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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 NU

题目

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);

相似考题

1.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。

2.（ 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）

3.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;

4.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

更多“Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEESColumn name Data t ”相关问题

  • 第1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.


    参考答案：C

  • 第2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);


    参考答案：F

  • 第3题：

    Examinethestatement:SQL>CREATETABLESPACEuser_data2>EXTENTMANAGEMENTLOCAL3>SEGMENTSPACEMANAGEMENTAUTO;Whichtwpassumptionsmustbetrueforthisstatementtoexecutesuccessfully?()

    A.OracleManagedFilesareusedforthisinstance.

    B.TheUSER_DATAtablespaceismanagedusingFET$/UET$tables.

    C.TheCOMPATIBLEinitializationparametermustbe9.0.0orhigher.

    D.SpacewithinsegmentsintheUSER_DATAtablespaceismanagedwithfreelists.


    参考答案：A, C

  • 第4题：

    UserSCOTTwantstoexporthisobjectsusingOracleDataPumpandexecutesthefollowingcommand：$expdpscott/tigerdirectory=EXPORT_DIRdumpfile=scott.dmpinclude=tableinclude=view："like’%DEPARTMENTS%’"content=DATA_ONLYWhichtaskwouldthecommandaccomplish？（）

    A.A

    B.B

    C.C

    D.D

    E.E


    参考答案：C

  • 第5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


    正确答案：(BD)+=ABCDEG X(O)=BD；由D→EG可知X(1) =BDEG；再由BE→C可知X(2)→BDEGC；又有CG→BDCE→AG可知X(3)=BDEGCA因为X(3)中包含了所有的属性集即有(BD)+=X(3)=ABCDEG T—C(TNO．CNO)主码(TNOCNO)外码TNOCNO S—C(SNOCNO成绩)主码(SNOCNO)．外码SNOCNO (2) create trigger sql_tri on employees for
    (BD)+=ABCDEG X(O)=BD；由D→EG可知X(1) =BDEG；再由BE→C，可知X(2)→BDEGC；又有CG→BD，CE→AG，可知X(3)=BDEGCA，因为X(3)中包含了所有的属性集，即有(BD)+=X(3)=ABCDEG T—C(TNO．CNO)主码(TNO，CNO)，外码TNO，CNO S—C(SNO，CNO，成绩)主码(SNO，CNO)．外码SNO，CNO (2) create trigger sql_tri on employees for 解析：本题考查了后触发器的Transact—SQL语句。其语法格式为：
    CREATE TRIGGER触发器名称
    0N |表名|视图名|
    [WITH ENCRYPTION]
    AS
    SQL语句
    AFTER和FOR，指定触发器只有在引发的SQL语句中指定的操作都已成功执行，并且所有的约束检查也成功完成后，才执行此触发器，即后触发型触发器。

  • 第6题：

    函数substr(“DATASTRUCTURE”，5，9)的返回值为()。

    A.“STRUCTURE”
    B.“DATA”
    C.“DATASTRUCTURE”
    D.“ASTRUCTUR”

    答案：A
    解析：
    substr函数的作用是找到从第5个字符开始，长度为9的子串。

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