Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE Which UPDATE statement is valid?（）

A . INSERT INTO employees VALUES ( NULL, 'John', 'Smith'); B . INSERT INTO employees( first_name, last_name) VALUES( 'John', 'Smith'); C . INSERT INTO employees VALUES ( '1000', 'John', NULL); D . INSERT INTO employees (first_name, last_name, employee_id) VALUES ( 1000, 'John', 'Smith'); E . INSERT INTO employees (employee_id) VALUES (1000); F . INSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'John', ' ')

A .  displaying the total order value for sales representatives 161 and 163 B .  displaying the order total for sales representative 161 in the year 1999 C .  displaying the number of orders that have order mode online and order date in 1999 D .  displaying the number of orders whose order total is more than the average order total for all online order

A .https://assets.asklib.com/images/image2/2018080115372627375.jpg B .https://assets.asklib.com/images/image2/2018080115372921481.jpg C .https://assets.asklib.com/images/image2/2018080115373261310.jpg D .https://assets.asklib.com/images/image2/2018080115373612285.jpg

A . INSERT INTO employees VALUES (NULL, 'JOHN','Smith'); B . INSERT INTO employees( first_name, last_name) VALUES ('JOHN','Smith'); C . INSERT INTO employees VALUES ('1000','JOHN','NULL'); D . INSERT INTO employees(first_name,last_name, employee_id) VALUES ('1000, 'john','Smith'); E . INSERT INTO employees (employee_id) VALUES (1000); F . INSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'john',")

A . UPDATE employees SET first_name = 'John' SET last_name = 'Smith' WHERE employee_id = 180; B . UPDATE employees SET first_name = 'John', SET last_name = 'Smoth' WHERE employee_id = 180; C . UPDATE employee SET first_name = 'John' AND last_name = 'Smith' WHERE employee_id = 180; D . UPDATE employee SET first_name = 'John', last_name = 'Smith' WHERE employee_id = 180

A . INSERT INTO employees VALUES (NULL, 'John', 'smith'); B . INSERT INTO employees (first_name, last_name) VALUES ('John', 'smith'); C . INSERT INTO employees VALUES ('1000, 'John', 'smith'); D . INSERT INTO employees (first_name, last_name, employee_id) VALUES (1000, 'John', 'smith'); E . INSERT INTO employees (employee_id) VALUES (1000); F . INSERT INTO employees ( employee_id, first_name, last_name, ) VALUES (1000, 'John','')

A . SELECT employee_id, salary, tax_percent FROM employees e, tax t WHERE e.salary BETWEEN t.min_salary AND t.max_salary; B . SELECT employee_id, salary, tax_percent FROM employees e, tax t WHERE e.salary > t.min_salary, tax_percent C . SELECT employee_id, salary, tax_percent FROM employees e, tax t WHERE 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.

A . SELECT a.last_name, a.salary, a.department_id, b.maxsal FROM employees a, (SELECT department_id, max(salary)maxsal FROM employees GROUP BY department_id) b WHERE a.department_id = b.department_id AND a.salary < b.maxsal; B . SELECT a.last name, a.salary, a.department_id FROM employees a WHERE a.department_id IN (SELECT department_id FROM employees b GROUP BY department_id having salary = (SELECT max(salary) from employees)) C . SELECT a.last_name, a.salary, a.department_id FROM employees a WHERE a.salary = (SELECT max(salary) FROM employees b WHERE a.department _ id = b.department _ id); D . SELECT a.last_name, a.salary, a.department_id FROM employees a WHERE (a.department_id, a.salary) IN (SELECT department_id, a.salary) IN (SELECT department_id max(salary) FROM employees b GROUP BY department_id ORDER BY department _ id)

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

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.

A . You have no rows in the table. B . You have an employee with the name of James. C . You cannot roll back to the same savepoint more than once. D . Your last update fails to update any rows because employee ID 180 was already eleted.

A .https://assets.asklib.com/images/image2/2018080115514154580.jpg B .https://assets.asklib.com/images/image2/2018080115514431184.jpg C .https://assets.asklib.com/images/image2/2018080115515173356.jpg D .https://assets.asklib.com/images/image2/2018080115515828127.jpg

A .https://assets.asklib.com/images/image2/2018080116405368499.jpg B .https://assets.asklib.com/images/image2/2018080116405773977.jpg C .https://assets.asklib.com/images/image2/2018080116410160098.jpg D .https://assets.asklib.com/images/image2/2018080116410455360.jpg

A . INSERT INTO employees (employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01/01/01); B . INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01 january 01'); C . INSERT INTO employees(employee_id, first_name, last_name, Hire_date) VALUES (1000, 'John', 'smith', To_ date ('01/01/01)); D . INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01-Jan-01')

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 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 then 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.

是 否

<img src='https://img2.soutiyun.com/shangxueba/ask/17133001-17136000/17135988/ea49d0916fc0345195a85a9b2816cb50.jpg' /> 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 nocommon column between the two tables.

<img src='https://img2.soutiyun.com/ask/uploadfile/2019-03-27/7fe8306013ab0f36b81d7271dc0c8478.jpg' /> A.<img src='https://img2.soutiyun.com/ask/uploadfile/2019-02-04/cfff1591b450909431e9f7ae7dc05f14.jpg' /> B.<img src='https://img2.soutiyun.com/ask/uploadfile/2019-02-04/c75e1c68e2e140258467ca466b7e474d.jpg' /> C.<img src='https://img2.soutiyun.com/ask/uploadfile/2019-02-04/c87a2c32bbbd462ee3f4b85cd7438fec.jpg' /> D.<img src='https://img2.soutiyun.com/ask/uploadfile/2019-02-04/659c7750c67c9d05542c4c9f4dd01ed9.jpg' />

<img src='https://img2.soutiyun.com/ask/uploadfile/2019-03-27/24b8b2127f6e518cc48ee2eb3e995f55.jpg' /> A．<img src='https://img2.soutiyun.com/ask/uploadfile/2019-02-04/9c9ecd3bc18e9e5ea56864899f6fd379.jpg' /> B．<img src='https://img2.soutiyun.com/ask/uploadfile/2019-02-04/ce2012a411732ad75a4283de3ff8e2ef.jpg' /> C．<img src='https://img2.soutiyun.com/ask/uploadfile/2019-02-04/07b2ad486cb8e688542e91608c75a123.jpg' /> D．<img src='https://img2.soutiyun.com/ask/uploadfile/2019-02-04/a73cb726d1955a9d3c0133877ce50fd8.jpg' />

A.SELECT e.last_name, d. department_name, d.location_id FROM employees e NATURAL JOIN departments D USING department_id ; B.SELECT last_name, department_name, location_id FROM employees NATURAL JOIN departments WHERE e.department_id =d.department_id; C.SELECT e.last_name, d.department_name, d.location_id FROM employees e NATURAL JOIN departments d; D.SELECT e.last_name, d.department_name, d.location_id FROM employees e JOIN departments d USING (department_id );

A.nIn 1+1 T-MPLS path protection structure, each working connection has a specific protection connection; working connection and protection connection are bridged at source end of protecting domain. B.nIn 1+1 T-MPLS path protection structure, if a unidirectional fault (transmission direction from node A to Z) occurs to working connection, it will be detected at sink node Z of protecting domain and then node Z selector will switch to protection connectio C.n1+1 T-MPLS path protection is of unidirectional switching type -- that is to say, only the affected end needs to switch to the protection path and selectors at the two ends are relatively independent. D.nIn 1+1 T-MPLS path protection structure, to avoid single-point fault, working connection and protection connection should run along isolated routes. 此题为多项选择题。

A.UPDATE employees SET first_name = ‘John‘ SET last_name =‘Smith‘ WHERE employee_id = 180; B.UPDATE employees SET first_name = ‘John‘, SET last_name =‘Smith‘ WHERE employee_id = 180; C.UPDATE employees SET first_name = ‘John‘ AND last_name =‘Smith‘ WHERE employee_id = 180; D.UPDATE employees SET first_name = ‘John‘, last_name =‘Smith‘ WHERE employee_id = 180;

A. CREATE INDEX NAME _IDX (first_name, last_name); B. CREATE INDEX NAME _IDX (first_name, AND last_name) C. CREATE INDEX NAME_IDX ON (First_name, last_name); D. CREATE INDEX NAME_IDX ON employees (First_name, AND last_name); E. CREATE INDEX NAME_IDX ON employees (First_name, last_name); F. CREATE INDEX NAME_IDX FOR employees (First_name, last_name);

A. DELETE FROM employees WHERE employee_id = (SELECT employee_id FROM employees); B. DELETE * FROM employees WHERE employee_id = (SELECT employee_id FROM new_ employees); C. DELETE FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE name = (‘Carrey‘)‘ D. DELETE * FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE last_ name = (‘Carrey‘)‘