Study Guide for: Oracle Database SQL Certified Expert Exam Guide (Exam 1Z0-047)

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1 Study Guide for: Oracle Database SQL Certified Expert Exam Guide (Exam 1Z0-047) Study Material for: Student :10:53 Examine the following SQL statement: CREATE TABLE INSTRUCTORS ( ID NUMBER(20), NAME VARCHAR2(30), STATUS CHAR, DEPARTMENT NUMBER(7,-3), OFFICE_HOURS INTERVAL DAY TO SECOND, START_DATE TIMESTAMP WITH LOCAL TIME ZONE ); What is true of the statement? C: The precision for DAY in OFFICE_HOURS is 2. C is correct. Without any specification for precision for DAY, INTERVAL DAY TO SECOND defaults to two digits. A, B, and D are incorrect. The STATUS column's datatype does not need a precision, CHAR without precision defaults to one character. The DEPARTMENT column's datatype is correct, the negative 3 indicates the level of detail at which a value will be rounded. The START_DATE datatype will not store the time zone offset; for that, you would need to see the TIMESTAMP WITH TIME ZONE datatype. Chapter 2, Using DDL Statements to Create and Manage Tables: 2.04 List the Data Types That Are Available for Columns OBJECTIVE: List the data types that are available for columns. You are tasked with creating a CHECK constraint to ensure that any incoming data is in the 5 plus 4 ZIP code mode. In other words, you want numbers to look like this: You want to reject any numbers that do not look like this. Which of the following defines the pattern? D: '^[[:digit:]]{5}-[[:digit:]]{4}$' LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 1

2 D is correct. The character class [:digit:] is the ideal way to reference numerics. The ^ and $ operators anchor the pattern to the beginning and end of the string, preventing extraneous leading or trailing characters. A, B, and C are incorrect. The pattern [1-9] omits zeros. Chapter 17, Regular Expression Support: Regular Expressions and CHECK Constraints OBJECTIVE: Regular expressions and CHECK constraints. Click the Exhibit button. Examine the illustration. Which of the following are valid queries against this data model? (Choose two.) B: SELECT D.TITLE, I.LAST_NAME, C.START_DATE FROM INSTRUCTORS I JOIN CLASSES C ON I.INSTRUCTOR_ID = C.INSTRUCTOR_ID JOIN DEPARTMENTS D ON I.DEPARTMENT_ID = D.DEPARTMENT_ID; C: SELECT D.TITLE, I.LAST_NAME, C.START_DATE FROM INSTRUCTORS I NATURAL JOIN CLASSES C NATURAL JOIN DEPARTMENTS D; B and C are correct. The syntax for joining multiple tables consists of the keyword JOIN and its associated ON clause repeated as required. For natural joins, no ON clause is required, just the NATURAL JOIN clause is repeated. A and D are incorrect. The use of two columns in the USING clause is incorrect. The run-on JOIN clauses without the associated ON clauses are not correct. Chapter 8, Displaying Data from Multiple Tables: 8.01 Write SELECT Statements to Access Data from More Than One Table Using Equijoins and Non-Equijoins OBJECTIVE: Write SELECT statements to access data from more than one table using equijoins and nonequijoins. Examine the following data listing for table INSTRUCTORS: ID LAST_NAME VACATION PAY_RATE DEPARTMENT_ID Smith Jones Cohen Green LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 2

3 5 Riley Simpson Bryant Now examine the following SELECT statement (line numbers added): 01 SELECT DEPARTMENT_ID, LAST_NAME 02 FROM INSTRUCTORS 03 WHERE VACATION < OR PAY_RATE < AND DEPARTMENT_ID NOT IN (1,2,7); Which of the following are true for this statement? (Choose two.) B: It will return two rows. B is correct. A quick glance at the WHERE clause, and you should be able to tell that the WHERE clause has an AND comparison at the end. The rules of operator precedence require that AND be processed before OR. This means that the expressions PAY_RATE < 60 AND DEPARTMENT_ID NOT IN (1, 2, 7) are evaluated and compared first, so the best solution is to review the rows of data for this comparison first. Only the second row returns a TRUE for this comparison. Given that the remaining comparison uses an OR, you can now glance at the VACATION column and see that two values return a TRUE, one of which is row 2, and the other is row 6. These are the rows returned by the SELECT statement. A, C, and D are incorrect. The OR comparison is evaluated last, not first. Also, NULL values, even though they are NOT IN the list of values shown for DEPARTMENT_ID, result in a FALSE return nevertheless. NULL errs on the side of FALSE. Is NULL, the unknown number, not within the list of values of 1, 2, and 7? Maybe it is, maybe it isn't-we can't tell. In such situations, SQL errs on the side of FALSE. OBJECTIVE: Limit the rows that are retrieved by a query. Examine the following SQL code: CREATE TABLE REGIONS ( REGION_ID NUMBER(7) UNIQUE, REGION_NAME VARCHAR2(30), MANAGER_ID NUMBER(11) ); CREATE TABLE OFFICES ( OFFICE_ID NUMBER(11) PRIMARY KEY, REGION_ID NUMBER(11,2), CONSTRAINT OFRE FOREIGN KEY (REGION_ID) REFERENCES REGIONS (REGION_ID) ); What will be the result of the preceding SQL statements? D: The constraint OFRE will be created successfully. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 3

4 D is correct. The foreign key will be created successfully. A, B, and C are incorrect. Two tables will be created, but the first has three columns, and the second only has two columns. The syntax of a FOREIGN KEY constraint requires that the referenced table's column have a UNIQUE constraint. This means that the referenced column may instead have a PRIMARY KEY constraint, which is a combination of UNIQUE and NOT NULL, and in fact this is often what is done in practice, but only the UNIQUE constraint is required. Even though the level of detail that can be stored in the OFFICES table's REGION_ID column is greater than what can be stored in the parent table REGIONS's REGION_ID column, the constraint will still work, since the smaller value datatype is going into the larger value datatype, and no precision will be lost. Chapter 2, Using DDL Statements to Create and Manage Tables: 2.05 Explain How Constraints Are Created at the Time of Table Creation OBJECTIVE: Explain how constraints are created at the time of table creation. Click the Exhibit button. Examine this illustration. You are tasked to write a query that will perform the following: * Query the PAY_RATE for each instructor. PAY_RATE contains the annual salary for each instructor. * Determine the base PAY_RATE if it were increased by 5 percent. * Throw in a one-time annual bonus of $200 for the year. * Show the monthly rate of pay for the increased annual PAY_RATE, including the once-per-year bonus. Which of the following queries will accomplish this task? B: SELECT (200 + PAY_RATE * 1.05)/12 NEW_SALARY FROM INSTRUCTORS; B is correct. The rules of operator precedence ensure that the portions of the expression within the parentheses will be executed first. Within the parentheses is one operation of addition, and one of multiplication. The rules of operator precedence dictate that the multiplication will be executed first, and then the addition. A, C, and D are incorrect. None of these reflect the intended equation. Answer A adds the bonus to the pay rate but then multiplies the 1.05 multiplier (representing a 5 percent increase). In Answer C, the pay rate is increased and then converted to the monthly rate, and after that the bonus is added-to each monthly rate, which is not the intent. In Answer D, a similar bad formula is indicated. Chapter 4, Retrieving Data Using the SQL SELECT Statement: 4.02 List the Capabilities of SQL SELECT Statements OBJECTIVE: List the capabilities of the SQL SELECT statement. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 4

5 Click the Exhibit button. Examine the illustration. You are tasked to create a report that joins data from INSTRUCTORS, STAFF, and DEPARTMENTS, and groups the data by department and EXEMPT status. The query will use the same select list and join but will not need each grouped set of rows-for example, you'll only need groups for certain departments, but not others. Your database is large and the production system is taxed heavily during the day. What feature can you use to ensure that you only return the grouped data you require, and avoid unwanted and unnecessary processing? C: GROUPING SETS C is correct. GROUPING SETS can be used with GROUP BY to process only those grouped sets that you specify, and ignore the others. A, B, and D are incorrect. Chapter 13, Generating Reports by Grouping Related Data: Use GROUPING SETS to Produce a Single Result Set OBJECTIVE: Use GROUPING SETS to produce a single result set. Which of the following datatypes can be used with a PRIMARY KEY constraint? D: TIMESTAMP WITH LOCAL TIME ZONE D is correct. The TIMESTAMP WITH LOCAL TIME ZONE datatype can be used with a PRIMARY KEY constraint. A, B, and C are incorrect. CLOB, BLOB, and TIMESTAMP WITH TIME ZONE cannot be used with a PRIMARY KEY. Think about the size of CLOB and BLOB and you'll realize they don't lend themselves well to the creation of an index, which a PRIMARY KEY column requires. The same is true for the TIMESTAMP WITH TIME ZONE datatype, which may store a value containing of a time zone offset. Chapter 2, Using DDL Statements to Create and Manage Tables: 2.05 Explain How Constraints Are Created at the Time of Table Creation. Also see Chapter 10, Creating Other Schema Objects: Create and Maintain Indexes OBJECTIVE: Explain how constraints are created at the time of table creation. Examine the following SQL code: CREATE TABLE STUDENTS LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 5

6 ( STUDENT_ID NUMBER(13) PRIMARY KEY, F_NAME VARCHAR2(25), L_NAME VARCHAR2(35), SSN VARCHAR2(11), STATUS VARCHAR2(30), REG_DATE TIMESTAMP ); CREATE SEQUENCE STUDENT_ID_SEQUENCE; Which of the following are valid INSERT statements for this table? (Choose two.) B: INSERT INTO STUDENTS VALUES ( STUDENT_ID_SEQUENCE.NEXTVAL, (SELECT 'JOE' FROM DUAL), 'SCHMOE', NULL, 'ACTIVE', SYSDATE); D: INSERT INTO STUDENTS ( STUDENT_ID, F_NAME, L_NAME, REG_DATE) SELECT STUDENT_ID_SEQUENCE.NEXTVAL, 'JOE', 'SCHMOE', SYSTIMESTAMP FROM DUAL; B and D are correct. In Answer B the syntax is correct, including the use of the scalar subquery. There is no difficulty entering SYSDATE into a column of the TIMESTAMP datatype. Answer D is an INSERT with a subquery. A and C are incorrect. In Answer A, the two-column inline view that selects 'JOE' and 'SCHMOE' from DUAL is not acceptable in this context. It leaves the values list short one value. In Answer C, the keyword AS should not be used with the subquery. It also supplies too many values. Chapter 3, Manipulating Data: 3.02 Insert Rows into a Table, and Chapter 15, Manipulating Large Data Sets: Manipulate Data Using Subqueries OBJECTIVE: Insert rows into a table. Click the Exhibit button. Examine the Illustration. Which of the following SQL statements is valid? B: SELECT I.LAST_NAME, C.START_DATE FROM INSTRUCTORS I JOIN CLASSES C USING (INSTRUCTOR_ID) WHERE START_DATE BETWEEN SYSTIMESTAMP AND TO_DATE('01-JUN-10', 'DD-MON-RR'); B is correct. The USING specification for INSTRUCTOR_ID does not clash with the use of the table alias elsewhere in the SELECT statement, provided that the table alias does not LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 6

7 prefix the column on which the join criteria are based, which is specified in USING as INSTRUCTOR_ID. A, C, and D are incorrect. In Answer A, the use of a NATURAL JOIN prevents the table name prefix from being used on the join column, which is INSTRUCTOR_ID. In Answer C, there is no such thing as a LEFT INNER join. In Answer D, the use of a table alias on the join column INSTRUCTOR_ID clashes with the USING specification on the column. Chapter 4, Retrieving Data Using the SQL SELECT Statement: 4.01 Execute a Basic SELECT Statement, and Chapter 8, Displaying Data from Multiple Tables: 8.01 Write SELECT Statements to Access Data from More Than One Table Using Equijoins and NonEquijoins OBJECTIVE: Write SELECT statements to access data from more than one table using equijoins and nonequijoins. Examine the following SQL code: CREATE TABLE STUDENTS ( STUDENT_ID NUMBER(13) PRIMARY KEY, F_NAME VARCHAR2(25), L_NAME VARCHAR2(35) NOT NULL, SSN VARCHAR2(11) UNIQUE, STATUS VARCHAR2(30)); ALTER TABLE STUDENTS SET UNUSED COLUMN SSN; What is the result of the ALTER TABLE statement at the end of the code listing? D: You can now add a new column to the table with the name SSN. D is correct. Once a column has been set to UNUSED, the column name is now available for a new column that you might choose to add to the table. A, B, and C are incorrect. Setting a column to UNUSED is comparable to dropping it altogether for all practical purposes-the column is no longer visible when the table's structure is described with the DESC command, and you can no longer query on the column. It is as though it has been dropped, although it remains internally stored with the table, awaiting some future time when you might choose to finally drop it. Chapter 11, Managing Schema Objects, Drop Columns and Set Column UNUSED OBJECTIVE: Drop columns and set column UNUSED. Click the Exhibit button. Examine the following illustration. Someone in your organization enters the following code while working with a server located in the 'Asia/Macau' time zone: LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 7

8 INSERT INTO TRANSACTIONS VALUES (1, SYSTIMESTAMP); Later, someone in your office located in the 'Europe/Vienna' time zone queries the TRANSACTIONS table, as follows: SELECT EXTRACT (TIMEZONE_REGION FROM TRANS_ENTRY) FROM TRANSACTIONS WHERE TRANS_ID = 1; What is the output of the query? B: Europe/Vienna B is correct. The TIMESTAMP WITH LOCAL TIME ZONE does not store information about the source time zone or offset, but it normalizes the time for the local system that queries it. A, C, and D are incorrect. Chapter 6, Using Single-Row Functions to Customize Output: 6.04 Managing Data in Different Time Zones-Use Various Datetime Functions OBJECTIVE: Manage data in different time zones; use various datetime functions. Click the Exhibit button. Examine the illustration. Next, examine the data listing for the table COURSE_CATALOG: ID MAJOR COURSE CLASS CREDITS SEATS Business Accounting Business Accounting Business Accounting Business Finance Law Constitution Law Tort Engineering Physics Engineering Physics Engineering Physics Note that there are nine rows of data. Now examine the following statement: SELECT MAJOR, COURSE, COUNT(CLASS), SUM(CREDITS) FROM COURSE_CATALOG GROUP BY CUBE (MAJOR, COURSE); How many rows of output will this query produce? C: 14 C is correct. The CUBE operation will produce one row for each grouped set of MAJOR and COURSE values, which will result in 5 rows. It will add one summary row for each unique LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 8

9 COURSE value, of which there are 5. It will add one summary row for each unique MAJOR value, of which there are 3. Finally it will add one grand total summary row for the entire data listing, of which there is, well, one. Add those numbers together and you get 14. A, B, and D are incorrect. Chapter 13, Generating Reports by Grouping Related Data: Use the CUBE Operation to Produce Crosstabulation Values OBJECTIVE: Use the CUBE operation to produce crosstabulation values. Examine the following SQL statement: SELECT TO_TIMESTAMP('MAY 01 11','MON DD RR') + TO_YMINTERVAL('1-1') + TO_DSINTERVAL(INTERVAL '75' MINUTE) FROM DUAL; Which of the following is the result? D: 01-JUN AM D is correct. The function TO_YMINTERVAL converts the string '1-1' into a time interval of one year, one month. The TO_DSINTERVAL function converts the enclosed literal value to one hour, 15 minutes. The TIMESTAMP representation of the combined result of adding all the values together is shown in Answer D. A, B, and C are incorrect. Chapter 6, Using Single-Row Functions to Customize Output: 6.03 Describe the Use of Conversion Functions OBJECTIVE: Describe the use of conversion functions. Which of the following are valid CREATE TABLE statements? (Choose two.) A: CREATE TABLE REGIONS (REGION_ID NUMBER(7) PRIMARY KEY USING INDEX (CREATE INDEX RI_IND ON REGIONS(REGION_ID)), REGION_NAME VARCHAR2(30), MANAGER_ID NUMBER(11) NOT NULL); C: CREATE TABLE REGIONS (REGION_ID NUMBER(7) UNIQUE USING INDEX (CREATE INDEX RI_IND LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 9

10 ON REGIONS(REGION_ID)), REGION_NAME VARCHAR2(30), MANAGER_ID NUMBER(11) NOT NULL); A and C are correct. The USING INDEX clause can be used to create an index for only the PRIMARY KEY and UNIQUE constraints. B and D are incorrect. Chapter 11, Managing Schema Objects: Create Indexes Using the CREATE TABLE Statement OBJECTIVE: Create indexes using the CREATE TABLE statement. Identify the true statements about the CREATE SEQUENCE statement. (Choose two.) A: If CYCLE is omitted, NOCYCLE is the default. D: If CYCLE is specified, MAXVALUE may be negative. A and D are correct. NOCYCLE is the default and does not need to be specified in the CREATE SEQUENCE statement. If CYCLE is specified, however, then NOCYCLE does not apply. If CYCLE is specified, MAXVALUE may be negative, provided that MINVALUE is less than MAXVALUE. B and C are incorrect. If CYCLE is specified and MINVALUE is omitted, the default minimum value is 1 for ascending sequences. If CYCLE is specified, MINVALUE is not necessarily required and will default to 1 if omitted. MAXVALUE must be specified as higher than 1 in such a situation. Chapter 10, Creating Other Schema Objects: Create, Maintain, and Use Sequences OBJECTIVE: Create, maintain, and use sequences. Examine the following data listing for a table STUDENTS: ROWNUM ID NAME SCORE JONES 2 2 SMITH SIMPSON 25 Note the presence of the pseudocolumn ROWNUM. Note that row 3 is completely NULL. Also note that the SCORE for JONES is NULL. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 10

11 Now examine the following SELECT statement: SELECT COUNT(*), AVG(SCORE) FROM STUDENTS; What will be the result of this query? D: 5, 20 D is correct. Even though row 3 is completely NULL, the COUNT function will count it as a valid row when using the asterisk feature. However, AVG will ignore NULL values when computing averages. A, B, and C are incorrect. Chapter 7, Reporting Aggregated Data Using the Group Functions: 7.02 Describe the Use of Group Functions OBJECTIVE: Describe the use of group functions. A multitable INSERT statement: (Choose two.) B: Can have multiple WHEN conditions. D: Can insert one row into multiple tables. B and D are correct. A multitable INSERT can have many WHEN conditions. It can insert rows into one or more tables. A and C are incorrect. A multitable INSERT can be conditional, or unconditional. It is based on one subquery. Chapter 15, Manipulating Large Data Sets: Describe the Features of Multitable INSERTs and Use the Following Types of Multitable INSERTs: Unconditional, Conditional, and Pivot OBJECTIVE: Describe the features of multitable INSERTs, and use the following types of multitable INSERTs: unconditional, conditional, and pivot. Examine the following SQL code (line numbers added): 01 MERGE INTO STAFF S 02 USING (SELECT INSTRUCTOR_ID, 'INSTRUCTOR' NEW_TITLE, 03 FIRST_NAME, LAST_NAME, EMPNO 04 FROM INSTRUCTORS) I LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 11

12 05 ON (S.EMPNO = I.EMPNO) 06 WHEN MATCHED THEN UPDATE SET TITLE = NEW_TITLE 07 WHEN NOT MATCHED THEN INSERT (STAFF_ID, FIRSTNAME, LASTNAME) 08 VALUES ((INSTRUCTOR_ID+1000), FIRST_NAME, LAST_NAME); Which of the following statements is true of the MERGE statement? C: A WHERE clause could be added to the UPDATE operation after line 6 and before line 7. C is correct. The UPDATE operation accepts the optional WHERE clause. A, B, and D are incorrect. DELETE is allowed with MERGE. The inline view is also valid. The combined DML operations are processed in one pass through the database, which is the advantage to using MERGE. Chapter 15, Manipulating Large Data Sets: Merge Rows in a Table OBJECTIVE: Merge rows in a table. Click the Exhibit button. Examine the illustration. Now examine the following query: SELECT LEVEL, LPAD(' ', LEVEL*2) LASTNAME AS LASTNAME, PRIOR LASTNAME AS BOSS FROM STAFF START WITH STAFF_ID = 1 CONNECT BY PRIOR STAFF_ID = SUPERVISOR_ID; Examine the output listing from this query: LEVEL LASTNAME BOSS LeGrange 2 Chen LeGrange 2 Marie LeGrange 3 Cramerica Marie 2 Franken LeGrange 3 Merckle Franken 4 Linkletter Merckle 3 Charles Franken What would you have to change in the query to reverse the direction of this query from top-down through the hierarchy to bottom-up? C: START WITH LASTNAME = 'Linkletter' CONNECT BY STAFF_ID = PRIOR SUPERVISOR_ID; C is correct. Reversing the position of PRIOR is the key to reversing the direction in which SQL is traversing the self-join table. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 12

13 A, B, and D are incorrect. Chapter 16, Hierarchical Retrieval: Interpret the Concept of a Hierarchical Query OBJECTIVE: Interpret the concept of a hierarchical query. Click the Exhibit button. Examine the illustration. Which of the following SQL statements is likely to result in an error? B: SELECT (SELECT OFFICE_PHONE FROM STAFF), FIRST_NAME, LAST_NAME FROM INSTRUCTORS; B is correct. The statement being used as a scalar subquery can return multiple rows. It isn't syntactically incorrect, but it runs the risk of resulting in an execution error. A, C, and D are incorrect. Answer A uses an inline view in the FROM statement. Answer C uses an inline view, which can return multiple rows. Answer D uses a scalar subquery, as indicated by the WHERE clause that identifies a row by its primary key value. All are valid and well-formed statements. Chapter 9, Retrieving Data Using Subqueries: 9.06 Use Scalar Subqueries in SQL. Also Chapter 10, Creating Other Schema Objects: Create and Use Simple and Complex Views OBJECTIVE: Use scalar subqueries in SQL, and create simple and complex views, and retrieve data from views. Click the Exhibit button. Examine the illustration. You are tasked with creating a query that will show a list of instructors in the INSTRUCTORS table who have the same names as staff in the STAFF table. Which of the following queries will satisfy the requirement? (Choose two.) A: SELECT I.FIRST_NAME, I.LAST_NAME FROM INSTRUCTORS I WHERE I.FIRST_NAME I.LAST_NAME IN (SELECT FIRSTNAME LASTNAME FROM STAFF); C: SELECT I.FIRST_NAME, I.LAST_NAME FROM INSTRUCTORS I WHERE (I.FIRST_NAME, I.LAST_NAME) IN (SELECT FIRSTNAME, LASTNAME FROM STAFF); LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 13

14 A and C are correct. Answer A uses concatenation correctly in this context to compare all of the names in INSTRUCTORS to any and all of the names in STAFF. Answer C is an example of a multiple-column subquery. B and D are incorrect. Answer B uses correct syntax in a technical sense, but the logic of the query doesn't support the task. Instead it finds instructors whose first name happens to match any first names among the staff, and then it makes a separate comparison for matching last names. Answer D is the same as Answer C with one difference, and that is its choice of an equal sign as a comparison operator, instead of the keyword IN. The equal sign means that only exact matches will occur and the subquery is expected to return only one row, which is not likely. If it returns more rows than one, an execution error will result. Chapter 9, Retrieving Data Using Subqueries: 9.04 Write single-row and multiple-row subqueries OBJECTIVE: Write single-row and multiple-row subqueries. Click the Exhibit button. Examine the illustration. You are tasked with producing a report that shows the names of the instructors in each department who have the highest pay rate. Which of the following are you least likely to use in creating your query? A: A hierarchical query A is correct. You are least likely to use a hierarchical query. B, C, and D are incorrect. You are most likely to use a correlated subquery. The subquery will most likely use a GROUP BY and an aggregate function to determine the set of instructors with the highest pay rate for the correlated subquery. Chapter 9, Retrieving Data Using Subqueries: 9.07 Solve Problems with Correlated Subqueries OBJECTIVE: Solve problems with correlated subqueries. Examine the following SQL code: CREATE TABLE TRANSACTIONS ( TRANS_ID NUMBER(11), TRANS_ENTRY TIMESTAMP WITH LOCAL TIME ZONE, ENTERED_BY NUMBER(11), STATUS VARCHAR2(20) ); ALTER TABLE TRANSACTIONS SET UNUSED COLUMN ENTERED_BY; Which of the following statements will drop the unused columns? LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 14

15 A: ALTER TABLE TRANSACTIONS DROP UNUSED COLUMNS; A is correct. Once a column has been set to UNUSED, it can no longer be specified by name. The DROP UNUSED COLUMNS clause will drop all outstanding unused columns that might happen to exist within the table. B, C, and D are incorrect. Chapter 11, Managing Schema Objects: Drop Columns and Set Column UNUSED OBJECTIVE: Drop columns and set column UNUSED. Examine the following SQL code executed from within the user account ADMIN: CREATE TABLE REGIONS ( REGION_ID NUMBER(7) PRIMARY KEY, REGION_NAME VARCHAR2(30), MANAGER_ID NUMBER(11) NOT NULL); CREATE VIEW REGIONS_VW AS SELECT * FROM REGIONS; CREATE PUBLIC SYNONYM REG FOR ADMIN.REGIONS; DROP TABLE REGIONS; Which of the following statements correctly describes the results after the preceding statements are executed? C: One index is dropped. C is correct. When the table is created, the inclusion of the PRIMARY KEY constraint will create the constraint as well as an index to support the constraint. A, B, and D are incorrect. The public synonym exists independently of any object. The same is not true of a view, but the view will not be dropped automatically; instead, it will be automatically rendered INVALID. And while the PRIMARY KEY constraint creates its own index, the NOT NULL constraint does not create an index. Chapter 4, Retrieving Data Using the SQL SELECT Statement: 4.03 Describe How Schema Objects Work, and Chapter 11, Managing Schema Objects: Add Constraints OBJECTIVE: Describe how schema objects work. To sort rows in a query that uses set operators, which of the following is valid? A: ORDER BY position. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 15

16 A is correct. When set operators are present, you can sort by position or by expressions from the first SELECT statement at the top. B, C, and D are incorrect. Chapter 12, Using the Set Operators: Control the Order of Rows Returned OBJECTIVE: Control the order of rows returned when set operators are present. Click the Exhibit button. Examine the illustration. Then examine the following code: WITH STAFF_STATISTICS AS ( SELECT DEPARTMENT_ID, COUNT(STAFF_ID) CT FROM STAFF GROUP BY DEPARTMENT_ID ), ADMIN_STATS AS ( SELECT DEPARTMENT_ID, CT FROM STAFF_STATISTICS WHERE DEPARTMENT_ID IN (SELECT DEPARTMENT_ID FROM DEPARTMENTS) ) SELECT I.LAST_NAME, A.CT FROM ADMIN_STATS A JOIN INSTRUCTORS I ON A.DEPARTMENT_ID = I.DEPARTMENT_ID; Which of the following is true of the preceding code? D: The STAFF_STATISTICS subquery block is not visible beyond the code sample shown. D is correct. None of the subquery blocks exist beyond the query. A, B, and C are incorrect. None of the restrictions mentioned apply to the queries within the subquery blocks. Chapter 9, Retrieving Data Using Subqueries: 9.10 Use the WITH Clause OBJECTIVE: Use the WITH clause. Examine the following SQL statements (line numbers added): 01 COMMIT; 02 SAVEPOINT POINT_A; 03 CREATE TABLE CLASSES 04 (CLASS_ID NUMBER(13) PRIMARY KEY, 05 INSTRUCTOR_ID NUMBER(13) LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 16

17 06 ); 07 INSERT INTO CLASSES VALUES (1,1); 08 INSERT INTO CLASSES VALUES (2,2); 09 SAVEPOINT POINT_B; 10 INSERT INTO CLASSES VALUES (3,3); 11 ROLLBACK WORK TO POINT_A; 12 INSERT INTO CLASSES VALUES (4,4); 13 ROLLBACK WORK TO POINT_B; 14 COMMIT; Which of the following statements correctly describes the result of the preceding statements? C: The ROLLBACK in line 11 won't work because the CREATE TABLE statement in line 3 reset the SAVEPOINTs. C is correct. The ROLLBACK will fail because the CREATE TABLE statement is part of Data Definition Language (DDL) and therefore results in an implied commit event. (Actually, DDL causes a commit immediately before execution, and another immediately after.) The result: the SAVEPOINT in line 1 is reset and no longer exists. Therefore the ROLLBACK in line 11 won't execute and won't reset anything, leaving the way for the ROLLBACK in line 13 to execute successfully. The result: the first two rows are added to the table, and no more. A, B, and D are incorrect. Chapter 3, Manipulating Data: 3.05 Control Transactions OBJECTIVE: Control transactions. The following code is executed from within user DANNA: CREATE TABLE SCHOOLS ( SCHOOL_ID NUMBER(7) PRIMARY KEY, NAME VARCHAR2(20), DEAN_ID NUMBER(7)); GRANT SELECT ON SCHOOLS TO PUBLIC; CREATE SYNONYM SCH FOR SCHOOLS; DROP TABLE SCHOOLS; Five minutes later, user DANNA executes the following: FLASHBACK TABLE SCHOOLS TO BEFORE DROP; What objects have been recovered? C: One table, one index, one grant C is correct. When the table was created, the PRIMARY KEY resulted in the automatic creation of an index. When the table was dropped, the index and grant were all dropped and placed into the recycle bin. FLASHBACK TABLE recovers all of these objects. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 17

18 A, B, and D are incorrect. The synonym is not dropped with the table, so there was no synonym that required recovery. Chapter 11, Managing Schema Objects: Perform FLASHBACK Operations OBJECTIVE: Perform FLASHBACK operations. Click the Exhibit button. Examine the illustration. Now examine the following SQL statement: UPDATE STUDENTS SET REG_DATE = TO_DATE('01-MAR-10','DD-MON-RR'), SSN = WHERE STUDENT_ID = 101; Which of the following statements is true for the preceding statement? B: The UPDATE statement will fail for any row with a STUDENT_ID of 101 because the value assigned to SSN is too long. If the table has no rows with a STUDENT_ID of 101, the UPDATE statement will complete execution without effect and without error. B is correct. The WHERE clause is processed first. If no rows are found that match the WHERE condition, the SET clauses will not execute. All of the SET clauses must execute successfully for all of the rows returned by the WHERE condition, or else none of the changes to any of the rows will be successful. A, C, and D are incorrect. You cannot add a number whose precision is too large into a numeric datatype of lesser precision. However, numbers with too much detail in the scale specification may accept large values and round them off, resulting in a loss of data. Chapter 3, Manipulating Data: 3.03 Update Rows in a Table, and Chapter 5, Restricting and Sorting Data: 5.01 Limit the Rows That Are Retrieved by a Query OBJECTIVE: Update rows in a table. User MARIA executes the following code, including GRANTS to user ANTHONY (line numbers added): 01 CREATE TABLE COURSE_CATALOG 02 ( COURSE_CATALOG_ID NUMBER(7) PRIMARY KEY 03, DEPARTMENT_ID VARCHAR2(80) 04, COURSE VARCHAR2(80) ); 05 CREATE VIEW CC_VIEW AS 06 SELECT * FROM COURSE_CATALOG WHERE COURSE <> 'TEACHER TRAINING'; 07 CREATE PUBLIC SYNONYM COURSE_CATALOG FOR MARIA.COURSE_CATALOG; 08 GRANT SELECT ON CC_VIEW TO ANTHONY; LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 18

19 Next, user ANTHONY executes the following code: 09 CREATE SYNONYM COURSE_CATALOG FOR MARIA.CC_VIEW; Finally, ANTHONY executes the following statement: 10 SELECT * FROM COURSE_CATALOG; Which object is identified in the SELECT statement executed by user ANTHONY? B: The private synonym created in line 9 B is correct. The private synonym is the object that is first identified when the SELECT statement is processed. Name priority looks to the local user space first before looking to the database-level namespaces. A, C, and D are incorrect. There are no naming conflicts in the code; each name is created to be unique within its appropriate namespace. Chapter 10, Creating Other Schema Objects: Create Private and Public Synonyms OBJECTIVE: Create private and public synonyms. Click the Exhibit button. Examine the illustration and then examine the following query: SELECT LEVEL, LPAD(' ', LEVEL*2) LASTNAME AS LASTNAME_FORMATTED, PRIOR LASTNAME AS BOSS FROM STAFF START WITH STAFF_ID = 1 CONNECT BY PRIOR STAFF_ID = SUPERVISOR_ID; Which ORDER BY clause can be appended to the end of this SQL statement so that the output will retain the ordering of the hierarchical query clause and only sort rows by LASTNAME within each hierarchical level? C: ORDER SIBLINGS BY LASTNAME C is correct. ORDER SIBLINGS BY is the clause you should use to sort rows in any hierarchical query. A, B, and D are incorrect. Answers A and B will interrupt the hierarchical order of data. Answer D will cause a syntax error. Chapter 16, Hierarchical Retrieval: Create and Format Hierarchical Data LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 19

20 OBJECTIVE: Create a tree-structured report, and format hierarchical data. Examine the following data listing of a table COURSE_CATALOG. ID Department Course Class Business Accounting General Ledger 2 Business Accounting Accounts Payable 3 Business Accounting Accounts Receivable 4 Business Finance Ratios 5 Engineering Systems Design 6 Engineering Systems Administration 7 Engineering Network Testing You are tasked to examine a SQL query that shows only unique combinations of values of Department and Course. Which of the following SQL statements will perform the task? A: SELECT DISTINCT DEPARTMENT, COURSE FROM COURSE_CATALOG; A is correct. The keyword DISTINCT is used once at the beginning of the select list, and unique column combinations will be returned. B, C, and D are incorrect. Chapter 4, Retrieving Data Using the SQL SELECT Statement: 4.02 List the Capabilities of SQL SELECT Statements OBJECTIVE: List the capabilities of SQL SELECT statements. Click the Exhibit button. Examine the illustration. The engineering department instructors are said to teach longer classes than the rest of the school. You are tasked to challenge this claim and see if you can identify any instructors who do not teach in the engineering department, but who have classes that are as long as, if not longer than, any class taught by any of the instructors in engineering. Which of the following queries do you create? C: SELECT I.FIRST_NAME, I.LAST_NAME FROM INSTRUCTORS I JOIN CLASSES C ON I.INSTRUCTOR_ID = C.INSTRUCTOR_ID WHERE (C.END_DATE - C.START_DATE) >= ANY ( SELECT (C.END_DATE - C.START_DATE) FROM INSTRUCTORS I JOIN CLASSES C ON I.INSTRUCTOR_ID = C.INSTRUCTOR_ID WHERE I.DEPARTMENT_ID = (SELECT DEPARTMENT_ID FROM DEPARTMENTS WHERE TITLE = 'Engineering')) AND I.DEPARTMENT_ID NOT IN LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 20

21 (SELECT DEPARTMENT_ID FROM DEPARTMENTS WHERE TITLE = 'Engineering'); C is correct. The comparison operator > ANY ensures that the outer query lists instructors whose classes are as long as or longer than any one of the rows returned by the subquery. A, B, and D are incorrect. Answer A uses a comparison operator that looks for a single-row subquery, and chances are that the subquery will return more than one row. That means the query will most likely result in an execution error. Answer B uses the > ALL comparison operator, which ensures that you'll get instructors who have longer classes than the engineering department's classes, but you'll lose those instructors who might be longer than some of the engineering classes, not necessarily all, and this is closer to the requirement of the task. Chapter 9, Retrieving Data Using Subqueries: 9.04 Write Single-Row and Multiple-Row Subqueries OBJECTIVE: Write single-row and multiple-row subqueries. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 21