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);
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);
相关考题：
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、名称和职工总人数。请按SQL Server所采用的’rransact—SQL语法格式编写实现具有如下功能的后触发型触发器：每当在Employees表中插入一行数据时，相应部门的职工总人数就加1。
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
（ 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、名称和职工总人数。请 按 SQL Serve r 所采用 的 Tansact — SQ L 语法格式编写实现具有功能的后触发型触发器 ： 每当在Employee s表中插入一行数据时，相应部门的职工总人数就加 1 。（ 10 分）
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;
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UserSCOTTwantstoexporthisobjectsusingOracleDataPumpandexecutesthefollowingcommand：$expdpscott/tigerdirectory=EXPORT_DIRdumpfile=scott.dmpinclude=tableinclude=view：like’%DEPARTMENTS%’content=DATA_ONLYWhichtaskwouldthecommandaccomplish？（）A.AB.BC.CD.DE.E
Examine the data in the EMPLOYEES and DEPARTMENTS tables.EMPLOYEESLAST_NAME DEPARTMENT_ID SALARYGetz 10 3000Davis 20 1500Bill 20 2200Davis 30 5000Kochhar 5000DEPARTMENTSDEPARTMENT_ID DEPARTMENT_NAME10 Sales20 Marketing30 Accounts40 AdministrationYou want to retrieve all employees, whether or not they have matching departments in the departments table.Which query would you use?（）A. SELECT last_name, department_name FROM employees , departments(+);B. SELECT last_name, department_name FROM employees JOIN departments(+);C. SELECT last_name, department_name ON (e. department_ id = d. departments_id); FROM employees(+) e JOIN departments dD. SELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);E. SELECT last_name, department_name FROM employees(+) , departments ON (e. department _ id = d. department _id);F. SELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e. department _ id = d. department _id);
已知关系模式R的全部属性集U={A，B，C，D，E，G}及函数依赖集：F=(AB→C，C→A，BC→D，ACD→B，D→EG，BE→C，CG→BD，CE→AG}求属性集闭包(BD)+(2) 现有如下两个关系模式：Employees(Eid，Name，DeptNO)Departments(DeptNO，DeptName，TotalNumber)Employees关系模式描述了职工编号、姓名和所在部门编号；Departments关系模式描述了部门编号、名称和职工总
Examinethestatement:SQLCREATETABLESPACEuser_data2EXTENTMANAGEMENTLOCAL3SEGMENTSPACEMANAGEMENTAUTO;Whichtwpassumptionsmustbetrueforthisstatementtoexecutesuccessfully?() A.OracleManagedFilesareusedforthisinstance.B.TheUSER_DATAtablespaceismanagedusingFET$/UET$tables.C.TheCOMPATIBLEinitializationparametermustbe9.0.0orhigher.D.SpacewithinsegmentsintheUSER_DATAtablespaceismanagedwithfreelists.
函数substr(“DATASTRUCTURE”，5，9)的返回值为()。A.“STRUCTURE”B.“DATA”C.“DATASTRUCTURE”D.“ASTRUCTUR”