1z1-071 Testking, 1z1-071 Exam

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Oracle 1z1-071 Exam Syllabus Topics:

Section	Weight	Objectives
Manipulating Data	14%	- Control transactions - Update rows in a table - Describe each data manipulation language (DML) statement - Insert rows into a table - Delete rows from a table
Controlling User Access	5%	- Distinguish between granting privileges and roles - Differentiate system privileges from object privileges - Grant privileges on tables and on a view
Using Set Operators	5%	- Describe set operators - Use a set operator to combine multiple queries into a single query - Describe the order of precedence when the set operators are used
Reporting Aggregated Data Using Group Functions	15%	- Include or exclude grouped rows by using the HAVING clause - Describe the use of group functions - Group data by using the GROUP BY clause - Identify the available group functions
Using Subqueries to Solve Queries	15%	- Write single-row and multi-row subqueries - Describe the types of problems that subqueries can solve - List the types of subqueries - Define subqueries
Creating Other Schema Objects	10%	- Create synonyms for database objects - Create and maintain indexes - Create, maintain, and use sequences - Create, modify, and retrieve data from a view
Displaying Data from Multiple Tables Using Joins	15%	- Write SELECT statements to access data from more than one table using equijoins and non-equijoins - Cross join tables - Join a table to itself by using a self-join - View data that does not meet a join condition by using outer joins
Using Conversion Functions and Conditional Expressions	10%	- Apply the NVL, NULLIF, and COALESCE functions to data - Use conditional IF THEN ELSE logic in a SELECT statement - Use the TO_CHAR, TO_NUMBER, and TO_DATE conversion functions - Describe implicit and explicit data type conversion
Using Single-Row Functions to Customize Output	10%	- Manipulate numbers with the ROUND, TRUNC, and MOD functions - Perform arithmetic with date data - Manipulate strings with character function in the SELECT and WHERE clauses - Describe the differences between single-row and multiple-row functions - Manipulate dates with the date functions
Working with Oracle Database SQL	8%	- Describe the Oracle Database architecture - Start up and shut down a database - Explain the installation of Oracle Database software - Create a container database
Retrieving Data Using the SQL SELECT Statement	5%	- Differentiate between SQL statements and SQL*Plus commands - Execute a basic SELECT statement - List the capabilities of SQL SELECT statements
Restricting and Sorting Data	6%	- Limit the rows that are retrieved by a query - Use ampersand substitution to restrict and sort output at runtime - Sort the rows that are retrieved by a query - Use DEFINE and UNDEFINE commands
Using DDL Statements to Create and Manage Tables	14%	- Review the table structure - Alter table definitions - Categorize the main database objects - Create a simple table - Drop, rename, and truncate tables - Describe the data types that are available for columns

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Oracle Database SQL 1z1-071 Prüfungsfragen mit Lösungen (Q296-Q301):

296. Frage
View the exhibit and examine the structure of the EMPLOYEES table.

You want to display all employees and their managers having 100 as the MANAGER_ID.
You want the output in two columns: the first column would have the LAST_NAME of the managers and the second column would have LAST_NAME of the employees.
Which SQL statement would you execute?

A. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE m.manager_id = 100;
B. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON e.employee_id = m.manager_idWHERE m.manager_id = 100;
C. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE e.manager_id = 100;
D. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eWHERE m.employee_id = e.manager_id and AND e.manager_id = 100

Antwort: C

297. Frage
Examine the description of the CUSTONERS table

CUSTON is the PRIMARY KEY.
You must derermine if any customers'derails have entered more than once using a different costno,by listing duplicate name Which two methode can you use to get the requlred resuit?

A. SUBQUERY
B. seif join
C. FULL OUTER JOIN with seif join
D. RIGHT OUTER JOIN with seif join
E. LEFT OUTER JOIN with seif join

Antwort: A,B

298. Frage
Evaluate the following two queries:

Which statement is true regarding the above two queries?

A. Performance would degrade in query 2.
B. Performance would improve in query 2 only if there are null values in the CUST_CREDIT_LIMIT column.
C. There would be no change in performance.
D. Performance would improve in query 2.

Antwort: C

299. Frage
Which three statements are correct regarding indexes? (Choose three.)

A. For each DML operation performed on a table, the corresponding indexes are automatically updated if required.
B. A non-deferrable PRIMARY KEY or UNIQUE KEY constraint in a table automatically attempts to create a unique index.
C. Indexes should be created on columns that are frequently referenced as part of any expression.
D. When a table is dropped, corresponding indexes are automatically dropped.

Antwort: A,B,D

Begründung:
Explanation
http://viralpatel.net/blogs/understanding-primary-keypk-constraint-in-oracle/

300. Frage
Examine the structure of the employees table:

There is a parent/child relationship between EMPLOYEE_ID and MANAGER_ID.
You want to display the name, joining date, and manager for all the employees. Newly hired employees are yet to be assigned a department or a manager. For them, 'No Manager1 should be displayed in the manager column.
Which SQL query gets the required output?

A. SELECT e.last_name, e.hire_date, NVL(m.last_name, 'No Manager') Manager FROM employees e RIGHT OUTER JOIN employees m ON (e.manager_id = m.employee_id);
B. SELECT e.last_name/ e.hire_date/ NVL(m.last_______ name/ 'No Manager') Manager FROM employees e NATURAL JOIN employees m ON (e.manager_id = m.employee_id);
C. SELECT e.last_name, e.hire_date, NVL(m.last_name, 'No Manager') Manager FROM employees e LEFT OUTER JOIN employees m ON (e.manager_id = m.employee_id);
D. SELECT e.last_name, e.hire_date, NVL(m.last_name, 'No Manager') Manager FROM employees e JOIN employees m ON (e.manager_id = m.employee_id);

Antwort: C

Begründung:
We should use LEFT OUTER JOIN as we want to display employees which have no (have NULL values for) managers.

301. Frage
......

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