11月份邮政AIX／Oracle培训测试题
邮政计算机考试题及答案
邮政计算机考试题及答案一、单项选择题（每题2分，共20分）1. 邮政计算机系统中，用于处理邮件分拣的软件是：A. 邮件处理系统B. 邮件跟踪系统C. 邮件分发系统D. 邮件存储系统答案：A2. 邮政计算机系统中，用于追踪邮件状态的数据库是：A. 邮件分拣数据库B. 邮件跟踪数据库C. 邮件分发数据库D. 邮件存储数据库答案：B3. 邮政计算机系统中，用于存储邮件信息的数据库是：A. 邮件分拣数据库B. 邮件跟踪数据库C. 邮件存储数据库D. 邮件分发数据库答案：C4. 邮政计算机系统中，用于邮件分发的软件是：A. 邮件分拣软件B. 邮件跟踪软件C. 邮件分发软件D. 邮件存储软件5. 邮政计算机系统中，用于邮件存储的软件是：A. 邮件分拣软件B. 邮件跟踪软件C. 邮件存储软件D. 邮件分发软件答案：D6. 邮政计算机系统中，用于邮件处理的硬件设备是：A. 扫描仪B. 打印机C. 服务器D. 路由器答案：A7. 邮政计算机系统中，用于邮件跟踪的硬件设备是：A. 扫描仪B. 打印机C. 服务器D. 路由器答案：A8. 邮政计算机系统中，用于邮件分发的硬件设备是：A. 扫描仪B. 打印机C. 服务器D. 路由器答案：B9. 邮政计算机系统中，用于邮件存储的硬件设备是：B. 打印机C. 服务器D. 路由器答案：C10. 邮政计算机系统中，用于邮件分拣的硬件设备是：A. 扫描仪B. 打印机C. 服务器D. 路由器答案：D二、多项选择题（每题3分，共15分）1. 邮政计算机系统中，以下哪些是邮件处理的组成部分？A. 邮件分拣B. 邮件跟踪C. 邮件分发D. 邮件存储答案：ABCD2. 邮政计算机系统中，以下哪些是邮件跟踪的组成部分？A. 邮件分拣B. 邮件跟踪C. 邮件分发D. 邮件存储答案：BD3. 邮政计算机系统中，以下哪些是邮件分发的组成部分？A. 邮件分拣B. 邮件跟踪C. 邮件分发D. 邮件存储答案：CD4. 邮政计算机系统中，以下哪些是邮件存储的组成部分？A. 邮件分拣B. 邮件跟踪C. 邮件分发D. 邮件存储答案：AD5. 邮政计算机系统中，以下哪些是邮件处理的硬件设备？A. 扫描仪B. 打印机C. 服务器D. 路由器答案：ABCD三、判断题（每题1分，共10分）1. 邮政计算机系统中，邮件分拣软件和邮件分发软件是同一个软件。
Oracle测试题及答案.doc
insert into borrow_save
values(:o，:new.bno，:new.rdate);
end if;
end;
/
12.建立一个视图，显示“力01”班学生的借书信息(只要求显示姓名和书名)。
create view ca_view as
解决两个写事务冲突的办法就是采用“加锁”机制。例如A、B两个事务同吋对一个表的某航数据进行修改，A事务首先修改时对该数据加锁，禁止其他事务对该数据的修改。只有当A事务完成修改工作、并将锁打开后，其他事物才被允许修改。这时候等待的B事务才可以对该数据进行修改。锁可以防止两个写事务同时对一个数裾进行修改，所以也就不会出现写丢失的错误。
11.在borrow表上建立一个触发器，完成如下功能：如果读者借阅的书名是“数据库技术及应用”，就将该读者的借阅记录保存在borrow_save表中(注：borrow_save表结构同borrow表)。
create table borrow_save as
(select * from borrow)
借书卡card (eno卡号,name姓名，class班级)
图书books (bno书号，bname书名，author作者，price单价,quantity库存册数)借书记录borrow (eno借书卡号,bno书号，rdate还书曰期)
注：限定每人每种书只能借一本；库存册数随借书、还书而改变。
1.写出建立borrow表的SQL语句，要求定义主码完整性约束和引用完整性约束create table card
mark INTEGER;
BEGIN
BEGIN//嵌套块1
select氺into tin_rec from tin ;
Oracle测试题+答案
Oracle测试题+答案1) PL/SQL块中可以使用下列（）命令。
（选择两项）a) TRUNCATEb) DELETEc) SA VEPOINTd) ALTER TABLE2) 授予sa用户在SCOTT.EMP表中SAL列的更新权限的语句是（B）〔选择一项〕a) GRANT CHANGE ON SCOTT.EMP TO SAb) GRANT UPDA TE ON SCOTT.EMP(SAL) TO SAc) GRANT UPDA TE (SAL) ON SCOTT.EMP TO SAd) GRANT MODIFY ON SCOTT.EMP(SAL) TO SA3) EMP表有14条记录，则语句SELECT ‘Aptech’FROM EMP 的执行结果是（C）〔选择一项〕a) Aptechb) 无输出c) 14行Aptechd) 编译出错4) PL/SQL块中哪几部分是可选的（）〔选择二项〕a) Declareb) Beginc) Exceptiond) Constant5) 在创建序列的过程中，下列（）选项指定序列在达到最大值或最小值后，将继续从头开始生成值。
（选择一项）a) Cycleb) Nocyclec) Cached) Nocache6) 同义词有以下（）用途。
（选择三项）a) 简化SQL 语句b) 隐藏对象的名称和所有者c) 提供对对象的公共访问d) 显示对象的名称和所有者7) Oracle中用来释放锁的语句有（）〔选择二项〕a) commitb) Drop lockc) rollbackd) unlock8) 关于类型定义Number(9,2)说法正确的有（）〔选择一项〕a) 整数部分9位，小数部分2位，共11位b) 整数部分7位，小数部分2位，共9位c) 整数部分6位，小数点一位，小数部分2位，共9位d) 以上说法均不正确9) 下列哪种Lob类型用来存储数据库以外的操作系统文件（）〔选择一项〕a) CLOBb) BLOBc) CFILEd) BFILE10) Oracle中的三种系统文件分别是（）〔选择三项〕a) 数据文件b) 归档文件c) 日志文件d) 控制文件11) 下列哪项不是Oracle中常用的分区方法？（）〔选择一项〕a) 范围分区b) 散列分区c) 列表分区d) 条件分区12) 查看Test中名称为P1的分区中的记录的查询语句为（）〔选择一项〕a) Select * from Test Where PartitionName=’p1’b) Select * from Test(p1)c) Select * from Test Partition(p1);d) 以上均正确13) 创建序列时，若未指定Cycle选项，则当当前值大于MaxValue时将（）〔选择一项〕a) 从MinValue重新开始循环b) 重复使用MaxValue 值c) 执行出错d) 使用随机值14) 下列哪项是创建索引组织表所必需的（）〔选一项〕a) Primary Keyb) Order Byc) Group Byd) 以上均不是15) 若表的某字段值存在大量的重复，则基于该字段适合创建哪种索引？（）〔选一项〕a) 标准索引b) 唯一索引c) 位图索引d) 分区索引16) 在PL/SQL块中定义一个名为PI值为3.14的Real型常量的语法是（）〔选一项〕a) A.Pi Const Real＝3.14;b) B.Pi Real Const ＝3.14;c) C.Constant Pi Real：＝3.14d) D.Pi Constant Real:=3.1417) 22.当Select语句没有返回行时，将引发下列哪个异常？（）[选择一项]a) A.No_rows_foundb) B.No_data_foundc) C.No_Data_rows_foundd) D.Invalid_Number18) Oracle中提供的两种游标是（）〔选择二项〕a) A.隐式游标b) B.静态游标c) C.REF游标d) D.显式游标19) 若Emp表中有14条记录，则用户执行了以下操作，结果是：（）〔选择一项〕Cursor mycur is Select * From emp；。
Oracle期末考试试题及答案
Oracle期末考试试题及答案考 生 信 息 栏装 订 线一、选择题：（本题共20个小题，每小题2分，共40分）1当Oracle 服务器启动时，下列哪种文件不是必须的 ( ) 。
A. 数据文件 B. 控制文件 C. 日志文件 D. 归档日志文件 2.在Oracle 中，一个用户拥有的所有数据库对象统称为 ( )。
A. 数据库 B. 模式 C. 表空间 D. 实例 3．在 Oracle 数据库的逻辑结构中有以下组件： 1. 表空间 2. 数据块 3. 区 4. 段 这些组件从大到小依次是（ ）。
A. 1→2→3→4B.1→4→3→2C.1→3→2→4D. 4→1→3→2 4.下列哪个子句实现对一个结果集进行分组和汇总( )。
A.HAVING B.ORDER BY C.WHERE D.GROUP BY 5.在Oracle 数据库中，( )用户模式存储数据字典表和视图对象。
A. SYS B. SYSTEM C. SCOTT D. SYSDBA 6.以下不属于命名的PL/SQL 块的是( )。
A ．程序包 B ．过程 C ．游标 D ．函数7.在Oracle 中创建用户时，若未提及DEFAULT TABLESPACE 关键字，则Oracle 就将（ ）表空间分配给用户作为默认表空间。
A ．HR B ．SCOTT C ．SYSTEM D ．SYS8. 假设用户Lisa用Lisa以普通用户身份登录到系统，现需查看本用户下有哪些表，请写出相应的命令___________________________________________;9. 假设用户Lisa用Lisa以普通用户身份登录到系统，现需为Class 表的ID 列添加一个名为PK_CLASS_ID 的主键约束，请补充完整相应的命令ALTER TABLE Class________________________;10. 假设用户Lisa用Lisa以普通用户身份登录到系统，现需创建一个UPDATE 语句来修改本用户下ARTISTS 表中的数据，并且把每一行的T_ID 值都改成15，应该使用的SQL 语句是________________________ ;四、编程题：（本题共2小题，每题10分，共20分）1.请按要求完成以下程序的编写。
Oracle试题（含答案）
Oracle试题（含答案）_C++_CSD/ESD14051. 把⼯资⼤于1000的first_name,salary 显⽰出来，按⼯资排序，⼯资相同按first_name降序的SQL语句是?A.select first_name,salary from s_emp order by salary,first_nameB.select first_name,salary from s_emp order by salary,first_name ascC.select first_name,salary from s_emp order by salary desc,first_nameD. select first_name,salary from s_emp order by salary,first_name desc正确答案：D2. 已知数据库中有员⼯表s_emp和部门表s_dept，具体字段如下：员⼯表字段介绍:ID 员⼯编号LAST_NAME 员⼯姓FIRST_NAME 员⼯名USERID ⽤户编号START_DATE ⼊职⽇期COMMENTS 员⼯备注信息MANAGER_ID 员⼯的领导IDTITLE 员⼯的职位DEPT_ID 部门的编号SALARY 员⼯的⽉薪COMMISSION_PCT 提成部门表的字段介绍：ID 部门编号NAME 部门名REGION_ID 地区编号注：以下所有数据库相关题⽬，均使⽤这两个表把s_emp表中的每个⼈的全名和⼯资列出来的SQL语句是？A.select first_name|last_name ,salary from s_empB.select first_name||last_name ,salary from s_empC.select first_name|last_name ,salary from empD. select first_name||last_name ,salary from emp正确答案：B3. 如果想在代码中禁⽌使⽤goto，可以采⽤的是？A. #pragma GCC dependency gotoB. #pragma GCC poison gotoC. #pragma pack(2)D. #pragma GCC goto正确答案：B4.阅读如下代码：void* p = sbrk(0);int r = brk(p+4);brk(p+8);brk(p+4);请问⽬前占⽤的内存空间字节是？A. 16B. 8C. 4D. 0正确答案：C5. 查找和Smith⼀个title的所有员⼯的名字的SQL语句是？A. SELECT last_name from s_emp where title=(select title from s_emp wherelast_name=’Smith’)B. SELECT last_name from s_emp where title=(select title from s_emp wherelast_name=”Smith”)C. SELECT last_name from s_emp where title=(select last_name,title from s_emp where last_name=’Smith’)D. SELECT last_name from s_emp where title=(select title from s_emp last_name=’Smith’)正确答案：A6. 下列函数中不能处理错误的是？A. exitB. printfC. perrorD. strerror正确答案：A7. 关于信号，以下说法错误的是？A. 信号分为可靠信号和不可靠信号B. 信号的默认处理 ,80%的情况是退出进程C. 所有信号都可以忽略D. 当前⽤户只能给⾃⼰的进程发信号，不能给别的⽤户的进程发信号正确答案：C8. 下列关于进程描述符说法错误的是？A. getpid 获取进程ID。
Oracle数据库试题100题(附答案)
Oracle数据库试题100题(附答案)1.当Oracle服务器启动时，下列哪种文件不是必须的（D）。
A．数据文件B．控制文件C．日志文件D．归档日志文件2.在Oracle中，当用户要执行SELECT语句时，下列哪个进程从磁盘获得用户需要的数据（B）。
A．用户进程B．服务器进程C．日志写入进程（LGWRD．检查点进程（CKPT）3.在Oracle中，一个用户拥有的所有数据库对象统称为（B）。
A．数据库B．模式C．表空间D．实例4.在Oracle中，有一个教师表teacher的结构如下：ID NUMBER(5)NAME V ARCHAR2(25)EMAIL VARCHAR2(50)下面哪个语句显示没有Email地址的教师姓名（C）。
A．SELECT name FROM teacher WHERE email = NULL;B．SELECT name FROM teacher WHERE email NULL;C．SELECT name FROM teacher WHERE email IS NULL;D．SELECT name FROM teacher WHERE email IS NOT NULL;5.在Oracle数据库的逻辑结构中有以下组件：A 表空间B 数据块C 区D 段这些组件从大到小依次是（B）。
A．A→B→C→DB．A→D→C→BC．A→C→B→DD．D→A→C→B6.在Windows操作系统中，Oracle的（A）服务监听并接受来自客户端应用程序的连接请求。
A．OracleHOME_NAMETNSListenerB．OracleServiceSIDC．OracleHOME_NAMEAgentD．OracleHOME_NAMEHTTPServer7.在Oracle 中创建用户时，若未提及DEFAULT TABLESPACE关键字，则Oracle 就将（B）表空间分配给用户作为默认表空间。
AIX考试题和答案
一、填空题1、操作系统的kernel 和硬件交互，用户通过shell 和操作系统交互。
2、vi 编辑器可在多种UNIX系统平台下通用。
3、在系统中man 命令可以提供命令语法的帮助信息。
4、pwd 命令可以确认当前的工作目录。
5、AIX的文件系统具有倒树状结构，通常包括普通文件、目录、特殊文件三种文件类型。
6、改变文件的权限模式为rwxr-xr-x，对应的8进制模式代码是755 。
7、vi编辑器有文本编辑和命令行模式，编辑完成要保存并且退出可使用：x 、：wq 、shift+zz 三种方式8、当使用vi编辑器时，敲escape键将会进入模式。
9、给出如下的vi命令，请填到正确的位置：10、标准输入的系统变量名是0 ，标准输出的系统变量名是 1 ，标准错误输出是 2 。
11、echo $$该命令显示的是当前SHELL的进程ID号。
12、在UNIX下生存周期很长的特殊系统进程叫daemons或守护进程。
13、用户想将进程放到后台运行，用户退出后进程也不会被挂起，使用的命令是nohup ，缺省情况下输出和错误信息会放到nohup.out 文件中。
14、在你自己的用户环境中你可以通过$HOME或家目录位置下的.profile 文件来自定义。
15、用户要在根目录下查找名为file1的文件命令是find / -name file1 。
16、按要求使用grep命令查找file1中内容：1、查找以大写B开头匹配的字符：grep …^B‟ file12、查找以5结尾匹配的字符：grep …5$‟ file13、查找以D或H开头匹配的字符：grep …^[DH]‟ file14、查找以A开头紧跟单个任意字符再跟零或者任意个字符最后以零结尾匹配的字符：grep …^A.*0$‟ file117、将/tmp/aix目录下的内容tar到磁带rmt0上的命令为：tar –cvf /dev/rmt0 /tmp/aix/二、选择题(多选)1、下列哪个命令的语法是正确的。
oracle考试试题及答案.doc
oracle考试试题及答案Questions one by one, fill in the blanks (4 points per question exergy a total of 20 points)1,database management technology has gone through three stages: manual management, file system and database system2,database three level data structure is external mode, mode, internal model3,the Oracle database SGA database buffer by exergy exergy exergy redo log buffer shared pool.4,in the Oracle database integrity constraints exergy types are Primay key constraints・ Foreign key Unique constraint exergyCheck not need exergy exergy constraint constraint constraintDeclare cursor open the cursor cursor exergy exergy exergy extraction in PL/SQL, including 5 close myCursor cursor operationTwo, the true or false questions in every day 2 points 20 points total exergy rateThe basic objects stored in the database is 1, the data rate in T2, the database system is the core of DBMS in the T rate The characteristics of relationship between the 3 and the operation is set in the operating rate of T4,five basic operations in relational algebra, and is the difference,selection, projection, connection in the F rate5,Oracle process is the server process in the F rate6,the oraclet system SGA process server and all users in the process of sharing rate T7,the Oracle database system in the data block size in the T operation of the rate system8,Oracle database system exergy start database and the first step is to start a database instance in the T rateThe cursor 9, PL/SQL data can be changed in the F rate10, the database concept model is mainly used in database conceptual structure design in the F rateThree, use the title in the match each 7 points 35 points total exergy rateLogical independence and physical independence in 1, what is the database system data in the programDBMSProvides a two layer mapping mechanism in external mode threemode structureSchema, image, and modeInternal schema image・ The twoThe layer mapping mechanism guarantees the logicalindependence and physical independence of data in the database system・External modeThe schema image defines the correspondence between the external schema of the different users in the database and the logical schema of the database・When the database schema changes such as a relational database system to increase exergy change with the new relati on ship, the relationship between attribute data types can exergyExternal mode adjustmentThe relationship between image mode exergy guarantee a constant user oriented mode of each. The application is based on the data of the model prepared by the exergyWhich application is not required to ensure the independence of the logicof exergy exergy data and application of logical data independence・PatternThe internal schema image defines the global logical structure of the data in the database and the physical storage organization of those data in the systemCorrespondence・When changing the physical storage of data structures in the database when the internal model changes such as the definition and selection of a storage structure can adjust theThe constant so that the external schema of database system and individual applications do not have to change the database schema in the schema mapping relationship / hold mode・This will ensure that the physical data independence and the independence of the physical data between applications or databases ・2, the relational algebra equi jo ins difference is not a natural connection without contact yesterdayAnswer when the operator connected conditions included in the use of 〃二〃this connection is called equivalent connection. Connection operation General in the two table between for can also be in a table does nothave its own connection between connection operation such as from the ・Answer・The equivalent connections and self connections belong to the internal connection query3, what is the database database design is generally divided into what stage from1)Exergy databaseDatabaseIn according to the data structure to organize, store and manage data warehouse・2)Requirement design conceptual designlogic design physical designImplementation, operation and maintenanceFourBrief descriptionOracleComposition of exergy logical databaseA table space, segment, data blockFiveWell, try any one example of using method from cursorA exergyCreate tableCreate table test(Name char (30),Age char (40),Subject char (20),ID numeric (10))insert dataInsert, into, test, values (' hehe，,' haha，，‘ hh', 4)Define variablesDeclare @name char (30)Declare @age char (40)Declare ©subject char (20)Declare @id numericCreate a CursorDeclare himml cursorFor, select, [name], age, subject, ID, from, test open Open hiininlUse cursors to scroll throughFetch, himml, into, @name, @age, @subject, @id 一一Be careful@@FETCH_STATUS yesSQL SERVERInside variables andORACLEThe@@sqlstatusDiffer・While (@@FETCH_STATUS 二0)BeginPrintPrint @namePrint @agePrint ©subjectPrint @idFetch, hiininl, into, @name, @age,@subject, @idEndClose the cursor rate close the cursor result set in its entirety instead of exergyClose hiimnlClose the cursor cursor rate release the memory in and let the cursor name can be used again exergyDeallocate hiimnl five with employee tablesEMP (empno, ename, age, Sal, Tel, deptno), in which empno -------- name ------ n ame age - number of exergy exergy exergy exergy oftel ---- electric sal ------- age wagewordDeptno ----- Department number・Please program at SQL*PLUS in the morning following the following requirements・In every day 3 points total of 15 points in the 1 exergy rate, home telephone staff information query. In the SQL>SELECT FROM EMP WHERE Tel NOT * NULL; in the 2, query wages in 500 to 800 yuan betweenthe employee information in SQL>SELECT * FROM EMP WHERE BETWEEN 500 AND 800: in 3, according to the age in creasing order display employee nu mber, name, age, salary in the SQL>SELECT empno, ename, age, Sal FROM EMP ORDER BY age ASC; SQL>SELEC, 4As the average wage in the Department of SQL>SELECT AVG DOI (SAL) FROM EMP WHERE deptno二'D_01' ; in the 5, find the department number D OI over 40 years of age and wage of 400 yuan in the list of employees in the SQL>SELECT ename FROM EMP WHERE deptno二'D_01' AND age>40 A7D.An examination question twoTwo, fill in the blanks (each 2 points 30 points total exergy) please fill in the correct answers in the blanks every day. No fill and no "11.1.data model is usually composed of three elements of the data structure, data operation and data __________ constraint _・2.database systems, all types of user requests for database operations (data definition, query, update, and various controls) are made up ofA complex software to complete the exergy this software called DBMS3.in the SQL SELECT statement in the query to remove duplicate records of exergy exergy in the query results should be used —DISTINCT_・Key word・ 4. the use of SQL language SELECT statement for the query packet in the packet if he hoped to get rid of not meet the conditions should beUse the HAVING clause.5.relational database data manipulation language (DML) includes two types of operation in their search and update _■6.in relational database design in the database design is divided into requirement analysis, concept design, logic design, physical design, applicationProgram coding, debugging operation, database operation and maintenance in six stages・What stage of database design is the design relational schema?Exergy ____ task logic design ____7.operations can be divided into _ relational algebra _relational calculus and _______ two categories・The relationship between the 8. INF _ non _ main function to eliminate the dependence on the key attribute in the paradigm after grade to 2NF. 2NFThe relationship between — eliminate non main attributes on the keys of the transfer function can be _ dependent upon his paradigm level increased to 3NF.The three level structure of the 9. database through the concept of the pattern / image within the pattern to ensure ________________ independenee in the physical model of concept mapping / byAs in the ― logic _ independence guarantee.10.the meaning of SQL is _ structured query language _________ ・11.DBMS usually provide the authorization function to control permissions in the data of different users to access the database in its purpose is to numberAccording to the _____ security database・Three, short answer questions (6 points each item in a total of 24 points)Safety protection function 1. database provides four aspects which try to explain their meaning of exergyExergy security database is a rate caused by use of database protection prevent 订legal data leakage, change or damage・ SQL Server 2000The security mechanism consists of four layersThe first layer included operating system loginSecond layer server security management exergy exergySQL ServerLoginSpecial accountSAThird layer database security management database exergyexergy accessDatabase userFourth layer database objects in safety management of exergyexergy exergy database object tables and views in accessDatabase user gets roles2. the referential integrity rules in the purpose of it in the test example donburi・3. to Oracle DBMS for the SQL relational database language support is given in the case of grade three logic schematic 1) SQLLanguage support relational database three level logic structure consists of the outer layer and a memory, the concept of as shown in fig・・2)The concept of recording layer corresponds to the conceptual model is the basic table・The basic table is a table that itself actually exists ・A basic table is a it not by other forms of export table・ The basic table is usedCREATE TABLEStatement built・3)In the outer as seen by the user can be the basic table can also be view can also be the basic table view. A view is a virtual tableIt is composed of one or several basic forms of export table it does not exist in the physical memory directly on the table・ View is usedCREATE SQL VIEWlanguageSentence established.4)In the inner each basic table with a file storage is said by a group of the same type of stored records to indicate the value・DBAYou can manipulate physical storage files.4., briefly describe the DBMS database security control function, including what are the commonly used means?A database management system for data control function data securitycontrol function in order to ensure the safety and reliability of the data within the database to preventThe use of illegal cause data leakage and damage the data that avoid being peeped, tampering or ruining exergy data integrity control refers to the function of insuranceThe data card in the database correctly and effectively and to prevent the compatibility error data is not the semantic input or output.Four, database design (15 points)Suppose there is a relationship between the 1. to record each person，s identity card number, name and work unit・ Also contains every one of his / her childrenThe identity card number, name and place of birth and the he / she has every car brands and models・The real world from known facts thatSome people may have several cars but these cars may be the same type but may also is not the same type of exergySome people do not have the car if someone has the car included his every car has a car includedSome people may have several children but there are some people without children. The relationship model of the preliminary design of the are as followsR (identity cards, the name of the work unit of the C identity cards, the name of the C C was born in the car the model)The 〃C C〃identity cards, the name C was born "are the child，s identity card number, name and place of birth・Please send this pat tern into the pattern of the relationship between BCNF to determine the relationship between the main key. 7 points in a exergy exergy, the citizen identity cards in the name of the work unitThe type of car car exergy the identity card number inThe child identity cards, the exergy of C C C was born in the name of the identity card number in a certain school library2.assumptions to establish a database to save the readers, books and readers of record. In order to build theWe need to design a good database design from the conceptual model is shown and then the figure - the conceptual model intoa relational model・ pleaseDesign fTom La - map・The reader has readers attribute number, name, age, address and unit.Attributes of each book are ISBN, title, author and publisher・Each book for each reader borrowed date and should also have out of date ・ 8 points in exergyA reader reader the name address the exergy number in the unitThe author of the book ISBN Title Exergy in the press・The number of readers to borrow the books ISBN exergy date the date should be in five, calculation (the title 3 items within a total of 16 points)Clients with a commercial relational database the three basic table the table structure is as followsTable Article (commodity goods, the price of the stock in the commodity name)Table Customer (customer clients, the clients name the sex the age the phone)Orderitem order form (the number of the dients, the purchase price of goods number the date)Note that the answer to will give the answers written provisions of the local exergy answer requirements must be clearly not allowed to change the included writing programAnd optionally add sub queries・1.please create a GM_VIEW view of the retrieval clients using SQL language dients, clients and ordering goodsName, amount and date・(the number is equal to the purchase price * amount) 6 points in exergyCREAT VIEW GM_VIEW (clients, the clients name the commodity name the amount of the number of date) * AS SELECT _ clients, clients in a brand name in the purchase price in the amount of as in the date of FROM Artcle, Customer,OrderItemWHERE Customer・clients, =OrderItem・clients, and Article・commodity No.二OrderItem・ Article No. 2. please use the SQL language of female clients buy goods number, commodity name and the total number of out ・ 6 exergy rateSELECT _OrderItem・，commodity number AS, commodity number, Order Item ・ commodity name, AS commodity name, SUM (Order Item ・ quantity)The total quantity of AS is FROM, Orderitem, Artcle, Customer, WHERE _Artcle・,commodity number 二Orderitem・,commodity number AND, OrderItem・,commodity number 二Customer・, commodity number ANDCustomer・=，GROUP BY OrderItem・female gender in commodity trade name No.3.please use the SQL language ALTER TABEL command of a field in a field called the origin will increase the number of goods to table ArticleAccording to the type of the CHAR in the length of 30 in the rate of 4 points exergy command is as followsChar ALTER TABEL —Article ADD (30) — originItem 31, fill in the blanks (each 2 points in a total of 20 points)The SELECT statement for grouping query 1, using the SQL language in the packet will not meet if you want to remove the conditions should beUsing —HAVING..・ _ clause・In 2, in the design of relational database in database design is divided into requirement analysis, concept design, logic design, physical design, should beProgram coding, debugging run, database operation and maintenance in six stages・ What stage of database design is the design relational schema?The task of exergy _ logic designRelational operations in relational algebra and 3, including the selection, projection, __________ connection and division.4,the relationship model of entity integrity in referential integrity in user-defined integrity of three types of integrity・5,two yuan for entity set between A and B between the set in mapping base set must be one of the following four1. , one to one, contact2., one to many con tacts, more than3. to one, contact more than4., many pairs of contacts6,PL/SQL cursor the two types of explicit and implicit cursor cursor ・Two, single choice (3 points per item in a total of 15 points)1, in a relational database management system will create the view in the database three layer structure belongs to (A)A.external modeB. storage modeC. intra schemaD. conceptual schemaThe general characteristics of the 2, in the world of things in reality in the information world is called (A)A. entityB. entity keyThe C・ property D・ key 3 Relationship Model S J P SJP in the S in is students J curriculum P is ranking・ Each student takes in each courseThe performance has a certain rank each course ranking only one student in the column and No. The relationship model belongs to exergy (C)A, 2NF, B, 3NF, C, BCNF, D, 4NF4,the company has a department of a num bet of departments and employees each staff only belongs to a department can have a number of staffThe type of contact from staff to department is (C)A.many to many,B. , one to one,C., one to many, one toD., one to many5,the logical independence of data refers to (A)The concept of A. mode change external mode and not the applicationB.concept mode change mode notIn the C. mode concept mode not changeD. mode change external mode and not the applicationThe correct statement, query on wildcards in the 6. part (D)A・"匚 B・"represents a number of characters _〃can represent zero or more charactersC.〃—〃can not 〃%〃to use D・represents a characterThree, Jane answer1,the referential integrity rules purpose it donburi test example・2,Briefly describe the architecture features of Oracle database system1)contains at least one SYSTEM table space, and the DDL language2)various spatial data dictionary informationThe data stored in the table space, table space exergy is reflected in the form of multiple data files・3,what is the logic of program data independence and physical independence from 4, the DBMS of the database security control functionsincluding what means?5. Sketch the main steps of database conceptual design. (1)Data abstract conceptual model in the design of local exergy (2)The concept of local mode integrated into the global conceptual schema in(3) review 6, what is the function from the rollback segment7,cold and heat Be if eng explain back up different points and advantages from each of the 3 SCG in S#, model C#, grade, S# in the No.C for students course No. grade Exergy for a studentExamination results for a certain course・The average score were going to query the average score over 80 points in the course of the query resultsAccording to the average scores in ascending order average the same number in descending order according to the curriculum・ Write the SQL query・A Select C# AVG analysis (grade), From SCGGroup by C#Having AVG (grade) >80Order by 2, C# desc。
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11月份邮政AIX／Oracle培训测试题
第一部份：AIX部份，共十题
1.一台H70的机器安装了aix4.3,在安装应用系统上线2个月后的某一天，系统突然宕机。
用户按Reset 键后，面板LED显示:888-102-700-0c0.那么这些LED显示所表示的意义是：
A.机器发生了软件故障，dump部分完成。
B.机器发生了硬件故障，dump部分完成。
C.机器发生了硬件故障，dump全部完成。
D.机器发生了软件故障，dump全部完成。
答案：
2. 为了得到系统当前运行状态的一个永久快照，必须产生一个system dump，下面哪条命令可以达到此目的？
A. snap
B. sysdumpstart
C. sysdumpdev
D. crash
答案：
3.一台RS/6000 270机器重启后，系统管理员要检查机器系统的错误日志，请问用什么命令？
A.crash
B.errpt
C.errlogger
D.diag
答案：
4.AIX运行环境中，哪一个进程用于处理/dev/error中接收到的系统错误信息？
A./usr/lib/erredmon
B./usr/sbin/syslogd
C./usr/sbin/syncd
D./usr/sbin/getty
答案：
5.银行的储蓄前端抱怨机器太慢，这时系统管理员看见机器的硬盘闪烁的很快，他想在主控台发送一条命令查看硬盘操作是否忙碌，那么他要发送的命令是：
A.vmstat
B.iostat
stat
D.filemon
答案：
6.系统管理员正在监控机器的性能问题，以下哪个工具更适合分析系统性能方面的瓶颈问题？
A.vmstat
B.lspv
C.lslpp
D.svmon
答案：
7.一名服务工程师要获取RS6000机器中某一设备故障的SRN号(Service Request Number),那么系统管理员要使用以下哪个命令?
A.lssrc
B.diag
C.errpt
D.lsdev
答案：
8.以下哪个命令用来判断网络接口有没有激活？
stat -a
B.lsdev -Cc en0
C.lsdev -l adapter
D.lsdev -Cc if
答案：
9.双机(a机，b机)系统组成的cascading方式HA业务环境中,a机的网络地址组成为(a_boot,a_svc,a_stby),b机为(b_boot,b_svc,b_stby)。
在双机HA启动后，系统管理员
用哪个命令用判断HA是否启动？
A.lspv
B.lppchk
C.lssrc -g cluster
D.vmstat
答案：
10.双机(a机，b机)系统组成的cascading方式HA业务环境中,a机为主机，b机为备机，其共享阵列sharevg。
系统管理员因业务扩展需要，在a机上用smitty mklv 在sharevg中增加业务用的裸设备datalv20,请问HA要能正常切换，是否要重新同步？
A.要
B.不要
答案：
第二部份：Oracle
1.checkpoint 频率过大或过小对oracle数据库系统性能有什么影响?
2.redo log file设置太小对oracle数据库性能的影响主要表现在哪几个方面?
3.根据工作经验,列举oracle数据库管理方面最麻烦的几个问题。
4.redo log buffer设置是否合理,通过什么现象来判断?
5.对rollback segment的管理应该注意什么问题?
6.对于150个客户访问的数据库,做update,insert和delete操作等能马上完成,但select(无for
update)操作,查找结果要等待很长时间才能出来,请分析现象原因.。