless otherwise specified. You must select the best possible answer. (13 X marks for question 14 = 29 marks) 1. You are defining a cursor in your PL/SQL block. Which line in the following statement will produce an error? A. cursor action_cursor is B. select name, rate, action C. into action_record D. from action_table; E. There are no errors in this statement 2. Which of the following integrity constraints automatically create an index when defined? (Choose two) A. Foreign keys B. UNIQUE constraints C. NOT NULL constraints D. Primary keys 3. Omitting the where clause from a delete statement has which of the following effects? A. The delete statement will fail because there are no records to delete. B. The delete statement will prompt the user to enter criteria for the deletion. C. The delete statement will fail because of syntax error. D. The delete statement will remove all records from the table. 4. Which of the following is TRUE / FALSE? 1 Host variables are declared anywhere in the program 2 Host variables are declared in the DECLARE section A. Only 1 is TRUE B. Only 2 is TRUE C. Both 1 & 2are TRUE D. Both are FALSE Page 2 of 16

3 5. Martha writes the following query to display information from the Ordinates table : SELECT Ord_no, DISTINCT Ord_name, Salary FROM Ordinates; A. The query will display all values for Ord_no, unique values for Ord_name and unique values for Salary. B. The query will display only those rows in Ordinates that have a unique Ord_name. C. The query will display all values for Ord_No, unique values for Ord_name and all values of Salary. D. The query will produce an error. E. none of the above 6. In this PL/SQL statement, which of the following lines will produce an error? A. CURSOR capital IS B. SELECT city, state, zip C. INTO my_city D. FROM CITIES; 7. This is an example of what type of cursor? DECLARE v_dname department.dname%type; BEGIN SELECT dname INTO v_dname FROM department WHERE dno = 60; END; A. There is no cursor used in the code B. SELECT C. Explicit D. Implicit 8. After executing an update statement, the developer codes a PL/SQL block to perform an operation based on SQL%ROWCOUNT. What data is returned by the SQL%ROWCOUNT operation? A. A Boolean value representing the success or failure of the update B. A numeric value representing the number of rows updated C. A VARCHAR2 value identifying the name of the table updated D. A LONG value containing all data from the table Page 3 of 16

4 9. Which line in the code below returns an error? The structure of table EMP is : EMP(NAME VARCHAR2(20), salary NUMBER(5)). 1 SELECT name, salary, SUM(salary) total 2 FROM emp 3 GROUP BY name 4 HAVING SUM(salary) > ORDER BY name; A. Line 1 B. Line 2 C. Line 4 D. Line PL/SQL blocks MAY NOT contain what? A. SELECT statements with subqueries. B. Data definition language statements such as CREATE TABLE. C. Data manipulation language statements such as UPDATE. D. Arithmetic statements such as v_variable := 24; 11. What is the name of Oracle's implicit cursor? A. CUR B. SQL C. ORA D. IMP_CUR E. You cannot reference an implicit cursor 12. The DBA created a public synonym called S_FOR_ALL for table S and has granted you access to it. You, as a user, have created a table called S in your account as well. What is the result of executing this query : SELECT * FROM S; A. You obtain results from the table S of the DBA. B. You obtain results from table S of your schema. C. You get an error message as you have access to 2 tables of the same name. D. You obtain a Cartesian product of the tables S of your schema and S of the DBA s schema. e. None of the above Page 4 of 16

5 13. VARIABLE VAR_X NUMBER DECLARE VAR_Y NUMBER (2) := &&VAR_X2; BEGIN :VAR_X := VAR_Y + 3; END; A. VAR_X is a substitution variable whereas VAR_X2 is a bind variable B. VAR_X is a bind variable whereas VAR_X2 is a substitution variable C. VAR_Y is a substitution variable whereas VAR_X2 is a bind variable D. VAR_X is a substitution variable whereas VAR_X is a bind variable 14. A table called TEST has 1000 rows and has the following structure: ID NAME BDATE NUMBER(3) PRIMARY KEY VARCHAR2(30) DATE The number of blocks required to store these 1000 rows is 50. The DBA decides to create a secondary index IND_NAME on the field NAME. How many entries will the primary index and the secondary index IND_NAME have? A. Primary index will have 1000 ; Secondary index will have 50 B. Primary index will have 50 ; Secondary index will have 1000 C. Both Primary and Secondary index will have the same = 1000 D. Both Primary and Secondary index will have the same = 50 Page 5 of 16

6 Section II ( = 31 marks) 1. (6 marks). In the PL/SQL block shown below, determine the values for each of the following cases, after the block is run. DECLARE customer credit_rating BEGIN DECLARE BEGIN END; VARCHAR2(50) := 'APPLE'; VARCHAR2(50) := 'EXCELLENT'; customer NUMBER(7) := 201; name VARCHAR2(25) := 'SAMSUNG'; credit_rating :='GOOD'; END; / a. The value of customer in the nested block is: 201 b. The value of name in the nested block is: SAMSUNG c. The value of credit_rating in the nested block is: GOOD d. The value of customer in the main block is: APPLE e. The value of name in the main block is: N / A f. The value of credit_rating in the main block is: GOOD Page 6 of 16

7 2. (4 + 4 = 8 marks) Use the Functional Dependency diagram given below in Figure 2 to answer Question 2a and b. Figure 2: C1 C2 C3 C4 Figure 2 represents all functional dependencies in relation R with attributes C1, C2, C3, C4 and C5. C5 a. (4 marks) Identify the following components in the FD diagram given below. Write None if there is none. Primary Key: (C1, C2) Non-trivial Functional Dependencies: C1 -> C3, C3 -> C4 Non-Full Functional dependencies: C1 -> C3 Transitive dependencies: C3 -> C4 b. (4 marks) Prove that relation R is not in 3NF. R is not in 3NF because of 2 reasons : 1. R is not in 2NF as we have a non-full functional dependency. 2. R has a transitive dependency. Page 7 of 16

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10 3.(17 marks + 1 bonus) Tables S and SP, You are also given a table called supplier_summary with the following fields. Currently, there are no records in this table. seq_num NUMBER PRIMARY KEY supplier_name VARCHAR2(20) total_number_of_parts NUMBER(3) total_quantity NUMBER (5) minimum_quantity NUMBER (5) todays_date DATE Page 10 of 16

11 Write a PL/SQL block using explicit cursors to populate some statistics into the supplier_summary table using the tables S and SP given above. - Column supplier_name is the supplier name. - Column total_number_of_parts stores the total number of parts supplied by each supplier - Columns total_quantity and minimum_quantity store the total quantity and minimum quantity respectively for each supplier. - To populate the column seq_num, use a sequence called midterm_seq that starts with 1 and increments by 1 (Assume that the sequence is already created and stored in the Database). - Todays_date column stores today s date (this will fetch you a bonus mark). In addition to populating the table supplier_summary, your program must display messages on the screen as shown below : Data 1 cannot be inserted - too few parts Data 2 cannot be inserted - too few parts Data 3 cannot be inserted - too few parts Data 4 Successful insertion Data 5 cannot be inserted - too few parts PL/SQL procedure successfully completed. rituch@cs01> After successful execution of your program on the tables S and SP as shown on Page 8, supplier_summary table must have the following rows: (Adams is the only supplier supplier who supplies more than 3 parts(6 parts actually) the rest of them supply 1 or 2 and hence are not inserted in the table). Page 11 of 16

12 Solution: DECLARE CURSOR lab2_2 IS SELECT sname, COUNT(*) AS total_parts, SUM(qty) AS total_qty, MIN(qty) as min_qty FROM s, sp WHERE s.sno=sp.sno GROUP BY s.sname; rec_cursor_lab2_2 lab2_2%rowtype; i NUMBER := 1; BEGIN OPEN lab2_2; LOOP FETCH lab2_2 INTO rec_cursor_lab2_2 ; EXIT WHEN lab2_2%notfound; IF (rec_cursor_lab2_2.total_parts <= 2) THEN DBMS_OUTPUT.PUT_LINE ('Data ' i ' cannot be inserted - too few parts '); ELSE DBMS_OUTPUT.PUT_LINE ('Data ' i 'Successful insertion'); INSERT INTO supplier_summary VALUES(midterm_seq.nextval, rec_cursor_lab2_2.sname, rec_cursor_lab2_2.total_parts, rec_cursor_lab2_2.total_qty, rec_cursor_lab2_2.min_qty, SYSDATE); END IF; i := i+ 1; END LOOP; END; / Note : There can be more than solutions to this question only 1 is provided. Page 12 of 16

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15 Supplement Sheet : Syntax PL/SQL Block Structure DECLARE (optional) Variables, cursors, user-defined exceptions BEGIN (mandatory) EXCEPTION (optional) END; (mandatory) Declaring and Initializing PL/SQL Variables identifier [CONSTANT] datatype [NOT NULL] [:= DEFAULT expr]; Declaring Variables with the %TYPE Attribute identifier table.column_name%type; IF Statements IF condition THEN statements; [ELSIF condition THEN statements;] [ELSE statements;] END IF; CASE Expressions CASE selector WHEN expression1 THEN result1 WHEN expression2 THEN result2... WHEN expressionn THEN resultn [ELSE resultn+1] END; Loops : LOOP statement1;... EXIT [WHEN condition]; END LOOP; WHILE Loops WHILE condition LOOP statement1;... END; FOR Loops FOR counter IN [REVERSE] lower_bound..upper_bound LOOP statement1; statement2;... END LOOP; Page 15 of 16

16 Declaring a Cursor CURSOR cursor_name IS select_statement; Opening the Cursor OPEN cursor_name; Fetching Data from the Cursor FETCH cursor_name INTO variables..; Closing the Cursor CLOSE cursor_name; Cursor FOR Loops FOR record_name IN cursor_name LOOP statement1; statement2;... END LOOP; Cursors with Parameters CURSOR cursor_name [(parameter_name datatype,...)] IS select_statement; Exceptions EXCEPTION WHEN exception1 [OR exception2...] THEN statement1; statement2; RAISE_APPLICATION_ERROR Procedure Syntax: raise_application_error (error_number, message[, {TRUE FALSE}]); PRAGMA EXCEPTION_INIT(message, error_number) Some Oracle Errors : CURSOR_ALREADY_OPEN NO_DATA_FOUND TOO_MANY_ROWS VALUE_ERROR ORA ORA ORA ORA Oracle error : Integrity constraint violation. Occurs when attempting to insert a child record for which no parent record exists. Oracle error : Integrity constraint violation. Occurs when attempting to delete a parent record for which children records exist. ORA-00001: unique constraint violation / Primary key violation Page 16 of 16