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The report from the US Department of Education shows that______.
A. the number of the expulsions is not large.
B. the number of the expulsion is wrong.
C. there are soldiers hiding among the students.
D. guns are out of control in the US schools.
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更多 “ The report from the US Department of Education shows that______.A. the number of the expulsions is not large.B. the number of the expulsion is wrong.C. there are soldiers hiding among the students.D. guns are out of control in the US schools. ” 相关考题
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It is reported that the film __________ this Saturday in some large cities.
A.is to showB.is to be shownC.is showingD.shows
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The results of the examination ________ that you have all made great ________.
A.shows...progressesB.show...progressesC.show...progressD.shows...progress
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Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEESColumn name Data type RemarksEMPLOYEE_ID NUMBER NOT NULL, Primary KeyEMP_NAME VARCHAR2 (30)JOB_ID VARCHAR2 (20)SALARY NUMBERMGR_ID NUMBER References EMPLOYEE_ID COLUMNDEPARTMENT ID NUMBER Foreign key to DEPARTMENT ID column of the DEPARTMENTS tableDEPARTMENTSColumn name Data type RemarksDEPARTMENT_ID NUMBER NOT NULL, Primary Key DEPARTMENT_NAME VARCHAR2(30)MGR_ID NUMBER References MGR_ID column of the EMPLOYEES tableEvaluate this SQL statement:SELECT employee_id, e.department_id, department_name, salaryFROM employees e, departments dWHERE e. department_id = d.department_id;Which SQL statement is equivalent to the above SQL statement? （）A. SELECT employee_id, department_id, department_name, salary FROM employees WHERE department_id IN (SELECT department_id FROM departments);B. SELECT employee_id, department_id, department_name, salary FROM employees NATURAL JOIN departments;C. SELECT employee_id, d.department_id, department_name, salary FROM employees e JOIN departments d ON e.department _ id = d. department_id;D. SELECT employee_id, department_id, department_name, Salary FROM employees JOIN departments USING (e.department_id, d.department_id);
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ExhibitExamine the data in the EMPLOYEES and DEPARTMENTS tables.You want to retrieve all employees‘ last names, along with their manager‘s 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 department 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.managaer_id = m.employee_id) LEFT OUTER JOIN department d ON (e.department_id = d.department_id);C. SELECT e.last_name, m.last_name, department_name FROM employees e RIGT OUTER JOIN employees m on ( e.manager_id = m.employee_id) FULL OUTER JOIN department 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) RIGT OUTER JOIN department 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 department d ON (e.department_id = d.department_id);
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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?（）
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Examine the data in the EMPLOYEES and DEPARTMENTS tables.EMPLOYEESLAST_NAME DEPARTMENT_ID SALARYGetz 10 3000Davis 20 1500Bill 20 2200Davis 30 5000Kochhar 5000DEPARTMENTSDEPARTMENT_ID DEPARTMENT_NAME10 Sales20 Marketing30 Accounts40 AdministrationYou want to retrieve all employees, whether or not they have matching departments in the departments table.Which query would you use?（）A. SELECT last_name, department_name FROM employees , departments(+);B. SELECT last_name, department_name FROM employees JOIN departments(+);C. SELECT last_name, department_name ON (e. department_ id = d. departments_id); FROM employees(+) e JOIN departments dD. SELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);E. SELECT last_name, department_name FROM employees(+) , departments ON (e. department _ id = d. department _id);F. SELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e. department _ id = d. department _id);
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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) ;
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