单选题Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX tables. EMPLOYEES NOT NULL, Primary EMPLOYEE_ID NUMBER Key VARCHAR2 EMP_NAME (30) VARCHAR2 JOB_ID (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table DEPARTMENTS NOT NULL, DEPARTMENT_ID NUMBER Primary Key VARCHAR2 DEPARTMENT_NAME |30| References MGR_ID column MGR_ID NUMBER of the EMPLOYEES table TAX MIN_SALARY NUMBER MAX_SALARY NUMBER TAX_PERCENT NUMBER For which situation would you use a nonequijoin query?（）ATo find the tax percentage for each of the employees.BTo list the name, job id, and manager name for all the employees.CTo find the name, salary, and department name of employees who are not working with Smith.DTo find the number of employees working for the Administrative department and earning less then 4000.ETo display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned.
单选题
Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX tables. EMPLOYEES NOT NULL, Primary EMPLOYEE_ID NUMBER Key VARCHAR2 EMP_NAME (30) VARCHAR2 JOB_ID (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table DEPARTMENTS NOT NULL, DEPARTMENT_ID NUMBER Primary Key VARCHAR2 DEPARTMENT_NAME |30| References MGR_ID column MGR_ID NUMBER of the EMPLOYEES table TAX MIN_SALARY NUMBER MAX_SALARY NUMBER TAX_PERCENT NUMBER 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 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.
参考解析
解析：
暂无解析
相关考题：
Examine the structure 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 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.
Examine the structure of the EMPLOYEES table:You need to update the records of employees 103 and 115. The UPDATE statement you specify should update the rows with the values specified below:Which UPDATE statement meets the requirements?（）A.B.C.D.E.
Examine the structure of the EMPLOYEES table:What is the correct syntax for an inline view?() A.B.C.D.
Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX tables.EMPLOYEESNOT NULL, PrimaryEMPLOYEE_ID NUMBERKeyVARCHAR2EMP_NAME(30)VARCHAR2JOB_ID(20)SALARY NUMBERReferencesMGR_ID NUMBEREMPLOYEE_IDcolumnDEPARTMENT_ID NUMBER Foreign key toDEPARTMENT_IDcolumn ofthe DEPARTMENTStableDEPARTMENTSNOT NULL,DEPARTMENT_ID NUMBERPrimary KeyVARCHAR2DEPARTMENT_NAME|30|ReferencesMGR_ID columnMGR_ID NUMBERof theEMPLOYEES tableTAXMIN_SALARY NUMBERMAX_SALARY NUMBERTAX_PERCENT NUMBERFor 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 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.
Examine the structure if the EMPLOYEES and NEW EMPLOYEES tables:Which MERGE statement is valid?() A.B.C.D.
Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables.Which two SQL statements produce the name, department name, and the city of all the employees who earn more then 10000?() A.B.C.D.
Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables:Which UPDATE statement is valid?() A.B.C.D.
Examine the structure of the EMPLOYEES and DEPARTMENTS tables:Evaluate this SQL statement:Which SQL statement is equivalent to the above SQL statement?() A.B.C.D.
Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables.EMPLOYEESNOT NULL,EMPLOYEE_ID NUMBERPrimary KeyVARCHAR2EMP_NAME(30)VARCHAR2JOB_ID(20)SALARY NUMBERReferencesMGR_ID NUMBEREMPLOYEE_IDcolumnDEPARTMENT_ID NUMBER Foreign key toDEPARTMENT_IDcolumn of theDEPARTMENTStableDEPARTMENTSNOT NULL, PrimaryDEPARTMENT_ID NUMBERKeyVARCHAR2DEPARTMENT_NAME(30)References NGR_IDMGR_ID NUMBERcolumn ofthe EMPLOYEES tableForeign key toLOCATION_ID NUMBERLOCATION_IDcolumn of theLOCATIONS tableLOCATIONSNOT NULL, PrimaryLOCATION_ID NUMBERKeyVARCHAR2CITY|30)Which two SQL statements produce the name, department name, and the city of all the employees who earn more then 10000?（）A. SELECT emp_name, department_name, city FROM employees e JOIN departments d USING (department_id) JOIN locations 1 USING (location_id) WHERE salary 10000;B. SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 JOIN ON (e.department_id = d.department id) AND (d.location_id =1.location_id) AND salary 10000;C. SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE salary 10000;D. SELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE e.department_id = d.department_id AND d.location_id = 1.location_id AND salary 10000;E. SELECT emp_name, department_name, city FROM employees e NATURAL JOIN departments, locations WHERE salary 10000;
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.
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?（）
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 employeesB.to list the name, job_id, and manager name for all the employeesC.to find the name, salary, and the department name of employees who are not working with SmithD.to find the number of employees working for the Administrative department and earning less than 4000E.to display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned
在执行语句SELECT department_id FROM employees，departments WHERE employees.department_id= departments.department_id；时报错，原因是（）。A、没有给表employees和表departments加别名B、没有给列department_id加别名C、不能用employees.department_id=departments.department_id作为条件D、SELECT后面的department_id没有指定是哪个表
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.
单选题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?（）AOnly the row with department ID 40 is deleted in the DEPARTMENTS table.BThe statement fails because there are child records in the EMPLOYEES table with department ID 40.CThe 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.DThe 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.EThe row with department ID 40 is deleted in the DEPARTMENTS table. Also all the rows in the EMPLOYEES table are deleted.FThe statement fails because there are no columns specifies in the DELETE clause of the DELETE statement.
单选题Examine the structure of the EMP_DEPT_VU view: Column Name Type Remarks EMPLOYEE_ID NUMBER From the EMPLOYEES table EMP_NAME VARCHAR2(30) From the EMPLOYEES table JOB_ID VARCHAR2(20) From the EMPLOYEES table SALARY NUMBER From the EMPLOYEES table DEPARTMENT_ID NUMBER From the DEPARTMENTS table DEPT_NAME VARCHAR2(30) From the DEPARTMENTS table Which SQL statement produces an error?（）ASELECT*FROMemp_dept_vu；BSELECTdepartment_id，SUM（salary）FROMemp_dept_vuGROUPBYdepartment_id；CSELECTdepartment_id，job_id，AVG（salary）FROMemp_dept_vuGROUPBYdepartment_id，job_id；DSELECTjob_id，SUM（salary）FROMemp_dept_vuWHEREdepartment_idIN（10，20）GROUPBYjob_idHAVINGSUM（salary）；20000；ENoneofthestatementsproduceanerror；allarevalid
单选题Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMPLOYEE_ID NUMBER DEPARTMENT_ID NUMBER MANAGER_ID NUMBER LAST_NAME VARCHAR2(25) DEPARTMENTS DEPARTMENT_ID NUMBER MANAGER_ID NUMBER DEPARTMENT_NAME VARCHAR2(35) LOCATION_ID NUMBER You want to create a report displaying employee last names, department names, and locations. Which query should you use to create an equi-join?（）ASELECT last_name, department_name, location_id FROM employees , department ;BSELECT employees.last_name, departments.department_name, departments.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;CSELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE manager_id = manager_id;DSELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;
单选题Examine the structures of the EMPLOYEES and TAX tables. EMPLOYEES EMPLOYEE_ID NUMBER NOT NULL, Primary Key EMP_NAME VARCHAR2 (30) JOB_ID VARCHAR2 (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT _ID column of the DEPARTMENT table TAX MIN_SALARY NUMBER MAX_SALARY NUMBER TAX_PERCENT NUMBER Percentage tax for given salary range You need to find the percentage tax applicable for each employee. Which SQL statement would you use?（）ASELECT employee_id, salary, tax_percent FROM employees e, tax t WHERE e.salary BETWEEN t.min _ salary AND t.max_salaryBSELECT employee_id, salary, tax_percent FROM employees e, tax t WHERE e.salary t.min_salary, tax_percentCSELECT employee_id, salary, tax_percent FROM employees e, tax t WHERE MIN(e.salary) = t.min_salary AND MAX(e.salary) = t.max_salaryDYou cannot find the information because there is no common column between the two tables.
单选题Examine the structure of the EMP_DEPT_VU view: Column Name Type Remarks EMPLOYEE_ID NUMBER From the EMPLOYEES table EMP_NAME VARCHAR2(30) From the EMPLOYEES table JOB_ID VARCHAR2(20) From the EMPLOYEES table SALARY NUMBER From the EMPLOYEES table DEPARTMENT_ID NUMBER From the DEPARTMENTS table DEPT_NAME VARCHAR2(30) From the DEPARTMENTS table Which SQL statement produces an error?（）ASELECT * FROM emp_dept_vu;BSELECT department_id, SUM(salary) FROM emp_dept_vu GROUP BY department _ id;CSELECT department_id, job_id, AVG(salary) FROM emp_dept_vu GROUP BY department _ id, job_id;DSELECT job_id, SUM(salary) FROM emp_dept_vu WHERE department_id IN (10,20) GROUP BY job_id HAVING SUM (salary) 20000ENone of the statements produce an error; all are valid.
多选题Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables. EMPLOYEES NOT NULL, EMPLOYEE_ID NUMBER Primary Key VARCHAR2 EMP_NAME (30) VARCHAR2 JOB_ID (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table DEPARTMENTS NOT NULL, Primary DEPARTMENT_ID NUMBER Key VARCHAR2 DEPARTMENT_NAME (30) References NGR_ID MGR_ID NUMBER column of the EMPLOYEES table Foreign key to LOCATION_ID NUMBER LOCATION_ID column of the LOCATIONS table LOCATIONS NOT NULL, Primary LOCATION_ID NUMBER Key VARCHAR2 CITY |30) Which two SQL statements produce the name, department name, and the city of all the employees who earn more then 10000?（）ASELECT emp_name, department_name, city FROM employees e JOIN departments d USING (department_id) JOIN locations 1 USING (location_id) WHERE salary 10000;BSELECT emp_name, department_name, city FROM employees e, departments d, locations 1 JOIN ON (e.department_id = d.department id) AND (d.location_id =1.location_id) AND salary 10000;CSELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE salary 10000;DSELECT emp_name, department_name, city FROM employees e, departments d, locations 1 WHERE e.department_id = d.department_id AND d.location_id = 1.location_id AND salary 10000;ESELECT emp_name, department_name, city FROM employees e NATURAL JOIN departments, locations WHERE salary 10000;
单选题Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEESColumn name Data type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key EMP_NAME VARCHAR2 (30) JOB_ID VARCHAR2 (20) SALARY NUMBER MGR_ID NUMBER References EMPLOYEE_ID COLUMN DEPARTMENT ID NUMBER Foreign key to DEPARTMENT ID column of the DEPARTMENTS table DEPARTMENTSColumn name Data type Remarks DEPARTMENT_ID NUMBER NOT NULL, Primary Key DEPARTMENT_NAME VARCHAR2(30) MGR_ID NUMBER References MGR_ID column of the EMPLOYEES table Evaluate this SQL statement: SELECT employee_id, e.department_id, department_name, salary FROM employees e, departments d WHERE e. department_id = d.department_id; Which SQL statement is equivalent to the above SQL statement? （）ASELECT employee_id, department_id, department_name, salary FROM employees WHERE department_id IN (SELECT department_id FROM departments);BSELECT employee_id, department_id, department_name, salary FROM employees NATURAL JOIN departments;CSELECT employee_id, d.department_id, department_name, salary FROM employees e JOIN departments d ON e.department _ id = d. department_id;DSELECT employee_id, department_id, department_name, Salary FROM employees JOIN departments USING (e.department_id, d.department_id);
单选题Examine the structure of the EMP_DEPT_VU view: Column Name Type Remarks EMPLOYEE_ID NUMBER From the EMPLOYEES table EMP_NAME VARCHAR2(30) From the EMPLOYEES table JOB_ID VARCHAR2(20) From the EMPLOYEES table SALARY NUMBER From the EMPLOYEES table DEPARTMENT_ID NUMBER From the DEPARTMENTS table DEPT_NAME VARCHAR2(30) From the DEPARTMENTS table Which SQL statement produces an error?（）ASELECT * FROM emp_dept_vu;BSELECT department_id, SUM(salary) FROM emp_dept_vu GROUP BY department _ id;CSELECT department_id, job_id, AVG(salary) FROM emp_dept_vu GROUP BY department _ id, job_id;DSELECT job_id, SUM(salary) FROM emp_dept_vu WHERE department_id IN (10,20) GROUP BY job_id HAVING SUM (salary) 20000ENone of the statements produce an error; all are valid.
单选题Click the Exhibit button and examine the data in the EMPLOYEES and DEPARTMENTS tables.You want to retrieve all employees, whether or not they have matching departments in the departments table. Which query would you use?（）ASELECT last_name, department_name FROM employees NATURAL JOIN departments;BSELECT last_name, department_name FROM employees JOIN departments ;CSELECT last_name, department_name FROM employees e JOIN departments d ON (e.department_id = d.department_id);DSELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);ESELECT last_name, department_name FROM employees FULL JOIN departments ON (e.department_id = d.department_id);FSELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);
单选题Examine the data in the EMPLOYEES and DEPARTMENTS tables. EMPLOYEES LAST_NAME DEPARTMENT_ID SALARY Getz 10 3000 Davis 20 1500 Bill 20 2200 Davis 30 5000 Kochhar 5000 DEPARTMENTS DEPARTMENT_ID DEPARTMENT_NAME 10 Sales 20 Marketing 30 Accounts 40 Administration You want to retrieve all employees, whether or not they have matching departments in the departments table. Which query would you use?（）ASELECT last_name, department_name FROM employees , departments(+);BSELECT last_name, department_name FROM employees JOIN departments(+);CSELECT last_name, department_name ON (e. department_ id = d. departments_id); FROM employees(+) e JOIN departments dDSELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);ESELECT last_name, department_name FROM employees(+) , departments ON (e. department _ id = d. department _id);FSELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e. department _ id = d. department _id);
单选题Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEESColumn name Data type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key EMP_NAME VARCHAR2 (30) JOB_ID VARCHAR2 (20) SALARY NUMBER MGR_ID NUMBER References EMPLOYEE_ID COLUMN DEPARTMENT ID NUMBER Foreign key to DEPARTMENT ID column of the DEPARTMENTS table DEPARTMENTSColumn name Data type Remarks DEPARTMENT_ID NUMBER NOT NULL, Primary Key DEPARTMENT_NAME VARCHAR2(30) MGR_ID NUMBER References MGR_ID column of the EMPLOYEES table Evaluate this SQL statement: SELECT employee_id, e.department_id, department_name, salary FROM employees e, departments d WHERE e. department_id = d.department_id; Which SQL statement is equivalent to the above SQL statement?（）ASELECT employee_id, department_id, department_name, salary FROM employees WHERE department_id IN (SELECT department_id FROM departments);BSELECT employee_id, department_id, department_name, salary FROM employees NATURAL JOIN departments;CSELECT employee_id, d.department_id, department_name, salary FROM employees e JOIN departments d ON e.department _ id = d. department_id;DSELECT employee_id, department_id, department_name, Salary FROM employees JOIN departments USING (e.department_id, d.department_id);
单选题Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMPLOYEE_ID NUMBER DEPARTMENT_ID NUMBER MANAGER_ID NUMBER LAST_NAME VARCHAR2(25) DEPARTMENTS DEPARTMENT_ID NUMBER MANAGER_ID NUMBER DEPARTMENT_NAME VARCHAR2(35) LOCATION_ID NUMBER You want to create a report displaying employee last names, department names, and locations. Which query should you use?（）ASELECT e.last_name, d. department_name, d.location_id FROM employees e NATURAL JOIN departments D USING department_id ;BSELECT last_name, department_name, location_id FROM employees NATURAL JOIN departments WHERE e.department_id =d.department_id;CSELECT e.last_name, d.department_name, d.location_id FROM employees e NATURAL JOIN departments d;DSELECT e.last_name, d.department_name, d.location_id FROM employees e JOIN departments d USING (department_id );
单选题Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX tables. EMPLOYEES NOT NULL, Primary EMPLOYEE_ID NUMBER Key VARCHAR2 EMP_NAME (30) VARCHAR2 JOB_ID (20) SALARY NUMBER References MGR_ID NUMBER EMPLOYEE_ID column DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the DEPARTMENTS table DEPARTMENTS NOT NULL, DEPARTMENT_ID NUMBER Primary Key VARCHAR2 DEPARTMENT_NAME |30| References MGR_ID column MGR_ID NUMBER of the EMPLOYEES table TAX MIN_SALARY NUMBER MAX_SALARY NUMBER TAX_PERCENT NUMBER For which situation would you use a nonequijoin query?（）ATo find the tax percentage for each of the employees.BTo list the name, job id, and manager name for all the employees.CTo find the name, salary, and department name of employees who are not working with Smith.DTo find the number of employees working for the Administrative department and earning less then 4000.ETo display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned.