A.CREATE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20);
B.CREATE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20) WITH READ ONLY;
C.CREATE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20) WITH CHECK OPTION;
D.CREATE FORCE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20);
E.CREATE FORCE VIEW emp_vu AS SELECT * FROM employees WHERE department_id IN (10,20) NO UPDATE;
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Evaluate this SQL statement:
SELECT e.employee_id, (.15* e.salary) + (.5 * e.commission_pct)
+ (s.sales_amount * (.35 * e.bonus)) AS CALC_VALUE
FROM employees e, sales s
WHERE e.employee_id = s.emp_id;
What will happen if you remove all the parentheses from the calculation?()
A.The value displayed in the CALC_VALUE column will be lower.
B.The value displayed in the CALC_VALUE column will be higher.
C.There will be no difference in the value displayed in the CALC_VALUE column.
D.An error will be reported.
Click the Exhibit button to examine the structures of the EMPLOYEES and TAX tables.
You need to find the percentage tax applicable for each employee. Which SQL statement would you use?()
A.SELECT employee_id, salary, tax_percent FROM employees e JOIN tax t ON e.salary BETWEEN t.min_salary AND t.max_salary;
B.SELECT employee_id, salary, tax_percent FROM employees e JOIN tax t WHERE e.salary > t.min_salary AND < t.max_salary;
C.SELECT employee_id, salary, tax_percent FROM employees e JOIN tax t ON (MIN(e.salary) = t.min_salary AND MAX(e.salary) = t.max_salary);
D.You cannot find the information because there is no common column between the two tables.
Click the Exhibit button and examine the data in the EMPLOYEES and DEPARTMENTS tables.
You want to retrieve all employees' last names, along with their managers' last names and their department names. Which query would you use?()
A.SELECT last_name, manager_id, department_name FROM employees e FULL OUTER JOIN departments d ON (e.department_id = d.department_id);
B.SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);
C.SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);
D.SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);
E.SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);
F.SELECT last_name, manager_id, department_name FROM employees e JOIN departments d ON (e.department_id = d.department_id) ;
Click the Exhibit button to examine the structures of the EMPLOYEES, DEPARTMENTS, and TAX tables.
For which situation would you use a nonequijoin query?()
A.to find the tax percentage for each of the employees
B.to list the name, job_id, and manager name for all the employees
C.to find the name, salary, and the department name of employees who are not working with Smith
D.to find the number of employees working for the Administrative department and earning less than 4000
E.to display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned
You need to create a table named ORDERS that contains four columns:
1.an ORDER_ID column of number data type
2.a CUSTOMER_ID column of number data type
3.an ORDER_STATUS column that contains a character data type
4.a DATE_ORDERED column to contain the date the order was placed
When a row is inserted into the table, if no value is provided for the status of the order, the value PENDING should be used instead.
Which statement accomplishes this?()
A.CREATE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status NUMBER(10) DEFAULT 'PENDING', date_ordered DATE );
B.CREATE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) = 'PENDING', date_ordered DATE );
C.CREATE OR REPLACE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) DEFAULT 'PENDING', date_ordered DATE );
D.CREATE OR REPLACE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) = 'PENDING', date_ordered DATE );
E.CREATE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) DEFAULT 'PENDING', date_ordered DATE );
F.CREATE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) DEFAULT 'PENDING', date_ordered VARCHAR2 );
Examine the SQL statement that creates ORDERS table:
CREATE TABLE orders
(SER_NO NUMBER UNIQUE,
ORDER_ID NUMBER, ORDER_DATE DATE NOT NULL,
STATUS VARCHAR2(10)
CHECK (status IN ('CREDIT', 'CASH')),
PROD_ID NUMBER
REFERENCES PRODUCTS(PRODUCT_ID),
ORD_TOTAL NUMBER,
PRIMARY KEY (order_id, order_date));
For which columns would an index be automatically created when you execute the above SQL statement? ()
A.SER_NO
B.ORDER_ID
C.STATUS
D.PROD_ID
E.ORD_TOTAL
F.composite index on ORDER_ID and ORDER_DATE
You own a table called EMPLOYEES with this table structure:
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2(25)
LAST_NAME VARCHAR2(25)
HIRE_DATE DATE
What happens when you execute this DELETE statement?
DELETE employees;()
A.You get an error because of a primary key violation.
B.The data and structure of the EMPLOYEES table are deleted.
C.The data in the EMPLOYEES table is deleted but not the structure.
D.You get an error because the statement is not syntactically correct.
The CUSTOMERS table has these columns:
CUSTOMER_ID NUMBER(4) NOT NULL
CUSTOMER_NAME VARCHAR2(100) NOT NULL
STREET_ADDRESS VARCHAR2(150)
CITY_ADDRESS VARCHAR2(50)
STATE_ADDRESS VARCHAR2(50)
PROVINCE_ADDRESS VARCHAR2(50)
COUNTRY_ADDRESS VARCHAR2(50)
POSTAL_CODE VARCHAR2(12)
CUSTOMER_PHONE VARCHAR2(20)
The CUSTOMER_ID column is the primary key for the table.
You need to determine how dispersed your customer base is. Which expression finds the number of different countries represented in the CUSTOMERS table?()
A.COUNT(UPPER(country_address))
B.COUNT(DIFF(UPPER(country_address)))
C.COUNT(UNIQUE(UPPER(country_address)))
D.COUNT DISTINCT UPPER(country_address)
E.COUNT(DISTINCT (UPPER(country_address)))
A.ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
B.ALTER TABLE customers MODIFY CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
C.ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn NOT NULL;
D.ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn IS NOT NULL;
E.ALTER TABLE customers MODIFY name CONSTRAINT cust_name_nn NOT NULL;
F.ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name NOT NULL;
Click the Exhibit button and examine the data from the EMP table.
The COMMISSION column shows the monthly commission earned by the employee.
Which two tasks would require subqueries or joins in order to be performed in a single step? ()
A.listing the employees who earn the same amount of commission as employee 3
B.finding the total commission earned by the employees in department 10
C.finding the number of employees who earn a commission that is higher than the average commission of the company
D.listing the departments whose average commission is more than 600
E.listing the employees who do not earn commission and who are working for department 20 in descending order of the employee ID
F.listing the employees whose annual commission is more than 6000
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Evaluate the SQL statement: TRUNCATE TABLE DEPT; Which three are true about the SQL statement? ()
You need to design a student registration database that contains several tables storing academic information.The STUDENTS table stores information about a student. The STUDENT_GRADES table storesinformation about the student's grades. Both of the tables have a column named STUDENT_ID. The STUDENT_ID column in the STUDENTS table is a primary key.You need to create a foreign key on the STUDENT_ID column of the STUDENT_GRADES table thatpoints to the STUDENT_ID column of the STUDENTS table. Which statement creates the foreign key?()
Which three are true? ()
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