1z0-071 Studienmaterialien: Oracle Database SQL & 1z0-071 Zertifizierungstraining

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Oracle 1z0-071 Exam Syllabus Topics:

Section	Weight	Objectives
Working with Time Zones and Date/Time Data		- TIMESTAMP, INTERVAL data types - Time zone conversion functions
Using Subqueries to Solve Queries		- Correlated subqueries for update/delete - Single-row and multi-row subqueries
Using Single-Row Functions to Customize Output	20%	- Character, number, and date functions - Nested functions usage
Managing Indexes, Synonyms and Sequences		- Create and use synonyms - Create and use sequences - Create and maintain indexes
Conversion Functions and Conditional Expressions	10%	- Implicit and explicit data type conversion - NVL, NULLIF, COALESCE and CASE expressions - TO_CHAR, TO_NUMBER, TO_DATE functions
Managing Tables and Constraints		- Temporary and external tables - Constraints: primary key, foreign key, unique, check, not null - Data types, creating/altering/dropping tables
Managing Views		- Updatable and read-only views - Create, modify and drop views
Relational Database Concepts	10%	- ERD and SQL statement mapping - Theoretical and physical aspects of relational databases - Relationship between database and SQL
Displaying Data from Multiple Tables	15%	- Cartesian products - Self-joins and outer joins - Equijoins and non-equijoins
Manipulating Data and Managing Transactions		- Transaction control: COMMIT, ROLLBACK, SAVEPOINT - Multi-table inserts - INSERT, UPDATE, DELETE, MERGE statements
Using SET Operators		- UNION, UNION ALL, INTERSECT, MINUS operators - Ordering results with set operations
Restricting and Sorting Data	15%	- ORDER BY clause and sorting rules - Operator precedence - WHERE clause and filtering conditions - Substitution variables, DEFINE and VERIFY commands
Using Data Dictionary Views		- Retrieve metadata and object information
Retrieving Data Using SQL SELECT Statement	20%	- SELECT statement syntax and usage - Arithmetic operations and NULL handling - Column aliases, concatenation, literals and DISTINCT
Controlling User Access		- Roles, granting and revoking privileges - System and object privileges
Aggregating Data Using Group Functions	10%	- Handling NULLs in group functions - SUM, AVG, COUNT, MAX, MIN functions - GROUP BY and HAVING clauses

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Durch Oracle 1z0-071 Zertifizierungsprüfung wird sich viel Wandel bei Ihnen vollziehen. Beispielsweise werden Ihr Beruf und Leben sicher viel verbessert, weil die Oracle 1z0-071 Zertifizierungsprüfung sowieso eine ziemlich wichtige Prüfung ist. Aber so einfach ist es nicht, diese Prüfung zu bestehen.

Oracle Database SQL 1z0-071 Prüfungsfragen mit Lösungen (Q89-Q94):

89. Frage
Examine the description of the PRODUCT_INFORMATION table:

Which query retrieves the number of products with a null list price?

A. SELECT COUNT (list_price) FROM product_information WHERE list_price = NULL;
B. SELECT COUNT (list_price) FROM product_information WHERE list_price IS NULL;
C. SELECT COUNT (DISTINCT list_price) FROM product_information WHERE list_price IS NULL;
D. SELECT COUNT(NVL(list_price, 0)) FROM product_information WHERE list_price IS NULL;

Antwort: D

90. Frage
Examine these statements:
CREATE TABLE dept (
deptno NUMBER PRIMARY KEY,
diname VARCHAR2(10) ,
mgr NUMBER ,
CONSTRAINT dept_fkey FOREIGN KEY(mgr) REFERENCES emp (empno));
CREATE TABLE emp (
Empno NUMBER PRIMARY KEY,
Ename VARCHAR2 (10) ,
deptno NUMBER,
CONSTRAINT emp_fkey FOREIGN KEY (deptno) REFERENCES dept (deptno) DISABLE);
ALTER TABLE emp MODIFY CONSTRAINT emp_fkey ENABLE;
Which two are true?

A. The Deptno column in the emp table will be able to contain nulls values.
B. All three statements execute successfully in the order shown
C. The DEFT FKEY constraint definition must be removed from the CREATE TABLE DEF statement.and be added with an AITER TABLE statement once both tables are created, for the two CREATE TABLE statements
to execute successfully in the order shown.
D. Both foreign key constraint definitions must be removed from the CREATE TABLE statements, andbe added with ALTER TABLE statements once both tables are created, for the two CREATE TABLEstatements to
execute successfully in the order shown.
E. The CREATE TABLE EMP statement must precede the CREATE TABLE DEPT statement for all threestatements to execute successfully.
F. The MGR column in the DEPT table will not be able to contain NULL values.

Antwort: A,C

91. Frage
View the Exhibit and examine the structure of the ORDERS table.

You must select ORDER_ID and ORDER_DATE for all orders that were placed after the last order placed by CUSTOMER_ID 101.
Which query would give you the desired result?

A. SELECT order_id, order_date FROM orders
WHERE order_date > ALL
(SELECT order_date FROM orders WHERE customer_id = 101);
B. SELECT order_id, order_date FROM orders
WHERE order_date > ALL
(SELECT MAX(order_date) FROM orders ) AND customer_id = 101;
C. SELECT order_id, order_date FROM orders
WHERE order_date >
ANY
(SELECT order_date FROM orders WHERE customer_id = 101);
D. SELECT order_id, order_date FROM orders
WHERE order_date > IN
(SELECT order_date FROM orders WHERE customer_id = 101);

Antwort: A

92. Frage
See the Exhibit and examine the structure of the PROMOTIONS table:

Using the PROMOTIONS table, you need to find out the average cost for all promos in the range
$0-2000 and $2000-5000 in category A.
You issue the following SQL statements:

What would be the outcome?

A. It generates an error because CASE cannot be used with group functions
B. It generates an error because NULL cannot be specified as a return value
C. It executes successfully and gives the required result
D. It generates an error because multiple conditions cannot be specified for the WHEN clause

Antwort: C

Begründung:
CASE Expression
Facilitates conditional inquiries by doing the work of an IF-THEN-ELSE statement:
CASE expr WHEN comparison_expr1 THEN return_expr1
[WHEN comparison_expr2 THEN return_expr2
WHEN comparison_exprn THEN return_exprn
ELSE else_expr]
END

93. Frage
View the exhibit and examine the ORDERS table.
ORDERS
Name
Null?
Type
ORDER ID
NOT NULL
NUMBER(4)
ORDATE DATE
DATE
CUSTOMER ID
NUMBER(3)
ORDER TOTAL
NUMBER(7,2)
The ORDERS table contains data and all orders have been assigned a customer ID. Which statement would add a NOT NULL constraint to the CUSTOMER_ID column?

A. ALTER TABLE ordersMODIFY customer_id CONSTRAINT orders_cust_nn NOT NULL (customer_id);
B. ALTER TABLE ordersMODIFY CONSTRAINT orders_cust_id_nn NOT NULL (customer_id);
C. ALTER TABLE ordersADD CONSTRAINT orders_cust_id_nn NOT NULL (customer_id);
D. ALTER TABLE ordersADD customer_id NUMBER(6)CONSTRAINT orders_cust_id_nn NOT NULL;

Antwort: A

94. Frage
......

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