1z1-071関連資格試験対応 & 1z1-071資格復習テキスト

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Oracle 1z1-071 Exam Syllabus Topics:

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

>> 1z1-071関連資格試験対応 <<

1z1-071資格復習テキスト & 1z1-071認定テキスト

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Oracle Database SQL 認定 1z1-071 試験問題 (Q20-Q25):

質問 # 20
View the Exhibit and examine PRODUCTS and ORDER_ITEMS tables.

You executed the following query to display PRODUCT_NAME and the number of times the product has been ordered:
SELECT p.product_name, i.item_cnt
FROM (SELECT product_id, COUNT (*) item_cnt
FROM order_items
GROUP BY product_id) i RIGHT OUTER JOIN products p
ON i.product_id = p.product_id;
What would happen when the above statement is executed?

A. The statement would not execute because the GROUP BY clause cannot be used in the inline view.
B. The statement would execute successfully to produce the required output.
C. The statement would not execute because inline views and outer joins cannot be used together.
D. The statement would not execute because the ITEM_CNT alias cannot be displayed in the outer query.

正解：B

質問 # 21
Evaluate these commands which execute successfully:

Which two statements are true about the ORD_ITEMS table and the ORD_SEQ sequence?

A. Sequence ORD_SEQ is guaranteed not to generate duplicate numbers.
B. Column ORD_NO gets the next number from sequence ORD_SEQ whenever a row is inserted into ORD_ITEMS and no explicit value is given for ORD_NO.
C. Sequence ORD_SEQ cycles back to 1 after every 5000 numbers and can cycle 20 times.
D. If sequence ORD_SEQ is dropped then the default value for column ORD_NO will be NULL for rows inserted into ORD_ITEMS.
E. Any user inserting rows into table ORD_ITEMS must have been granted access to sequence ORD_SEQ.

正解：A、E

質問 # 22
View the Exhibit and examine the description of the DEPARTMENTS and EMPLOYEES tables.

To retrieve data for all the employees for their EMPLOYEE_ID, FIRST_NAME, and DEPARTMENT NAME, the following SQL statement was written:
SELECT employee_id, first_name, department_name
FROM employees NATURAL JOIN departments;
The desired output is not obtained after executing the above SQL statement. What could be the reason for this?

A. The NATURAL JOIN clause is missing the USING clause.
B. The EMPLOYEES and DEPARTMENTS tables have more than one column with the same column name and data type.
C. The table prefix is missing for the column names in the SELECT clause.
D. The DEPARTMENTS table is not used before the EMPLOYEES table in the FROM clause.

正解：B

質問 # 23
Which three statements are true about performing Data Manipulation Language (DML) operations on a view in an Oracle Database?

A. Views cannot be used to add or modify rows in an underlying table if the defining query of the view contains aggregating functions.
B. The WITH CHECK clause has no effect when deleting rows from the underlying table through the view.
C. Insert statements can always be done on a table through a view.
D. Views cannot be used to add rows to an underlying table if the table has columns with NOT NULL constraints lacking default values which are not referenced in the defining query of the view.
E. Views cannot be used to query rows from an underlying table if the table has a PRIMARY KEY and the PRIMARY KEY columns are not referenced in the defining query of the view.
F. Views cannot be used to add or modify rows in an underlying table if the defining query of the view contains the DISTINCT keyword.

正解：A、D、E

質問 # 24
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. RIGHT OUTER JOIN with seif join
B. FULL OUTER JOIN with seif join
C. seif join
D. LEFT OUTER JOIN with seif join
E. SUBQUERY

正解：C、E

解説：
To determine if customer details have been entered more than once using a different custno, the following methods can be used:
SUBQUERY: A subquery can be used to find duplicate names by grouping the names and having a count greater than one.
SELECT custname FROM customers GROUP BY custname HAVING COUNT(custname) > 1; Self Join: A self join can compare rows within the same table to find duplicates in the custname column, excluding matches with the same custno.
SELECT c1.custname FROM customers c1 JOIN customers c2 ON c1.custname = c2.custname AND c1.custno != c2.custno; These methods allow us to find duplicates in the custname column regardless of the custno. Options A, B, and E are not applicable methods for finding duplicates in this context.
Reference:
Oracle Documentation on Group By: Group By
Oracle Documentation on Joins: Joins

質問 # 25
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あらゆる人にとって、時間は非常に大切です。1z1-071試験に対して、いろいろな資料があります。そのような資料を勉強するには、長い時間がかかります。でも、1z1-071問題集を利用すれば、短い時間で1z1-071試験に合格できます。そして、1z1-071問題集は安くて、便利です。誰でも、1z1-071問題集を選択すれば、試験に合格する可能性が大きいです。もし、1z1-071問題集を勉強すれば、もし、将来にITエリートになります。

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