CREATE INDEX NAME _IDX (first_name, last_name);
CREATE INDEX NAME _IDX (first_name, AND last_name)
CREATE INDEX NAME_IDX ON (First_name, last_name);
CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);
CREATE INDEX NAME_IDX ON employees (First_name, last_name);
CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
第1题:
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.
第2题:
Examine the data of the EMPLOYEES table.EMPLOYEES (EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID)Which statement lists the ID, name, and salary of the employee, and the ID and name of the employee‘s manager, for all the employees who have a manager and earn more than 4000?()
A.
B.
C.
D.
E.
第3题:
Examine the data in the EMPLOYEES and DEPARTMENTS tables:Also examine the SQL statements that create the EMPLOYEES and DEPARTMENTS tables:On the EMPLOYEES table, EMPLOYEE_ID is the primary key.MGR_ID is the ID of managers and refers to the EMPLOYEE_ID.DEPT_ID is foreign key to DEPARTMENT_ID column of the DEPARTMENTS table.On the DEPARTMENTS table, DEPARTMENT_ID is the primary key.Examine this DELETE statement:What happens when you execute the DELETE statement?()
A. Only the row with department ID 40 is deleted in the DEPARTMENTS table.
B. The statement fails because there are child records in the EMPLOYEES table with department ID 40.
C. The row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 110 and 106 are deleted from the EMPLOYEES table.
D. The row with department ID 40 is deleted in the DEPARTMENTS table. Also the rows with employee IDs 106 and 110 and the employees working under employee 110 are deleted from the EMPLOYEES table.
E. The row with department ID 40 is deleted in the DEPARTMENTS table. Also all the rows in the EMPLOYEES table are deleted.
F. The statement fails because there are no columns specifies in the DELETE clause of the DELETE statement.
第4题:
Examine the data in the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMP_NAME DEPT_ID MGR_ID JOB_ID SALARY EMPLOYEE_ID 101 Smith 20 120 SA_REP 4000 102 Martin 10 105 CLERK 2500 103 Chris 20 120 IT_ADMIN 4200 104 John 30 108 HR_CLERK 2500 105 Diana 30 108 IT_ADMIN 5000 106 Smith 40 110 AD_ASST 3000 108 Jennifer 30 110 HR_DIR 6500 110 Bob 40 EX_DIR 8000 120 Ravi 20 110 SA*DIR 6500 DEPARTMENTS DEPARTMENT_ID DEPARTMENT_NAME 10 Admin 20 Education 30 IT 40 Human Resources Also examine the SQL statements that create the EMPLOYEES and DEPARTMENTS tables: CREATE TABLE departments (department_id NUMBER PRIMARY KEY, department _ name VARCHAR2(30)); CREATE TABLE employees (EMPLOYEE_ID NUMBER PRIMARY KEY, EMP_NAME VARCHAR2(20), DEPT_ID NUMBER REFERENCES departments(department_id), MGR_ID NUMBER REFERENCES employees(employee id), MGR_ID NUMBER REFERENCES employees(employee id), JOB_ID VARCHAR2(15). SALARY NUMBER); ON the EMPLOYEES, On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. DEPT_ID is foreign key to DEPARTMENT_ID column of the DEPARTMENTS table. On the DEPARTMENTS table, DEPARTMENT_ID is the primary key. Examine this DELETE statement: DELETE FROM departments WHERE department id = 40; What happens when you execute the DELETE statement?()
第5题:
You get an error because of a primary key violation.
The data and structure of the EMPLOYEES table are deleted.
The data in the EMPLOYEES table is deleted but not the structure.
You get an error because the statement is not syntactically correct.
第6题:
You get an error because of a primary key violation.
The data and structure of the EMPLOYEES table are deleted.
The data in the EMPLOYEES table is deleted but not the structure.
You get an error because the statement is not syntactically correct.
第7题:
UPDATE employees SET first_name = 'John' SET last_name = 'Smith' WHERE employee_id = 180;
UPDATE employees SET first_name = 'John', SET last_name = 'Smoth' WHERE employee_id = 180;
UPDATE employee SET first_name = 'John' AND last_name = 'Smith' WHERE employee_id = 180;
UPDATE employee SET first_name = 'John', last_name = 'Smith' WHERE employee_id = 180;
第8题:
UPDATE employees SET first_name = 'John' SET last_name ='Smith' WHERE employee_id = 180;
UPDATE employees SET first_name = 'John', SET last_name ='Smith' WHERE employee_id = 180;
UPDATE employees SET first_name = 'John' AND last_name ='Smith' WHERE employee_id = 180;
UPDATE employees SET first_name = 'John', last_name ='Smith' WHERE employee_id = 180;
第9题:
INSERT INTO employees VALUES (NULL, 'JOHN','Smith');
INSERT INTO employees( first_name, last_name) VALUES ('JOHN','Smith');
INSERT INTO employees VALUES ('1000','JOHN','NULL');
INSERT INTO employees(first_name,last_name, employee_id) VALUES ('1000, 'john','Smith');
INSERT INTO employees (employee_id) VALUES (1000);
INSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'john',);
第10题:
INSERT INTO employees (employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01/01/01);
INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01 january 01');
INSERT INTO employees(employee_id, first_name, last_name, Hire_date) VALUES (1000, 'John', 'smith', To_ date ('01/01/01));
INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01-Jan-01');
第11题:
The SELECT statement is syntactically accurate.
The SELECT statement does not work because there is no HAVING clause.
The SELECT statement does not work because the column specified in the GROUP BY clause is not in the SELECT list.
The SELECT statement does not work because the GROUP BY clause should be in the main query and not in the subquery.
第12题:
Examine the data in the EMPLOYEES table.On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. The JOB_ID column is a NOT NULL column.Evaluate this DELETE statement:Why does the DELETE statement fail when you execute it?()
A. There is no row with dept_id 90 in the EMPLOYEES table.
B. You cannot delete the JOB_ID column because it is a NOT NULL column.
C. You cannot specify column names in the DELETE clause of the DELETE statement.
D. You cannot delete the EMPLOYEE_ID column because it is the primary key of the table.
第13题:
A.
B.
C.
D.
第14题:
Click the Exhibit button and examine the data in the EMPLOYEES table.On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. The JOB_ID column is a NOT NULL column.Evaluate this DELETE statement:DELETE employee_id, salary, job_idFROM employeesWHERE dept_id = 90;Why does the DELETE statement fail when you execute it?()
A.There is no row with dept_id 90 in the EMPLOYEES table.
B.You cannot delete the JOB_ID column because it is a NOT NULL column.
C.You cannot specify column names in the DELETE clause of the DELETE statement.
D.You cannot delete the EMPLOYEE_ID column because it is the primary key of the table.
第15题:
An error is generated.
You will have two identical tables in the HR schema with different names.
You create a table called employees in the HR schema based on you EMP table.
You create an alternative name for the employees table in the HR schema in your own schema.
第16题:
CREATE VIEW emp_Vu AS SELECT employee_id, emp_name, department_id FROM employees WHERE mgr_id IN (102, 120);
CREATE VIEW emp_Vu AS SELECT employee_id, emp_name, job_id department_id FROM employees WHERE mgr_id IN (102, 120);
CREATE VIEW emp_Vu AS SELECT department_id, SUM(sal) TOTALSAL FROM employees WHERE mgr_id IN (102, 120) GROUP BY department_ id;
CREATE VIEW emp_Vu AS SELECT employee_id, emp_name, job_id, DISTINCT department_id FROM employees;
第17题:
CREATE INDEX NAME _IDX (first_name, last_name);
CREATE INDEX NAME _IDX (first_name, AND last_name)
CREATE INDEX NAME_IDX ON (First_name, last_name);
CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);
CREATE INDEX NAME_IDX ON employees (First_name, last_name);
CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
第18题:
You cannot use the EMP_ID_SEQ sequence to populate the JOB_ID column.
The EMP_ID_SEQ sequence is invalidated when you modify the EMPLOYEE_ID column.
The EMP_ID_SEQ sequence is not affected by modifications to the EMPLOYEES table.
Any other column of NUMBER data type in your schema can use the EMP_ID_SEQ sequence.
The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEES table.
The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEE_ID column.
第19题:
You cannot use the EMP_ID_SEQ sequence to populate the JOB_ID column.
The EMP_ID_SEQ sequence is invalidated when you modify the EMPLOYEE_ID column.
The EMP_ID_SEQ sequence is not affected by modifications to the EMPLOYEES table.
Any other column of NUMBER data type in your schema can use the EMP_ID_SEQ sequence.
The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEES table.
The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEE_ID column.
第20题:
SELECT * FROM emp_dept_vu;
SELECT department_id, SUM(salary) FROM emp_dept_vu GROUP BY department _ id;
SELECT department_id, job_id, AVG(salary) FROM emp_dept_vu GROUP BY department _ id, job_id;
SELECT job_id, SUM(salary) FROM emp_dept_vu WHERE department_id IN (10,20) GROUP BY job_id HAVING SUM (salary) > 20000
None of the statements produce an error; all are valid.
第21题:
You cannot use the EMP_ID_SEQ sequence to populate the JOB_ID column.
The EMP_ID_SEQ sequence is invalidated when you modify the EMPLOYEE_ID column.
The EMP_ID_SEQ sequence is not affected by modifications to the EMPLOYEES table.
Any other column of NUMBER data type in your schema can use the EMP_ID_SEQ sequence.
The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEES table.
The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEE_ID column.
第22题:
To find the tax percentage for each of the employees.
To list the name, job id, and manager name for all the employees.
To find the name, salary, and department name of employees who are not working with Smith.
To find the number of employees working for the Administrative department and earning less then 4000.
To display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned.