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Oracle 1z0-071 Exam Syllabus Topics:
| Section | Objectives |
|---|
| SQL Functions | - Conversion and conditional functions
- 1. TO_CHAR, TO_NUMBER, CASE expressions
- Single-row functions
- 1. Character, numeric, date functions
|
| Relational Database Concepts | - Tables, rows, columns, and relational model basics - Data types and database objects overview
|
| SQL Retrieval | - SELECT statements and filtering data
- 1. ORDER BY and row limiting
- 2. WHERE clause and conditions
- Joins and set operations
- 1. UNION, INTERSECT, MINUS
- 2. INNER, OUTER, CROSS joins
|
| Data Manipulation and Schema Objects | - DDL operations
- 1. CREATE, ALTER, DROP tables and constraints
- DML operations
- 1. INSERT, UPDATE, DELETE, MERGE
|
| Subqueries and Advanced SQL | - Single-row and multi-row subqueries - Correlated subqueries and nested queries
|
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Oracle Database SQL 認定 1z0-071 試験問題 (Q110-Q115):
質問 # 110
Examine these two queries and their output:
SELECT deptno, dname FROM dept;

SELECT ename, job, deptno FROM emp ORDER BY deptno;

Now examine this query:
SELECT ename, dname
FROM emp CROSS JOIN dept WHERE job = 'MANAGER'
AND dept.deptno IN (10, 20) ;
正解:C
解説:
In a CROSS JOIN (also known as a Cartesian join), each row from the first table (emp) is joined to every row of the second table (dept). When we apply the filter for job = 'MANAGER' and dept.deptno IN (10, 20), we are restricting the results to managers in departments 10 and 20.
From the given data:
* There are 2 managers in department 10 (CLARK and KING).
* There is 1 manager in department 20 (JONES).
With a cross join, each of these emp records will join with each dept record where deptno is 10 or 20. Since there are two departments (10 and 20) in the dept table, each employee will match with two departments.
Therefore, the result set will be:
* CLARK with accounting (dept 10) and research (dept 20)
* KING with accounting (dept 10) and research (dept 20)
* JONES with accounting (dept 10) and research (dept 20)
That gives us a total of 6 rows, which means the correct answer is B: 6.
質問 # 111
Which two queries execute successfully?
- A. SELECT prod_id, exp_date FROM products
UNION ALL
SELECT prod_id, NULL FROM new_products; - B. SELECT k FROM products
MINUS
SELECT prod_id FROM new_products; - C. SELECT prod_id FROM products
UNION ALL
SELECT prod_id, prod_name FROM new_products; - D. SELECT prod_id, prod_name FROM products
INTERSECT
SELECT 100, prod_name FROM newproducts; - E. SELECT * FROM products
UNION
SELECT * FROM new_products;
正解:A、E
解説:
For SQL queries involving set operations like UNION, UNION ALL, INTERSECT, and MINUS, the data types and number of columns in the SELECT statements being combined must be the same. Here is a breakdown of each query option:
* Option A: This query successfully combines results from two tables using UNION ALL, which combines all rows from both SELECT statements, including duplicates. Both queries select two columns (prod_id and exp_date, with exp_date being NULL in the second query). This is valid as the number of columns and their data types match.
* Option B: This query attempts to use INTERSECT, which returns only the rows common to both SELECT statements. However, the first SELECT statement chooses prod_id and prod_name, while the second only substitutes a constant value for prod_id without specifying from which table it selects prod_name. If "newproducts" (presumably a typo and should be "new_products") does not have the exact columns as "products", or if the data types do not match, it will fail.
* Option C: This query uses UNION, which removes duplicates from the combined dataset. The use of SELECT * assumes that both "products" and "new_products" tables have the same columns in the same order with matching data types. This query will succeed if those conditions are met.
* Option D: This query uses MINUS to subtract results of the second SELECT from the first. However, it fails because 'k' does not explicitly match a column name in "products", suggesting a potential error unless 'k' is indeed a valid column.
* Option E: This query fails because the first SELECT statement fetches only one column (prod_id) while the second fetches two columns (prod_id and prod_name). UNION ALL requires the same number of columns with compatible types in all SELECT statements involved.
質問 # 112
Examine these statements:

Which two are true?
- A. The DEPTNO column in the EMP table will be able to contain null values.
- B. The create TABLE EMP statement must precede the create table DEPT statement for all three statements to execute successfully.
- C. Both foreign key constraint definition must be removed from the CREATE statements, and be added with ALTER TABLE statements once both tables are created, for the two CREATE TABLE statements to execute successfully in the order shown.
- D. The DEPT_FKEY constraint definition must be removed from the CREATE TABLE DEPT statement. and the added with an D. ALTER statement once both tables are created, for the two CREATE TABLE statements to execute successfully in the order shown.
- E. All three statements execute successfully in the order shown.
- F. The MGR column In the DEPT table will not be able to contain null values.
正解:A、D
質問 # 113
Examine the description or the CUSTOMERS table:

For Customers whose income level has a value, you want to display the first name and due amount as 5% of their credit limit. Customers whose due amount is null should not be displayed.
Which query should be used?
- A. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers WHERE cust_income_level != NULL AND due_amount != NULL;
- B. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers WHERE cust_income_level IS NOT NULL AND due_amount IS NOT NULL;
- C. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers WHERE cust_income_level IS NOT NULL AND cust_credit_limit IS NOT NULL;
- D. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers WHERE cust_income_level <> NULL AND due_amount <> NULL;
- E. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMOUNT FROM customers WHERE cust_income_level != NULL AND cust_credit_level != NULL;
正解:C
質問 # 114
Which two are true about external tables that use the ORACLE _DATAPUMP access driver?
- A. Creating an external table creates a dump file that can be used only by an external table in the same database.
- B. Creating an external table creates a dump file that can be used by an external table in the same or a different database.
- C. When creating an external table, data can be selected from another external table or from a table whose rows are stored in database blocks.
- D. Creating an external table creates a directory object.
- E. When creating an external table, data can be selected only from a table whose rows are stored in database blocks.
正解:C
解説:
External tables using the ORACLE_DATAPUMP access driver have specific characteristics:
C . Creating an external table using the ORACLE_DATAPUMP access driver allows you to select data from other tables, including another external table or a regular table whose rows are stored in database blocks.
A, B, D, and E are incorrect. Specifically:
A is incorrect because creating an external table does not automatically create a directory object; the directory object must exist prior to or be created separately.
B is incorrect as it limits the creation to tables stored in database blocks, which is not a restriction.
D is incorrect because creating an external table does not create a dump file; it reads from an existing dump file created by Data Pump.
E is also incorrect because the dump file is not created as part of the external table creation process.
Reference:
Oracle Database SQL Language Reference, 12c Release 1 (12.1): "CREATE TABLE" for External Tables
質問 # 115
......
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