1z0-071日本語 & 1z0-071出題範囲

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Oracle 1z0-071 Exam Syllabus Topics:

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

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Oracle Database SQL 認定 1z0-071 試験問題 (Q377-Q382):

質問 # 377
View the Exhibit and examine the structure of the PRODUCTS table.

You must display the category with the maximum number of items.
You issue this query:

What is the result?

A. It generates an error because = is not valid and should be replaced by the IN operator.
B. It generate an error because the subquery does not have a GROUP BY clause.
C. It executes successfully and gives the correct output.
D. It executes successfully but does not give the correct output.

正解：B

質問 # 378
In which three situations does a new transaction always start? (Choose three.)

A. when issuing a CREATE TABLEstatement after a SELECTstatement was issued in the same session
B. when issuing a SELECT FOR UPDATEstatement after a CREATE TABLE AS SELECTstatement was issued in the same session
C. when issuing a CREATE INDEXstatement after a CREATE TABLEstatement completed successfully in the same session
D. when issuing the first Data Manipulation Language (DML) statement after a COMMIT or ROLLBACK statement was issued in the same session
E. when issuing a DML statement after a DML statement failed in the same session
F. when issuing a TRUNCATEstatement after a SELECT statement was issued in the same session

正解：B、D、E

解説：
Explanation/Reference: https://docs.oracle.com/cd/B19306_01/server.102/b14220/transact.htm

質問 # 379
Examine the following query:
SQL> SELECT prod_id, amount_sold
FROM sales
ORDER BY amount_sold
FETCH FIRST 5 PERCENT ROWS ONLY;
What is the output of this query?

A. It results in an error because the ORDER BYclause should be the last clause.
B. It displays 5 percent of the products with the lowest amount sold.
C. It displays 5 percent of the products with the highest amount sold.
D. It displays the first 5 percent of the rows from the SALEStable.

正解：B

解説：
Explanation
Explanation/Reference:
References:
https://oracle-base.com/articles/12c/row-limiting-clause-for-top-n-queries-12cr1

質問 # 380
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 eWHERE m.employee_id = e.manager_id and AND e.manager_id = 100
C. 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;
D. 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;

正解：D

質問 # 381
View the Exhibit and examine the description of the EMPLOYEES table.

You want to calculate the total renumeration for each employee. Total renumeration is the sum of the annual salary and the percentage commission earned for a year. Only a few employees earn commission.
Which SQL statement would you execute to get the desired output?

A. SELECT first_name, salary (salary + NVL (commission_pct, 0)*salary)*12 "Total" FROM EMPLOYEES;
B. SELECT first_name, salary*12 + NVL(salary,0)*commission_pct, "Total" FROM EMPLOYEES;
C. SELECT first_name, salary, salary*12+(salary*NVL2 (commission_pct,
salary,salary+commission_pct))"Total"
FROM EMPLOYEES;
D. SELECT first_name, salary, salary*12+salary*commission_pct "Total" FROM EMPLOYEES;

正解：C

質問 # 382
......

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