湘潭大学oracle试卷A

2023年Oracle认证考试真题1. 数据库设计与规范化（25分）在关系数据库的设计和规范化过程中，你是负责设计一个学生信息管理系统的数据库。

请按照以下要求完成相关任务：a. 根据需求分析得到的实体和关系，绘制ER图，并对ER图进行适当的优化和调整；b. 基于ER图，将实体和关系转化为关系模式，确定属性及其关联关系；c. 对关系模式进行规范化，确保数据库满足第三范式。

2. SQL查询与优化（35分）本题需要你运用SQL语言进行数据查询和性能优化。

请根据以下要求完成相关任务：a. 根据给定的表结构和数据，编写SQL查询语句，实现以下功能：- 查询学生表中的所有记录；- 查询平均成绩大于80分的学生姓名和其对应的课程名；- 查询课程表中课程名以“数据库”开头的记录。

b. 对于上述查询语句，分别给出优化方案，提高查询性能。

3. 数据库备份与恢复（20分）在数据库管理过程中，备份与恢复是非常重要的环节。

请根据以下要求完成相关任务：a. 使用Oracle提供的工具或语句进行数据库备份，要求备份包含完整的数据库结构和数据；b. 恢复数据库时，请说明你会使用的方法和步骤，确保数据能够完整恢复。

4. 数据安全与权限管理（20分）数据库安全和权限管理是企业中保护数据的重要措施。

请根据以下要求完成相关任务：a. 列举至少三种数据库安全控制措施，并简要说明其实现原理；b. 设计一个用户角色并赋予相应的权限，确保用户只能访问其所需的数据。

5. 总结与建议（重点评分项）请根据你在考试中的实际经历，结合自身专业知识和经验，对Oracle数据库认证考试进行总结与建议。

包括但不限于以下方面：a. 考试难度及相关学习准备建议；b. 考试题型及题量特点，对备考提出建议；c. 经验分享，例如学习方法、重点复习内容等。

以上是2023年Oracle认证考试真题的内容要求。

在实际考试中，请根据题目要求和时间安排合理分配答题时间。

祝你考试顺利！。

Oracle基础测试试卷姓名：学校/专业: 成绩：一．选择题（20题，共60分）1、Oracle数据库中为新创建的表分配的初始空间通常为多大？（ B ）A、一个块B、一个区C、一个段D、一个表空间2、关于存储过程参数，正确的说法是（ B ）。

A、存储过程的输出参数可以是标量类型，也可以是表类型B、存储过程输入参数可以不输入信息而调用过程C、可以指定字符参数的字符长度（函数的()或者过程的(number/varchar2)）D、以上说法都不对3、有一产品表（编号，名称，价格，数量，所属分类），下列语法不正确的是（ D ）。

A、select * from 产品表where价格>1000B、select sum(价格) from 产品表group by 所属分类having max(价格)>1000C、select所属分类,sum(价格) from 产品表where 价格>1000 group by 所属分类D、select所属分类,sum(价格) from 产品表where max(价格)>1000 group by 所属分类4、下列哪些是Oracle的伪列。

（ACD ）(多选)A、ROWIDB、ROW_NUMBER()C、LEVELD、ROWNUME、COLUMN5、在SQL*Plus环境中，删除触发器trg_emp的命令是（ A ）。

A、DROP TRIGGER trg_empB、DELETE TRIGGER trg_empC、REMOVE TRIGGER trg_empD、ALTER TRIGGER trg_emp REMOVE6、单行（ B ）函数能够得到字符串的一部分？A、INSERTB、SUBSTRC、LPADD、LEAST7、利用游标来修改数据时，所用的。

FOR UPDATE充分利用了事务的哪个特性？（ D ）A、原子性B、一致性C、永久性D、隔离性8、下列说法不正确的是（）。

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期末考试卷Oracle期末考试卷一、选择题（每题2分，共20分）1. Oracle数据库的默认端口号是什么？A. 1521B. 3306C. 1433D. 54322. 在Oracle数据库中，以下哪个命令用于查看当前数据库的所有用户？A. SELECT USER FROM DUAL;B. SELECT DISTINCT USERNAME FROM DBA_USERS;C. SELECT * FROM ALL_USERS;D. SELECT * FROM USER_USERS;3. 以下哪个不是Oracle数据库的体系结构组件？A. 数据库B. 实例C. 表空间D. 索引4. Oracle数据库的哪个组件负责管理数据的存储和检索？A. SQLB. PL/SQLC. RDBMSD. DBMS5. 在Oracle数据库中，以下哪个命令用于创建一个新的表？A. CREATE TABLE;B. CREATE DATABASE;C. CREATE INDEX;D. CREATE VIEW;6. Oracle数据库中，哪个命令用于删除一个表？A. DROP TABLE;B. REMOVE TABLE;C. DELETE TABLE;D. ERASE TABLE;7. Oracle数据库中，哪个命令用于查看表的结构？A. DESCRIBE table_name;B. EXPLAIN table_name;C. SHOW table_name;D. SELECT table_name;8. 在Oracle数据库中，以下哪个命令用于添加新的列到一个已存在的表？A. ADD COLUMN column_name column_type;B. INSERT COLUMN column_name column_type;C. CREATE COLUMN column_name column_type;D. APPEND COLUMN column_name column_type;9. Oracle数据库中，哪个命令用于更新表中的记录？A. UPDATE table_name SET column_name = new_value WHERE condition;B. MODIFY table_name SET column_name = new_value WHERE condition;C. CHANGE table_name SET column_name = new_value WHERE condition;D. ALTER table_name SET column_name = new_value WHERE condition;10. 在Oracle数据库中，以下哪个命令用于删除表中的记录？A. ERASE FROM table_name WHERE condition;B. REMOVE FROM table_name WHERE condition;C. DELETE FROM table_name WHERE condition;D. DROP FROM table_name WHERE condition;二、简答题（每题5分，共30分）1. 简述Oracle数据库的体系结构。

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。

Oracle数据库应用开发考试试题(doc 12页)Oracle1. （）是Oracle维护数据库中其他文件的列表、数据库名称和系统改变号（SCN）的文件。

（第二章）A. 控制文件。

B. 参数文件。

C. 数据文件。

D. 可执行文件。

答案. A2. Oracle数据库被划分为（）的逻辑存储单元，该逻辑存储单元本身由操作系统磁盘文件和原始分区组成。

（第二章）A. 表空间。

B. 联机重做日志文件。

C. 归档重做日志文件。

D. 回滚段。

答案. A3. （）是包含Oracle数据库中所有数据的数据库对象。

（第二章）A. 视图。

B. 索引。

C. 存储过程。

D. 表。

答案. D4. （）是为了加速对特定表数据的访问而创建的数据段。

（第二章）A. 视图。

B. 索引。

C. 存储过程。

D. 触发器。

答案. B5. 下面是有关date数据类型和timestamp 数据类型的描述，请问哪一种描述是正确的？（第二章）A. 两者都只能存储日期数据，不能存储时间数据。

B. 两者都只能存储时间数据，不能存储日期数据。

C. date数据类型比timestamp数据类型的存储时间精度高。

D. date数据类型比timestamp数据类型的存储时间精度低。

答案. D6. 可以自动执行操作的存储对象是（）。

（第二章）A. 程序包。

B. 函数。

C. 过程。

D. 触发器。

答案. D7. 当Oracle创建数据库时，所有为数据库的数据字典所建立的基本表和视图都存储在（）模式中。

（第二章）A. SYS。

B. SYSTEM。

C. USER。

D. TEMP。

答案. A8. （）是Oracle数据库对象的别名，可以强化对象的安全性。

（第二章）A. 触发器。

B. 视图。

C. 表。

D. 同义词。

答案. D9. （）是具有只读类型的表和视图集，存储了数据库的物理和逻辑结构信息。

（第二章）A. 表空间。

B. 盘区。

C. 数据字典。

D. 方案。

答案. C10. （）是一组Oracle进程和SGA。

Oracle试卷A答案海洋⼤学试卷诚信考试承诺书本⼈重承诺：我已阅读且透彻理解了“海洋⼤学学⽣考场规则”和“海洋⼤学学⽣违反校纪校规处理规定”，承诺在考试中⾃觉遵守，如有违反，按有关条款接受处理。

承诺⼈签名：⽇期：考⽣：学号：专业班名：选择题1.修改⽤户时，不能更改⽤户的什么属性（ D ）A 密码B 表空间C 临时表空间D 名称2.撤销权限使⽤（ A ）语句A revokeB dropC deleteD alter3.查找⾓⾊中的⽤户，使⽤哪个数据字典（ A ）A user_role_privsB role_sys_privsC role_tab_privsD dba_users4.假设有存储过程add_student，其创建语句的头部容如下:create procedureadd_student(stu_id IN NUMBER,stu_name IN VARCHAR2)……，请问下列调⽤该存储过程的语句中，不正确的是（ B ）A EXEC add_student(1001,”CANDY”);B exec add_student(‘CANDY’,1001);C exec add_student(stu_id=>1001,stu_name=>’CANDY’);D exec add_student(stu_name=>’CANDY’,stu_id=>1001);5.如果在包规mypackage中没有声明某个过程myprocedure，⽽在创建包体时包含了该过程，那么对该过程叙述正确的是（ D ）A 包体将⽆法创建成功，因为在包体中含有包规中没有声明的元素B 该过程影响包体的创建，因为它属于包的私有元素C 可以通过mypackage.myprocedure调⽤该过程D ⽆法在包体外使⽤该过程6.修改过程应该使⽤下列哪种语句（）A alter procedure 语句B drop procedure 语句C create procedure 语句D create or replace procedure 语句7.研究视图创建语句：create view dept as selectdepartment_id,employee_id,last_name from employees where department_id=30 with check option;是什么导致如下语句失败？（C）A 除⾮另外指定，否则视图会被创建为with read onlyB 视图太复杂⽽不允许DML操作C with check option 会拒绝任何修改department_id的语句D 该语句会成功8.如下PL_SQL程序块：SQL> declarei BINARY_INTERGE:=1beginwhile i>=1loopi:=i+1;DBMS_OUTPUT.PUTLINE(i);End loop;END执⾏上述PL_SQL语句，结果（ D ）A 输出从1开始，每次递增1的数B 输出从2开始，每次递增1的数C 输出2D 该循环将陷⼊死循环9.使⽤游标的什么属性可以获取SELECT语句当前检索到的⾏数（ D ）A %FOUNDB %NOTFOUNDC %ISOPEND %ROWCOUNT10.如果更改SCOTT⽤户的权限，应该在哪种模式下？（B）A SCOTTB SYSTEMC DBSNMPD ROOT11.Oracle数据库的⽂件后缀（ A ）A DBFB MDFC LDFD LOG12.解析后的SQL语句会缓存在SGA的哪个区域中（ C ）A JAVA池B ⼤型池C 共享池D 数据缓存区13.系统全局区不包括下⾯哪些区域（ C ）A Java池 B数据缓冲区 C 软件代码区域 D ⽇志缓冲区14.如果希望控制列的显⽰格式，那么可以使⽤下⾯的哪个命令（D ）A SHOWB DEFINEC SPOOLD COLUMN15.数据的逻辑独⽴性是指 ( A)A.概念模式改变，外模式和应⽤程序不变B.概念模式改变，模式不变C.模式改变，概念模式不变D.模式改变，外模式和应⽤程序不变16.数据库结构的描述和定义是DBMS运⾏的基本依据，它们存储于（ B ）A．封锁表 B．数据字典C．索引 D．⽇志17.下列权限中，哪⼀个不是数据库的访问权限？（ D ）A．Read权限 B．Resource权限考试⽤书C．Update权限 D．Lock权限18.关系模型有三类完整性约束：实体完整性、参照完整性和⽤户定义的完整性。