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第1题:
Examine the structure of the EMPLOYEES and DEPARTMENTS tables:EMPLOYEESEMPLOYEE_ID NUMBERDEPARTMENT_ID NUMBERMANAGER_ID NUMBERLAST_NAME VARCHAR2(25)DEPARTMENTSDEPARTMENT_ID NUMBERMANAGER_ID NUMBERDEPARTMENT_NAME VARCHAR2(35)LOCATION_ID NUMBERYou want to create a report displaying employee last names, department names, and locations. Which query should you use to create an equi-join?()
A. SELECT last_name, department_name, location_id FROM employees , department ;
B. SELECT employees.last_name, departments.department_name, departments.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;
C. SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE manager_id = manager_id;
D. SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;
第2题:
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;
第3题:
已知关系模式R的全部属性集U={A,B,C,D,E,G}及函数依赖集:
F=(AB→C,C→A,BC→D,ACD→B,D→EG,BE→C,CG→BD,CE→AG}求属性集闭包(BD)+
(2) 现有如下两个关系模式:
Employees(Eid,Name,DeptNO)
Departments(DeptNO,DeptName,TotalNumber)
Employees关系模式描述了职工编号、姓名和所在部门编号;Departments关系模式描述了部门编号、名称和职工总
第4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5题:
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 d
D. 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);
第6题:
如果存在一张表格表示员工信息 Employees(ID,Name,Gender,ReportsTo),其中 ReportsTo 为外码,用于表示该员工的直接上级领导,那么要查询出结果(ID,Name,LeaderName),SQL 是Select ID, Name, u.Name As LeaderName From Employees e JOIN Empoyees u ON e.ID = u.ReportsTo