单选题Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) Which statement produces the number of different departments that have employees with last name Smith?()ASELECT COUNT (*) FROM employees WHERE last _name='smith';BSELECT COUNT (dept_id) FROM employees WHERE last _name='smith';CSELECT DISTINCT (COUNT (dept_id) FROM employees WHERE last _name='smith';DSELECT COUNT (DISTINCT dept_id) FROM employees WHERE last _name='smith';ESELECT UNIQE (dept_id) FROM employees WHERE last _name='smith';

单选题
Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) Which statement produces the number of different departments that have employees with last name Smith?()
A

SELECT COUNT (*) FROM employees WHERE last _name='smith';

B

SELECT COUNT (dept_id) FROM employees WHERE last _name='smith';

C

SELECT DISTINCT (COUNT (dept_id) FROM employees WHERE last _name='smith';

D

SELECT COUNT (DISTINCT dept_id) FROM employees WHERE last _name='smith';

E

SELECT UNIQE (dept_id) FROM employees WHERE last _name='smith';


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Examine the structure of the EMPLOYEES table:Column name Data type RemarksEMPLOYEE_ID NUMBER NOT NULL, Primary KeyLAST_NAME VARCNAR2(30)FIRST_NAME VARCNAR2(30)JOB_ID NUMBERSAL NUMBERMGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBERYou need to create an index called NAME_IDX on the first name and last name fields of the EMPLOYEES table. Which SQL statement would you use to perform this task? ()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);

Examine the description of the EMPLOYEES table:EMP_ID NUMBER(4) NOT NULLLAST_NAME VARCHAR2(30) NOT NULLFIRST_NAME VARCHAR2(30)DEPT_ID NUMBER(2)JOB_CAT VARCHARD2(30)SALARY NUMBER(8,2)Which statement shows the maximum salary paid in each job category of each department? ()A. SELECT dept_id, job_cat, MAX(salary) FROM employees WHERE salary MAX (salary);B. SELECT dept_id, job_cat, MAX(salary) FROM employees GROUP BY dept_id,job_cat;C. SELECT dept_id, job_cat, MAX(salary) FROM employees;D. SELECT dept_id, job_cat, MAX(salary) FROM employees GROUP BY dept_id;E. SELECT dept_id, job_cat, MAX(salary) FROM employees GROUP BY dept _ id job _ cat salary;

Examine the data in the EMPLOYEES and DEPARTMENTS tables:EMPLOYEESEMP_NAME DEPT_ID MGR_ID JOB_ID SALARYEMPLOYEE_ID101 Smith 20 120 SA_REP 4000102 Martin 10 105 CLERK 2500103 Chris 20 120 IT_ADMIN 4200104 John 30 108 HR_CLERK 2500105 Diana 30 108 IT_ADMIN 5000106 Smith 40 110 AD_ASST 3000108 Jennifer 30 110 HR_DIR 6500110 Bob 40 EX_DIR 8000120 Ravi 20 110 SA*DIR 6500DEPARTMENTSDEPARTMENT_ID DEPARTMENT_NAME10 Admin20 Education30 IT40 Human ResourcesAlso 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 REFERENCESdepartments(department_id),MGR_ID NUMBER REFERENCESemployees(employee id),MGR_ID NUMBER REFERENCESemployees(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:DELETEFROM departmentsWHERE department id = 40;What happens when you execute the DELETE statement?()

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

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Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) Which statement produces the number of different departments that have employees with last name Smith?()A、SELECT COUNT(*) FROM employees WHERE last_name='Smith';B、SELECT COUNT(dept_id) FROM employees WHERE last_name='Smith';C、SELECT DISTINCT(COUNT(dept_id)) FROM employees WHERE last_name='Smith';D、SELECT COUNT(DISTINCT dept_id) FROM employees WHERE last_name='Smith';E、SELECT UNIQUE(dept_id) FROM employees WHERE last_name='Smith';

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?()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.

Examine the structure of the EMPLOYEES table: Column name Data type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key LAST_NAME VARCNAR2(30) FIRST_NAME VARCNAR2(30) JOB_ID NUMBER SAL NUMBER MGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBER You need to create an index called NAME_IDX on the first name and last name fields of the EMPLOYEES table. Which SQL statement would you use to perform this task? ()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);

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Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) Which statement produces the number of different departments that have employees with last name Smith?()A、SELECT COUNT (*) FROM employees WHERE last _name='smith';B、SELECT COUNT (dept_id) FROM employees WHERE last _name='smith';C、SELECT DISTINCT (COUNT (dept_id) FROM employees WHERE last _name='smith';D、SELECT COUNT (DISTINCT dept_id) FROM employees WHERE last _name='smith';E、SELECT UNIQE (dept_id) FROM employees WHERE last _name='smith';

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单选题Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) JOB_CAT VARCHAR2(30) SALARY NUMBER(8,2) Which statement shows the department ID, minimum salary, and maximum salary paid in that department, only of the minimum salary is less then 5000 and the maximum salary is more than 15000?()ASELECT dept_id, MIN(salary(, MAX(salary) FROM employees WHERE MIN(salary) 5000 AND MAX (salary) 15000;BSELECT dept_id, MIN(salary), MAX(salary) FROM employees WHERE MIN(salary) 5000 AND MAX(salary) 15000 GROUP BY dept_id;CSELECT dept_id, MIN(salary), MAX(salary) FROM employees HAVING MIN(salary) 5000 AND MAX (salary) 15000;DSELECT dept_id, MIN(salary), MAX(salary) FROM employees GROUP BY dept_id HAVING MIN (salary) 5000 AND MAX(salary)ESELECT dept_id, MIN(salary), MAX(salary) FROM employees GROUP BY dept_id, salary HAVING MIN (salary) 5000 AND MAX (salary) 15000;

单选题The EMPLOYEES table contains these columns: LAST_NAME VARCHAR2 (25) SALARY NUMBER (6,2) COMMISSION_PCT NUMBER (6) You need to write a query that will produce these results: 1. Display the salary multiplied by the commission_pct. 2. Exclude employees with a zero commission_pct. 3. Display a zero for employees with a null commission value. Evaluate the SQL statement: SELECT LAST_NAME, SALARY*COMMISSION_PCT FROM EMPLOYEES WHERE COMMISSION_PCT IS NOT NULL; What does the statement provide?()AAll of the desired resultsBTwo of the desired resultsCOne of the desired resultsDAn error statement

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多选题Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) Which three statements inserts a row into the table? ()AINSERT INTO employees VALUES (NULL, 'JOHN','Smith');BINSERT INTO employees( first_name, last_name) VALUES ('JOHN','Smith');CINSERT INTO employees VALUES ('1000','JOHN','NULL');DINSERT INTO employees(first_name,last_name, employee_id) VALUES ('1000, 'john','Smith');EINSERT INTO employees (employee_id) VALUES (1000);FINSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'john',);

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