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 :49:30 Examine the following data listing for table WORKERS: WORKER_ID LAST_NAME SUPERVISOR_ID Madeoff 2 Gotaway 1 3 Didit 1 4 Tookit 2 5 Skipped 2 6 Gone 2 7 Bundy 3 You are tasked to produce a report listing of this table's data in a hierarchy, excluding worker Didit and all those who report to worker Didit. Which of the following queries will accomplish this task? B: SELECT WORKER_ID, LPAD(' ',LEVEL) LAST_NAME NAME FROM WORKERS START WITH WORKER_ID = 1 CONNECT BY PRIOR WORKER_ID = SUPERVISOR_ID AND LAST_NAME <> 'Didit'; B is correct. Using the AND operator to extend the logic of CONNECT BY and exclude worker Didit is correct. In addition, PRIOR should be used on the side of the CONNECT BY comparison with the row containing the higher-level value in the hierarchy. A, C, and D are incorrect. You cannot simply use the WHERE clause to exclude worker Didit, as this will merely exclude that individual worker, but not worker Bundy, who reports to worker Didit. Furthermore, PRIOR should be used on the side of the CONNECT BY comparison with the higher-level value in the hierarchy. See Chapter 16, Hierarchical Retrieval: Create and Format Hierarchical Data, and Exclude Branches from the Tree Structure OBJECTIVE: Create a tree-structured report; Format hierarchical data; exclude branches from the tree structure. A scalar subquery can be used (choose two): A: In a WHERE clause of an UPDATE statement C: In a VALUES list of an INSERT statement LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 1

2 A and C are correct. Scalar subqueries are allowed in an UPDATE statement's WHERE clause, and also in an INSERT statement's VALUES list. B and D are incorrect. Scalar subqueries are not allowed in a number of situations, including a CHECK constraint, the GROUP BY and HAVING clauses of a SELECT statement, the WHEN condition of CASE, and more. See Chapter 9, Retrieving Data Using Subqueries: 9.06 Use Scalar Subqueries in SQL OBJECTIVE: Use scalar subqueries in SQL. Examine the following SQL code: CREATE OR REPLACE DIRECTORY ORDER_RECS AS 'G:\ext\txt\order_recs'; If the directory structure specified in the statement does not already exist, what will result from the attempt to execute the statement? C: The statement will succeed but will not create the directory; furthermore, if the ORDER_RECS object is used to create an external table, the CREATE TABLE statement will fail. C is correct. The DIRECTORY object can be created regardless of whether the file system subdirectory exists. SQL will not test for the presence of the file system's subdirectory until you attempt to use the DIRECTORY object in a CREATE TABLE statement to build an EXTERNAL TABLE. A, B, and D are incorrect. The statement will not fail with an error. SQL will not attempt to create the file system's subdirectory for you. You must create it, and you must do so before you use the DIRECTORY object to create an EXTERNAL TABLE. See Chapter 11, Managing Schema Objects: 11:07 Create and Use External Tables OBJECTIVE: Create and use external tables. Consider a series of two or more SELECT statements that are combined into one using set operators. Which of the following is true of the resulting SELECT statement? C: It can include a hierarchical query in at least one of the SELECT statements. C is correct. You may combine one or more hierarchical with one or more non-hierarchical queries with set operators. A, B, and D are incorrect. Set operators are allowed in subqueries. GROUP BY clauses are also accepted. Each of the combined SELECT statements must have the same number of LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 2

3 expressions in the select list. See Chapter 12, Using the Set Operators: 12:02 Use a Set Operator to Combine Multiple Queries into a Single Query OBJECTIVE: Use a set operator to combine multiple queries into a single query. Examine the following SQL code: CREATE TABLE SUPPLIERS (SUPPLIER_ID NUMBER(9) CONSTRAINT PK_SUPPLIERS PRIMARY KEY, SUPPLIER_NAME VARCHAR2(20)); CREATE TABLE ORDERS (ORDER_ID NUMBER(9) PRIMARY KEY, SUPPLIER_ID NUMBER(9)); ALTER TABLE ORDERS ADD CONSTRAINT ORDERS_FK_1 FOREIGN KEY (SUPPLIER_ID) REFERENCES SUPPLIERS (SUPPLIER_ID); Which of the following statements will disable the primary key constraint in the SUPPLIERS table? C: ALTER TABLE SUPPLIERS DISABLE PRIMARY KEY CASCADE; C is correct. The CASCADE keyword is required because of the ORDERS_FK_1 constraint in the ORDERS table that references the SUPPLIERS table. A, B, and D are incorrect. The DISABLE option by itself will not work given that there's a referential constraint involved. The CASCADE CONSTRAINTS keywords are not relevant; only CASCADE is used with DISABLE. You cannot directly disable the constraint; instead, you alter the table and disable the constraint accordingly. See Chapter 11, Managing Schema Objects: 11:03 Add Constraints OBJECTIVE: Add constraints. Click the Exhibit button. Examine the Exhibit. Next, match the SELECT statement with the type of join. 1. Inner join 2. Outer join 3. Non-equijoin 4. Cross-join a. SELECT STORE_NAME, LAST_NAME FROM STORES S JOIN EMPLOYEES E ON S.STORE_ID > E.STORE_ID; b. SELECT STORE_NAME, LAST_NAME FROM STORES NATURAL JOIN EMPLOYEES; LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 3

4 c. SELECT STORE_NAME, LAST_NAME FROM STORES, EMPLOYEES; d. SELECT STORE_NAME, LAST_NAME FROM STORES S RIGHT JOIN EMPLOYEES E ON S.STORE_ID = E.STORE_ID; A: 1-b, 2-d, 3-a, 4-c A is correct. The RIGHT JOIN keywords indicate an OUTER join. The NATURAL join is, by default, also an INNER join. A join that uses the greater-than or less-than sign is a non-equijoin. Finally, a join that uses no criteria at all is a cross join, or Cartesian product. B, C, and D are incorrect. The RIGHT JOIN is not an inner join, but an outer join. The simple JOIN is an inner join, not an outer join. The lack of any join criteria at all is a Cartesian product, and the greater-than or less-than operators indicate the non-equijoin. See 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 / View Data That Generally Does Not Meet a Join Condition by Using Outer Joins, and 8.03 Generate a Cartesian Product of All Rows from Two or More Tables OBJECTIVE: Write SELECT statements to access data from more than one table using equijoins and nonequijoins; view data that generally does not meet a join condition by using outer joins; generate a Cartesian product of all rows from two or more tables. Click the Exhibit button. Examine the Exhibit. Now examine the following code: CREATE VIEW VW_ADDRESSES (LINE_1, LINE_2, LINE_3) AS SELECT STREET, SUITE, CITY ', ' STATE ' ' ZIP_BASE '-' ZIP_PLUS FROM ADDRESSES; Which of the following is true for this SQL code? D: None of the above. D is correct. The statement will succeed and create a new view with three columns named LINE_1, LINE_2, and LINE_3. A, B, and C are incorrect. The statement will not fail, it is syntactically correct. No data will be copied-view objects do not copy data. See Chapter 10, Creating Other Schema Objects: 10:01 Create and Use Simple and Complex Views OBJECTIVE: Create simple and complex views; retrieve data from views. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 4

5 Click the Exhibit button. Examine the Exhibit. Now examine the following SQL code (line numbers added): 01 CREATE SEQUENCE C_ID; MERGE INTO CONTACTS C1 04 USING EMPLOYEES E1 05 ON (C1.FIRST_NAME C1.LAST_NAME = E1.FIRST_NAME E1.LAST_NAME) 06 WHEN MATCHED THEN 07 UPDATE 08 SET C1.ADDRESS_ID = E1.ADDRESS_ID 09 WHEN NOT MATCHED THEN 10 INSERT 11 (C1.CONTACT_ID, C1.FIRST_NAME, C1.LAST_NAME, C1.ADDRESS_ID) 12 VALUES 13 (C_ID.NEXTVAL, E1.FIRST_NAME, E1.LAST_NAME, E1.ADDRESS_ID) 14 ; Which of the following statements is true? D: The sequence object reference in line 13 is acceptable and will produce a desirable result. D is correct. The use of the sequence generator is correct here. A, B, and C are incorrect. The MERGE statement is not required to include a DELETE clause, but if included, it belongs within the WHEN MATCHED clause, after UPDATE, and not the WHEN NOT MATCHED clause. The ON clause in this example is acceptable. See Chapter 15, Manipulating Large Data Sets: Merge Rows in a Table OBJECTIVE: Merge rows in a table. The data dictionary view USER_CONSTRAINTS has a column R_CONSTRAINT_NAME. What can be said of this column? D: None of the above. D is correct. The R_CONSTRAINT_NAME column is not any of the displayed options. It contains the name of a referred constraint. For a foreign key constraint, the R_CONSTRAINT_NAME column will show the name of a primary key constraint that the foreign key references. A, B, and C are incorrect. The R_CONSTRAINT_NAME column does not store the name of the constraint type; that is stored in the column CONSTRAINT_TYPE. There is no column that tracks the previous name of a renamed constraint. And while it's true that the R_CONSTRAINT_NAME column is only used for foreign key constraints, it is not the constraint name, which is stored in the column CONSTRAINT_NAME. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 5

6 See Chapter 14, Managing Objects with Data Dictionary Views: Query Various Data Dictionary Views OBJECTIVE: Query various data dictionary views. Examine the following SQL code: CREATE TABLE EMPLOYEES ( EMPLOYEE_ID NUMBER(9) PRIMARY KEY, NOTES CLOB, RESUME CLOB, FIRST_NAME VARCHAR2(25), LAST_NAME VARCHAR2(35) CONSTRAINT LN_REQUIRED NOT NULL ); Which of the following statements correctly describe the preceding statement? D: It will successfully create the table. D is correct. The syntax is correct, and it will successfully create the table. A, B, and C are incorrect. You can specify names for NOT NULL constraints. What you cannot do is create a NOT NULL constraint with the out-of-line syntax, and that is not what is being used here-this is in-line syntax. You may create more than one CLOB datatype. You may place them in any location within the table's column list. See Chapter 2, Using DDL Statements to Create and Manage Tables: 2.04 List the Data Types That Are Available for Columns OBJECTIVE: Use DDL statements to create and manage tables. In the data dictionary view FLASHBACK_TRANSACTION_QUERY, the column UNDO_SQL contains SQL code to reverse the effects of (choose the best answer): B: A single SQL statement B is correct. The UNDO_SQL column shows information required to undo a single SQL statement. A, C, and D are incorrect. A SQL transaction is defined as all of the statements from one commit event to another. While that might encompass a single SQL statement, it may also encompass multiple SQL statements, and the UNDO_SQL column only shows the code to reverse the effects of a single SQL statement. The VERSIONS_BETWEEN and AS OF statements are not relevant to the UNDO_SQL column of the FLASHBACK_TRANSACTION_QUERY view. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 6

7 See Chapter 15, Manipulating Large Data Sets: Track the Changes to Data over a Period of Time OBJECTIVE: Track the changes to data over a period of time. Examine the following SQL code: CREATE TABLE COURSES (COURSE_ID NUMBER(3), DAYS VARCHAR2(30)); ALTER TABLE COURSES ADD CONSTRAINT CK_DAYS CHECK (DAYS <= 90); INSERT INTO COURSES VALUES (1,100); Which of the following statements is true of the code? B: The INSERT statement will fail due to a constraint violation. B is correct. The INSERT statement will fail due to the violation of the constraint, since you cannot enter a value greater than 90. A, C, and D are incorrect. Even though the constraint uses a mathematical equation on a column that accepts text data, the attempt to create the constraint will succeed. See Chapter 11, Managing Schema Objects: 11:03 Add Constraints OBJECTIVE: Add constraints. Which of the following clauses is used to sort rows in a hierarchical query at the same level without disrupting the tree structure? C: ORDER BY SIBLINGS BY C is correct. ORDER SIBLINGS BY is used to sort rows within a given level, while preserving the hierarchical tree structure of the output. A, B, and D are incorrect. ORDER BY can work syntactically, but the effect is often undesirable, as the tree structure will be disrupted. ORDER BY LEVEL can also work syntactically but will produce a result that sorts rows according to level, which may be useful but will not preserve the tree structure. ORDER BY ROOT includes the word ROOT, which is not a particular keyword and would assume the presence of a column ROOT; it does not answer this question. See Chapter 16, Hierarchical Retrieval: Create and Format Hierarchical Data OBJECTIVE: Create a tree-structured report; format hierarchical data. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 7

8 Click the Exhibit button. Examine the Exhibit. Which of the following SELECT statements will fail with one or more syntax errors? D: SELECT AVG(SUM(DISCOUNT)) FROM SUPPLIERS; D is correct. The AVG of the SUM would only work if the SELECT statement had a valid GROUP BY clause. A, B, and C are incorrect. Answer A is fine, because you may nest an aggregate function within another aggregate function provided that a GROUP BY clause is present in the SELECT statement, and it is in this example. Answer B is valid, but it would not be valid if the COUNT function were omitted. In other words, if the SELECT statement specified ORDER_DATE by itself, then a syntax error would result, indicating that ORDER_DATE is "not a GROUP BY function". If the same use of TO_CHAR were applied to ORDER_DATE as it is used in the GROUP BY clause, then it would work. But in the code shown in Answer B, the aggregate function COUNT is applied to the ORDER_DATE column, which is accepted as a valid SELECT statement. Answer C is valid. The CASE function is a single-row function that processes text data and returns numeric data that feeds into the outer aggregate SUM function. See Chapter 6, Use Single-Row Functions to Customize Output: 6.02 Use Character, Number, and Date Functions in SELECT Statements; also see Chapter 7, Reporting Aggregated Data Using the Group Functions: 7.03 Group Data by Using the GROUP BY Clause OBJECTIVE: Use character, number, and date functions in SELECT statements; also group data by using the GROUP BY clause. Examine the following SQL code: CREATE SEQUENCE CTR MAXVALUE 30 CYCLE; What is the setting for MINVALUE? C: 1 C is correct. The default value for MINVALUE is 1. The sequence will increment to 30, then start over again at 1. A, B, and D are incorrect. The statement will succeed and a sequence will be created. The MINVALUE will not default to 0 or 30, the default is 1. See Chapter 10, Creating Other Schema Objects: 10:02 Create, Maintain, and Use Sequences LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 8

9 OBJECTIVE: Create, maintain, and use sequences. Click the Exhibit button. Examine the Exhibit. You are tasked with producing a report that lists data from the PRODUCTS and SUPPLIERS tables so that the data is grouped by SUPPLIERS.SUPPLIER_NAME and PRODUCTS.PERISHABLE. Furthermore, your output must include the count of each perishable and non-perishable item, as well as a total of all products for that particular supplier. Your report will not include a grand total, however. The output should look something like this: SUPPLIER_NAME P COUNT(*) Acme N 2 Acme Y 3 Acme 5 Joes N 2 Joes 2 Atrexa N 3 Atrexa Y 1 Atrexa 4 Which of the following queries will satisfy this task? A: SELECT S.SUPPLIER_NAME, P.PERISHABLE, COUNT(*) FROM PRODUCTS P JOIN SUPPLIERS S ON P.SUPPLIER_ID = S.SUPPLIER_ID GROUP BY S.SUPPLIER_NAME, ROLLUP(P.PERISHABLE); A is correct. The use of the ROLLUP operator in the second GROUP BY group will produce the correct set of subtotals, without producing a grand total line at the end. B, C, and D are incorrect. The use of ROLLUP on the SUPPLIER_NAME column will product a total for all suppliers, and this violates the task. But the use of ROLLUP on only the PERISHABLE column does accomplish the task. See Chapter 13, Generating Reports by Grouping Related Data: Use the ROLLUP Operation to Produce Subtotal Values OBJECTIVE: Use the ROLLUP operation to produce subtotal values. In a multitable INSERT statement: B: A table alias defined in a multitable INSERT statement subquery is not recognized throughout the rest of the INSERT statement. B is correct. If you create a table alias within the subquery of a multitable INSERT, its scope is that subquery and no further. A, C, and D are incorrect. The keyword ALL is used in both conditional and unconditional LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 9

10 multitable INSERT. Each INSERT statement is not required to specify an identical number of expressions. In a multitable INSERT, the SELECT statement at the end will specify the population from which each INSERT may draw values, but each INSERT is free to specify its own unique set of expressions. Any error in the INSERT will cause the statement to roll back all changes made to the database as a result of that particular execution of the statement. See Chapter 15, Manipulating Large Data Sets: Use the Following Types of Multitable INSERTs: Unconditional, Conditional, and Pivot OBJECTIVE: Use the following types of multitable INSERTs: unconditional, conditional, and pivot. Which of the following keywords will never appear in any valid self-join? B: SELF B is correct. The word SELF is not a valid keyword in SELECT statements. A, C, and D are incorrect. RIGHT, INNER, and JOIN are all valid keywords that may be used with a valid self-join query. See Chapter 8, Displaying Data from Multiple Tables: 8.02 Join a Table to Itself by Using a Self-Join OBJECTIVE: Join a table to itself by using a self-join. Examine the following output: Answer :00: The output is the result of which of the following statements? B: SELECT NUMTODSINTERVAL(120, 'MINUTE') AS "Answer" FROM DUAL; B is correct. The function NUMTODSINTERVAL takes two parameters. The first specifies a value, and the second specifies the interval unit that defines the first parameter. In this particular example, therefore, the value of 120 is specified as 'MINUTES', which must be enclosed in single quotes; the list of parameters must be enclosed in parentheses. The result displayed is a literal value of the datatype INTERVAL DAY TO SECOND. A, C, and D are incorrect. Quotation marks are required with the second parameter of the NUMTODSINTERVAL function, and its corresponding function NUMTOYMINTERVAL. This should not be confused with the syntax for displaying literal datetime intervals-for example, SELECT INTERVAL '120' MINUTE FROM DUAL will return a literal representation of the LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 10

11 same thing; note how quotation marks surround the value, and not the keyword MINUTE in this context. The same is true for the other datetime units of YEAR, MONTH, DAY, and SECOND. The NUMTODSINTERVAL function differs in that its quoted strings are input parameters to the function, and its output represents values of the INTERVAL DAY TO SECOND datatype, suitable for situations where the datatype is strictly required. See Chapter 6, Use Single-Row Functions to Customize Output: 6.02 Use Character, Number, and Date Functions in SELECT Statements OBJECTIVE: Use character, number, and date functions in SELECT statements. Which of the following statements about datatypes is true? C: NUMBER will round off numeric data according to its specified scale. C is correct. While the NUMBER datatype does not require a specified value for scale, any specified value for scale will determine the detail to which numbers will be rounded. A, B, and D are incorrect. The VARCHAR2 datatype does not pad text with blanks to ensure its length is equal to the specified precision-it's the CHAR datatype that does. The VARCHAR2 datatype varies in stored length, up to the specified precision. It's not the DATE datatype that stores fractional seconds, but the TIMESTAMP datatype. Finally, TIMESTAMP WITH LOCAL TIME ZONE does store fractional seconds. See Chapter 2, Using DDL Statements to Create and Manage Tables: 2.04 List the Data Types That Are Available for Columns, and also see Chapter 6, Use Single-Row Functions to Customize Output: 6.04 Manage Data in Different Time Zones-Use Various Datetime Functions OBJECTIVE: Use DDL statements to create and manage tables; manage data in different time zones. Consider the following SQL code: CREATE TABLE PRODUCTS (PRODUCT_ID NUMBER(9) PRIMARY KEY, PRODUCT_TITLE VARCHAR2(20), PERISHABLE CHAR(1), CONSTRAINT CHECK_PERISHABLE CHECK (PERISHABLE IN ('Y','N'))); SET CONSTRAINT CHECK_PERISHABLE DEFERRED; What can be said of these statements? D: None of the above. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 11

12 D is correct. The answer is not included in the options that are shown. The SET CONSTRAINT statement will fail because the CHECK_PERISHABLE constraint needs to be set to DEFERRABLE. The constraint itself must be set to DEFERRABLE; if not, the SET CONSTRAINT statement cannot defer it. A, B, and C are incorrect. The SET CONSTRAINT statement does not have a syntax error. But it will fail at execution because of the constraint, and not the statement's syntax. See Chapter 11, Managing Schema Objects: 11:03 Add Constraints OBJECTIVE: Add constraints. Click the Exhibit button. Examine the Exhibit and the data listing for table PRODUCTS that follows: PRODUCT_ID PRODUCT_TITLE ENTERED TV Stand 01-SEP-09 2 Floor lamp 02-SEP-09 3 Wood shelf 03-SEP-09 4 Cupboard 04-SEP-09 Next, examine the following code: SELECT CASE INSTR(PRODUCT_TITLE,'Wood') WHEN 0 THEN 'No furniture polish required' ELSE 'Recommend furniture polish' END FROM PRODUCTS WHERE PRODUCT_ID = 3; What will the SELECT statement return as output? D: 'Recommend furniture polish'. D is correct. The CASE function is correct and will produce the output specified by the ELSE clause of the CASE function. A, B, and C are incorrect. There is nothing incorrect about the CASE function syntax. CASE can handle numeric or text data. See Chapter 6, Use Single-Row Functions to Customize Output: 6.02 Use Character, Number, and Date Functions in SELECT Statements OBJECTIVE: Use character, number, and date functions in SELECT statements. A user CLEO executes the following SQL statement: CREATE TABLE TAB_PRODUCTS (PRODUCT_ID CHAR(10), PRODUCT_TITLE CHAR(20), CATALOG_NO CHAR(10)) LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 12

13 ORGANIZATION EXTERNAL (TYPE ORACLE_LOADER DEFAULT DIRECTORY IMPORT_PRODUCTS ACCESS PARAMETERS (RECORDS DELIMITED BY NEWLINE SKIP 3 FIELDS (PRODUCT_ID CHAR(10), PRODUCT_TITLE CHAR(20), CATALOG_NO CHAR(10)) ) LOCATION ('IMPORTED_PRODUCTS.ASC') ); Assuming the statement executes successfully and all associated files and objects exist as required, which of the following additional actions must user CLEO take in order to ensure that user TONY will have access to the table TAB_PRODUCTS? (Choose two.) B: GRANT SELECT ON TAB_PRODUCTS TO TONY; C: GRANT READ, WRITE ON DIRECTORY IMPORT_PRODUCTS TO TONY; B and C are correct. A DIRECTORY object is an integral part of an EXTERNAL TABLE. In order for a user to gain access to an EXTERNAL TABLE, the user must also be granted READ and WRITE access to the DIRECTORY object on which the EXTERNAL TABLE is based. A and D are incorrect. You do not grant direct access to the text file itself in any way, this is encompassed within the definition of the EXTERNAL TABLE. The EXTERNAL keyword is not relevant to any GRANT statement. See Chapter 11, Managing Schema Objects: 11:07 Create and Use External Tables OBJECTIVE: Create and use external tables. Click the Exhibit button. Examine the Exhibit. Now look at this code listing: SELECT PRODUCT_TITLE FROM PRODUCTS P WHERE NOT EXISTS (SELECT SUPPLIER_ID FROM SUPPLIERS S WHERE P.SUPPLIER_ID = S.SUPPLIER_ID); If there are no rows in the SUPPLIERS table, which of the following statements is true? C: The query will return all the rows in the PRODUCTS table, regardless of what may be in the SUPPLIER_ID column. C is correct. If the SUPPLIERS table has no rows, then all of the rows in PRODUCTS will be returned. A, B, and D are incorrect. It might be tempting to consider that the NULL rows will somehow reverse the results of the NOT EXISTS, particularly in the event of the NULL values for LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 13

14 SUPPLIER_ID. But SQL will not assume that a lack of information-meaning no rows in SUPPLIERS-compared to a lack of information-meaning no values in PRODUCTS.SUPPLIER_ID-will somehow combine to produce a positive result. The bottom line: none of the PRODUCTS rows can possibly have a match in an empty table, so all of the PRODUCTS rows do not exist in the SUPPLIERS table. See Chapter 9, Retrieving Data Using Subqueries: 9.09 Use the EXISTS and NOT EXISTS Operators OBJECTIVE: Use the EXISTS and NOT EXISTS operators. Click the Exhibit button. Examine the Exhibit. Now examine the following SQL code: SELECT GROUPING(P.PERISHABLE),S.SUPPLIER_NAME, P.PERISHABLE, COUNT(*) FROM PRODUCTS P JOIN SUPPLIERS S ON P.SUPPLIER_ID = S.SUPPLIER_ID GROUP BY ROLLUP(S.SUPPLIER_NAME, P.PERISHABLE); What will be the effect of the GROUPING function in line 1? C: To return a 1 for each subtotal of PERISHABLE and for each subtotal of SUPPLIER_NAME C is correct. The GROUPING function identifies superaggregate rows by returning a value of 1 in each of those rows. In all other rows, the GROUPING function returns a zero. Superaggregate rows, in this example, will include all subtotals of PERISHABLE and higher, which will also include SUPPLIER_NAME. A, B, and D are incorrect. GROUPING does not group rows but rather returns a value of 0 or 1 to identify aggregate and superaggregate rows. See Chapter 13, Generating Reports by Grouping Related Data: Use the GROUPING Function to Identify the Row Values Created by ROLLUP or CUBE OBJECTIVE: Use the GROUPING function to identify the row values created by ROLLUP or CUBE. LearnKey, Inc. Copyright 2006, All Rights Reserved. Page 14