Oracle 1Z0-051 問題集

質問 1：
Here is the structure and data of the CUST_TRANS table: Exhibit:

Dates are stored in the default date format dd-mm-rr in the CUST_TRANS table.
Which three SQL statements would execute successfully? (Choose three.)
A. SELECT * FROM cust_trans WHERE transdate = '01-01-07';
B. SELECT transdate + '10' FROM cust_trans;
C. SELECT * FROM cust_trans WHERE transdate='01-JANUARY-07';
D. SELECT transamt FROM cust_trans WHERE custno > '11';
E. SELECT custno + 'A' FROM cust_trans WHERE transamt > 2000;
正解：B,C,D

質問 2：
Which three statements are true regarding the data types in Oracle Database 10g/11g? (Choose three.)
A. A TIMESTAMP data type column stores only time values with fractional seconds
B. The value for a CHAR data type column is blank-padded to the maximum defined column width
C. The BLOB data type column is used to store binary data in an operating system file
D. The minimum column width that can be specified for a VARCHAR2 data type column is one
E. Only One LONG column can be used per table
正解：B,D,E
解説: (Topexam メンバーにのみ表示されます)

質問 3：
Which two statements are true regarding indexes? (Choose two.)
A. They can be created on tables and simple views.
B. You can create only one index by using the same columns.
C. They can be created on tables and clusters.
D. You can create more than one index by using the same columns if you specify distinctly
different combinations of the columns.
正解：C,D

質問 4：
View the Exhibit and examine the structure of the CUSTOMERS table. Evaluate the following SQL statement:

Which statement is true regarding the outcome of the above query?

A. It returns an error because WHERE and HAVING clauses cannot be used to apply conditions on the same column.
B. It returns an error because the BETWEEN operator cannot be used in the HAVING clause.
C. It executes successfully.
D. It returns an error because WHERE and HAVING clauses cannot be used in the same SELECT statement.
正解：C

質問 5：
What is true about sequences?
A. Once created, a sequence is automatically available to all users.
B. Once created, a sequence is linked to a specific table.
C. Only the DBA can control which sequence is used by a certain table.
D. Once created, a sequence belongs to a specific schema.
E. Once created, a sequence is automatically used in all INSERT and UPDATE statements.
正解：D

質問 6：
Examine the data in the CUST_NAME column of the CUSTOMERS table.
CUST_NAME
Renske Ladwig
Jason Mallin
Samuel McCain
Allan MCEwen
Irene Mikkilineni
Julia Nayer
You need to display customers' second names where the second name starts with "Mc" or "MC."
Which query gives the required output?
A. SELECT SUBSTR(cust_name, INSTR(cust_name,' ')+1)
FROM customers
WHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,' ')+1)) LIKE 'Mc%';
B. SELECT SUBSTR(cust_name, INSTR(cust_name,' ')+1)
FROM customers
WHERE SUBSTR(cust_name, INSTR(cust_name,' ')+1) LIKE INITCAP('MC%');
C. SELECT SUBSTR(cust_name, INSTR(cust_name,' ')+1)
FROM customers
WHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,' ')+1)) = INITCAP('MC%');
D. SELECT SUBSTR(cust_name, INSTR(cust_name,' ')+1)
FROM customers
WHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,' ')+1))='Mc';
正解：A

質問 7：
View the Exhibit and examine the description for the CUSTOMERS table.

You want to update the CUST_INCOME_LEVEL and CUST_CREDIT_LIMIT columns for the customer with the CUST_ID 2360. You want the value for the CUST_INCOME_LEVEL to have the same value as that of the customer with the CUST_ID 2560 and the CUST_CREDIT_LIMIT to have the same value as that of the customer with CUST_ID 2566.
Which UPDATE statement will accomplish the task?
A. UPDATE customers SET (cust_income_level,cust_credit_limit) = (SELECT cust_income_level, cust_credit_limit FROM customers WHERE cust_id=2560 OR cust_id=2566) WHERE cust_id=2360;
B. UPDATE customers SET (cust_income_level,cust_credit_limit) = (SELECT cust_income_level, cust_credit_limit
FROM customers
WHERE cust_id IN(2560, 2566)
WHERE cust_id=2360;
C. UPDATE customers SET cust_income_level = (SELECT cust_income_level FROM customers WHERE cust_id = 2560), cust_credit_limit = (SELECT cust_credit_limit FROM customers WHERE cust_id = 2566) WHERE cust_id=2360;
D. UPDATE customers SET (cust_income_level,cust_credit_limit) = (SELECT cust_income_level, cust_credit_limit FROM customers WHERE cust_id=2560 AND cust_id=2566) WHERE cust_id=2360;
正解：C
解説: (Topexam メンバーにのみ表示されます)