oracle4级考试习题,带答案选择题3

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oracle4级考试习题,带答案选择题2

oracle4级考试习题,带答案选择题22 (第1题/共80题)--------------------------------------------------------------------------------关于ORDER BY子句，下列哪个说法是不正确的？1.升序和降序的关键字是ASC和DESC 。

2.只能按一个列进行排序。

(正确答案)3.排序列可以用它们在SELECT子句中的次序号代替。

4.允许对多个列进行排序。

2 (第2题/共80题)--------------------------------------------------------------------------------下面哪个语句是错误的？1.SELECT empno FROM emp WHERE empno = 493945;2.SELECT empno FROM emp;3.SELECT * FROM emp WHERE empno = 493945;4.SELECT empno WHERE empno = 56949 AND ename = 'smith'; (正确答案)2 (第3题/共80题)--------------------------------------------------------------------------------下列哪个语句将产生这样的结果ENAME NEW_ENAME ---------- ------------------- SMITH SMITH-dog- ALLEN ALLEN-dog- WARD W ARD-dog-d JONES JONES-dog- MARTIN MARTIN-dog BLAKE BLAKE-dog- CLARK CLARK-dog- SCOTT SCOTT-dog- KING KING-dog-d TURNER TURNER-dog ADAMS ADAMS-dog- JAMES JAMES-dog- FORD FORD-dog-d MILLER MILLER-dog1.SELECT SUBSTR(ename, 1, 10) AS new_ename FROM emp;2..SELECT RPAD(ename, 10, '-dog') AS new_ename FROMemp; (正确答案)3.SELECT TRIM(trailing '-dog' from ename) AS new_ename FROM emp;4.SELECT LPAD(ename, 10, '-dog') AS new_ename FROM emp;2 (第4题/共80题)--------------------------------------------------------------------------------传递给函数的值是-97.342，函数返回的结果是-97，这个函数是？1.CEIL( ) (正确答案)2.ABS( )3.SQRT( )4.MOD( )2 (第5题/共80题)--------------------------------------------------------------------------------定单表ORDERS 包含三列：customer, order_date和order_amt。

oracle(4级一体化) 附答案

一体化考核模拟试卷《基本SQL操作》模块一、基础知识(每题2分，共50分)1．以下不是分组函数的为 D 。

A．MAX B．SUM C．A VG D．SQRT2．关于ORDERBY子句，下列说法中不正确的是D C 。

A．排序列可以用它们在SELECT子句中的次序号代替B．升序和降序的关键字是ASC和DESCC只能按一个列进行排序D．允许对多个列进行排序。

3．在UPDATE语句中．更改列值的关键字是 A 。

A．SET B，ON C ORDER BY D．WHERE4．事务处理 B 可以防止事务未完前其他用户更新数据。

A.提交B．锁 C.回滚D．备份5.下列 A B 约束对访问权限有要求·：A．UNIQUE B．FOREIGN KEYC．CHECK D．NOT NULL6．以下语句中不属于DDL的是 C B ：A．CREATE B．DELETE C．ALTER D．DROP7．Oracle的体系结构是D 类型。

A．服务器 B.客户机 C B／S D．C／S8．搜索—列中的部分匹配的数据，下列关键字中最合适的是 C B 。

A.EXISTSB.LIKE C．BETWEEN D．IN9．下面WHERE子句实现了查找以“黄浦”开头的地址的是 D 。

A.LIKE黄浦．．．，B．LIKE'黄浦_' C.LIKE'黄浦*' D. LIKE'黄浦%'10．一般来说．—个学校会有很多班级，每个班级有很多学生，因此班级与学生之间可以构成D C 的关系·IA.一对一B．多对一C．一对多 D.多对多11．在关系数据库二，主键的性质是 C 。

A．创建唯一索引,允许空值B．在一个表中可以有多个主键C.为标识表中唯一的实体D.只允许以表中第一个字段建立22812A．V ARCHAR2 D．DATETIME shi SQL13．将多个查询结果返回一个结果集合的运算符是 C 。

oracle练习题及答案

oracle练习题及答案Oracle练习题及答案Oracle是一种强大的关系数据库管理系统，被广泛应用于企业级应用程序和数据管理中。

为了帮助大家更好地掌握Oracle数据库的知识，以下是一些Oracle练习题及答案，希望能够帮助大家更好地理解和掌握Oracle数据库的知识。

1. 什么是Oracle数据库？Oracle数据库是一种关系数据库管理系统，由美国Oracle公司开发。

它是一种高性能、可靠性高的数据库系统，被广泛应用于企业级应用程序和数据管理中。

2. Oracle数据库的特点有哪些？Oracle数据库具有以下特点：高性能、高可用性、可伸缩性、安全性高、灵活性强、易管理等。

3. 如何创建一个新的数据库用户？在Oracle数据库中，可以使用以下SQL语句来创建一个新的数据库用户：```CREATE USER username IDENTIFIED BY password;```4. 如何查看Oracle数据库中所有的表？可以使用以下SQL语句来查看Oracle数据库中所有的表：```SELECT table_name FROM user_tables;```5. 如何在Oracle数据库中插入一条新的记录？可以使用以下SQL语句来在Oracle数据库中插入一条新的记录：```INSERT INTO table_name (column1, column2, column3, ...) VALUES (value1, value2, value3, ...);```6. 如何在Oracle数据库中更新一条记录？可以使用以下SQL语句来在Oracle数据库中更新一条记录：```UPDATE table_name SET column1 = value1, column2 = value2, ... WHERE condition;```7. 如何在Oracle数据库中删除一条记录？可以使用以下SQL语句来在Oracle数据库中删除一条记录：```DELETE FROM table_name WHERE condition;```通过以上Oracle练习题及答案的学习，相信大家对Oracle数据库有了更深入的了解。

Oracle选择题复习(带答案)(精品文档)

1．参数文件中，下列四个符号中哪个符号开头的行表示注释行___C___。

A、$B、@C、#D、!2. 手工创建一个新的数据库时，首先需要___D___。

A、启动一个实例B、启动SQL*PLUS，以SYSDBA身份连接OracleC、关闭当前实例D、创建一个参数文件3．关于控制文件，下列说法正确的是___A___。

A、建议至少有两个位于不同磁盘上的控制文件B、建议至少有两个位于同一磁盘上的控制文件C、建议保存一个控制文件D、一个控制文件，数据库不能运行。

4．DB_BLOCK_SIZE初始化参数的度量单位是___B___。

A、BLOCKB、BYTEC、PAGED、ROW5．下列表空间中，__D____表空间是运行一个数据库必须的一个表空间。

A、ROLLBACKB、TOOLSC、TEMPD、SYSTEM6. PL/SQL代码段中注释符号是___C___。

A、//B、\\C、--D、’7．___D___是Oracle中一个有效的后台服务器进程。

A、ARCHiverB、LGWR (Log Writer)C、DBWR (Dbwriter)D、以上所有项8. 数据库启动过程中何时读取参数文件___C___。

A、打开数据库时B、装载数据库时C、实例启动时D、以上所有项9. 在PL/SQL代码段的异常处理块中，捕获所有异常的关键词是___A___。

A、OTHERSB、ALLC、ExceptionD、ERRORS10. 关闭数据库时，首先会___C___。

A、关闭数据库实例B、卸载数据库C、关闭数据库D、关闭操作系统11. DBA可以使用下列哪一个命令查看当前归档状态__A____。

A．ARCHIVE LOG LIST B、FROMARCHIVE LOGSC、SELECT * FROM V$THREADD、SELECT * FROM ARCHIVE_LOG_LIST12. 当创建控制文件的时候，数据库一定要__B____。

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。

计算机4级技巧考试题及答案

计算机4级技巧考试题及答案计算机技巧考试题：第一部分：选择题1. 下列哪个选项是计算机网络的定义？A. 一种人和人之间的交流方式B. 一种人与计算机之间的交流方式C. 一种计算机与计算机之间的交流方式D. 一种人、计算机和计算机之间的交流方式2. 在Excel中，以下哪个公式可以求出A1到A5单元格的和？A. =SUM(A1:A5)B. =ADD(A1:A5)C. =TOTAL(A1:A5)D. =PLUS(A1:A5)3. 在HTML中，以下哪个标签可以定义段落？A. <p>B. <h1>C. <a>D. <div>4. 在Windows操作系统中，通过什么组合键可以实现复制选中的内容？A. Ctrl+CB. Ctrl+VC. Ctrl+XD. Ctrl+Z5. 以下哪个选项是数据库管理系统（DBMS）的定义？A. 一种用于创建和管理数据库的软件B. 一种用于创建和管理文件的软件C. 一种用于创建和管理网络的软件D. 一种用于创建和管理网站的软件第二部分：填空题1. DOS命令"dir"用于_______。

2. 在PowerPoint中，可以通过设置_______来为幻灯片添加过渡效果。

3. 在计算机编程中，"if-else"语句用于实现_______。

4. PHP是一种_______编程语言。

5. 在计算机网络中，IP地址是由_______位二进制数表示的。

第三部分：简答题1. 请简要解释HTTP和HTTPS之间的区别。

2. 什么是文件压缩？请列举两种常见的文件压缩格式。

3. 请说明什么是数据备份，为什么数据备份非常重要？4. 请简述数据库的概念，并列举两种常见的数据库软件。

5. 请解释什么是网络安全，列举两种常见的网络安全威胁。

第四部分：应用题1. 请列举至少两种常见的电子邮件客户端，并简要介绍它们的特点。

oracle4级考试习题,带答案oracle(4级考题)

一、单项选择题(下列每题选项中，只有1个选项是正确的，请将其代号填在横线的空白处)1．与语句SELECT emp—id Id，1ast—name Name，sa1ary Gz FROM emp；查询结果不相同的是。

A．SELECT emp—id id，1ast—name name，Sa1ary gz FROM emp；B SELECT emp—id ”Id”，last—name ”Name”，salary ”Gz”FROM emp；C．SELECT emp—id ID，1ast—name NAME，salary GZ FROM emp；D．SELECT emp—id iD，1ast—name nAME，salary gZ FROM emp；2．要给列指定别名，下面的描述中错误的是——。

A. 如果别名要区分大小写或包含空格，则要给别名加上双引号B．把别名放在每一个列后，别名和列名之间用空格分隔C．把别名放在每一个列后，别名和列名之间使用as分隔D 如果别名要区分大小写或包含空格，则要给别名加上单引号3．下列命令中可以执行SQL*Plus缓冲区中的一条SQI。

语句的是——。

A．START B．LIST C．／D．SPOOL4．以下语句中正确的是——。

A．SELECT * FROM cmp WHERE name＝’King’；B．SELECT * WHERE name＝King；C．SELECT * FROM name＝，King；D．SELECT * FROM emp WHERE name＝”King”；5．在进行条件查询时，如果只记得目标字符串的一部分，则在以下关键字中可以用于找到符合条件的数据的是——。

A．IN E．EXISTS C．IS D．LIKE6．以下子句中，对查询结果的行进行限定的是——。

A．WHERE B，ORDER BY C．SELECT D．FROM一、单项选择题(下列每题选项中，只有1个选项是正确的，请将其代号填在横线的空白处)1．以下函数中，结果为63的是——。

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PART A第1题:使用下列哪个数据字典可以查看因约束而创建的唯一性索引？(单选题)USER_INDEXESUSER_TAB_COLUMNSUSER_COLUMNSUSER_CONS_COLUMNS (正确答案)第2题:你运行了select 5+4 from DUAL的查询。

但你以前从未在DUAL表中插入过数据。

下列哪一条语句最好的描述了这一DUAL表？(单选题)包含了两个模式名的字典视图一行一列的广泛适用的表 (正确答案)含有两个索引名的字典试图两列无行的广泛适用的表第3题:DBA为什么要测试备份和恢复策略?(多选题)为验证备份和恢复过程 (正确答案)为熟悉崩溃的类型(正确答案)练习备份和恢复(正确答案)构造后备产品数据库来支持新版本第4题:你发出以下语句： SELECT DECODE(ACCTNO, 123456, 'CLOSED', 654321,'SEIZED', 590395, 'TRANSFER','ACTIVE') FROM BANK_ACCT; 如果ACCTNO的值是503952，这条语句将显示什么信息？ (单选题)ACTIVE (正确答案)TRANSFERSEIZEDCLOSED第5题:哪种崩溃通常是最严重的?(单选题)非介质崩溃语句失败实例崩溃介质崩溃 (正确答案)第6题:哪两种数据库配置影响可恢复性?(多选题)ARCHIVELOG模式 (正确答案)NOARCHIVELOG模式 (正确答案)MOUNT模式NOMOUNT模式第7题:哪种数据库配置最容易在数据库崩溃时导致数据丢失?(单选题) ARCHIVELOG模式NOARCHIVELOG模式 (正确答案)第8题:选择数据库崩溃类型(多选题)语句 (正确答案)进程 (正确答案)实例 (正确答案)NOMOUNT模式第9题:哪种才是经受得起灾难性崩溃的高可靠性解决方案?(单选题) 并行服务器 (正确答案)独立数据库第10题:哪两种是Oracle数据库最通常的磁盘容错选项?(多选题)RAID 0RAID 10 (正确答案)RAID 5 (正确答案)RAID 0 + 1第11题:下列哪一功能不是having子句的？(单选题)将返回值按序排放排除规范的数据组包含不规范的数据组 (正确答案)包含规范的数据组第12题:下列关于笛卡尔集的定义哪个是正确的?(单选题)Oracle数据库的一个特性组函数模糊理论的结果没有where子句的连接语句的结果 (正确答案)第13题:你从一个表列里得到一组值，并想对其进行一组操作.下列哪一选项的函数在对所有行的数据进行操作时没有成为一组？(单选题)avg( )sqrt( ) (正确答案)count( )stddev( )第14题:选择防止对那些没有执行完全数据库恢复的数据库不正确修改的方法(多选题) 备份经常使用的表空间 (正确答案)备份控制文件 (正确答案)Exports多份重做日志 (正确答案)第15题:你正在写对数据库的查询。

下列哪一查询充分利用了一个内联视图？(单选题) select * from EMP_VW where EMPID = (select EMPID from INVOICE whereINV_NUM = 5506934);select STNAME, B.DEPT_NO from EMP A, (select EMPID, DEPT_NO from DEPT) B where A.EMPID = B.EMPID;select * from EMP where EMPID IN (select EMPID from INVOICE whereINV_NUM > 23); (正确答案)第16题:使用下列哪个语句，可以删除因约束而自动创建的唯一性索引？(单选题)drop indexalter table drop primary key cascade (正确答案)alter table drop constraintdrop table第17题:下列哪一数据修改操作无法用回滚命令恢复？(单选题)updatetruncate (正确答案)deleteinsert第18题:哪三项是隐式游标的异常？(多选题)%found%too_many_rows (正确答案)%notfound (正确答案)%rowcount%rowtype第19题:你向另一用户授权，哪一权限允许他在你的表里建立行？(单选题) references (正确答案)indexselectdelete第20题:哪三句是正确的？(多选题)角色能向其它角色授权. (正确答案)能授权给角色 (正确答案)角色可赋予用户 (正确答案)角色可赋予同义词第21题:哪一事务处理防止事务未完前其他用户更新数据？(单选题)锁 (正确答案)提交回滚SavepointsPART B1 什么导致PL/SQL中WHILE循环结束？控制传递给EXIT语句Boolean变量或表达式值为NULLBoolean变量或表达式值为TRUEBoolean变量或表达式值为FALSE (正确答案)执行指定次数的循环后2 在PL/SQL块中创建一个变量，使用一条select语句就可取出customer表中所有的行和列数据，应该如何定义？DECLARE TYPE customer_table_type IS TABLE OF customer%ROWTYPE INDEX BY BINARY_INTEGER; customer_table customer_table_type; (正确答案)DECLARE TYPE customer_table_type IS TABLE OF customer%TYPE INDEX BY BINARY_INTEGER; customer_table customer_table_type;DECLARE customer_table_type customer%ROWTYPE INDEX BYBINARY_INTEGER; customer_table customer_table_type;DECLARE customer_table_type customer%ROWTYPE; customer_tablecustomer_table_type;3 下列哪个SQL操作不使用排序?group byselect * from EMP; (正确答案)order byselect count(*)create index建立角色tmpusr，将在stu表中插入数据的权限赋予tmpusr，使用以下的语句：create ___ tmpusr;___ ___ on ___ to ___; （如果表stu不存在则先使用语句create table stu (id varchar(20))建立stu表）4备份数据库激活为主产品数据库时最容易受到攻击的是?数据文件崩溃任何需要恢复的崩溃 (正确答案)用户访问长时间运行的查询５数据装库时，parameters for the SQL*Loader execution 不可装于哪一位置？Command lineControl fileDatafile (正确答案)Parameter file６Bitmap索引提高下面哪种情况的性能?在查询一个有50000条记录表的表列，该表列值只有四个不同的值时 (正确答案) 当被索引的列值更改时当每次仅删除一或两条记录时当一次插入上百条记录时７The initial insert of new data into the table will leave most of its large columns NULL, to be filled in later by subsequent updates.记录重未更新. PCTFREE 和PCTUSED取值多少合适?PCTUSED = 99, PCTFREE = 1PCTUSED = 40, PCTFREE = 30 (正确答案)PCTUSED = 40, PCTFREE = 10PCTUSED = 80, PCTFREE = 10８哪一子句可用于alter user语句而不可用于create user语句?identified bytemporary tablespaceprofiledefault role (正确答案)account lockpassword expire９分析V$ROLLSTAT视图的输出结果，下列哪个组合显示应该增加回滚段中区的大小?低WRAPS, 低 EXTENDS高WRAPS, 低 EXTENDS高 WRAPS, 高 EXTENDS (正确答案)低 WRAPS, 高 EXTENDS１０DBA用CREATE TABLESPACE data_01 ONLINE语句创建数据库，要使实例启动必须至少分配多少个回滚段？2 (正确答案)345１１关于回滚段的使用，下列哪种分配方法比较合适?给系统中每个事务分配回滚段给短事务分配小回滚段给长事务分配大回滚段 (正确答案)给长事务分配小回滚段１２决策支持系统有250个用户，系统24小时在线且用户遍布US和Japan，该系统中的数据库块应设置为多少比较合适?2K4K8K32K (正确答案)１３调整排序操作的性能时，增加SORT_AREA_SIZE的效果是?能使PGA的大小超过真实内存大小 (正确答案)提高排序直接写入的性能增加排序生成的重做日志条目的大小将警告日志的位置改到排序位置１４DBA需要修改表空间。

哪一权限可防止数据库打开时其他用户登录Oracle ？create sessionrestricted session (正确答案)connectmount１５create user 语句中的哪一子句防止用户的磁盘种类与字典对象相矛盾？identified bytemporary tablespace (正确答案)default tablespacedefault role１６为了用户能远程管理Oracle数据库的表空间，数据库中必须有哪种文件类型？Password file (正确答案)Initialization fileDatafileControl fileNothing, sysdba privileges are not required for these actions.１７主数据库和备份数据库中哪些init.ora参数是不一致的?CONTROL_FILES (正确答案)BACKUP SETOS COPYIMPORT１８下列哪些操作是维护独立数据库的操作？转换数据文件文件名 (正确答案)备份主数据库备份备份数据库 (正确答案)决定最近应用的归档日志 (正确答案)１９从哪个备份资源恢复需要使用RESTORE命令来处理备份内容?Image copyBackup set (正确答案)OS copyImport２０数据库里有30条回滚段, TRANSACTIONS_PER_ROLLBACK_SEGMENT 设为49 ，TRANSACTIONS 设为1000. 大量使用时，有多回滚段可被Oracle充分使用?506020 (正确答案)30２１create user 语句中的哪一子句防止用户的磁盘种类与字典对象相矛盾？identified bytemporary tablespace (正确答案)default tablespacedefault role２２The DBA has shut down the database and made copies of all datafiles manually as a benchmark backup for stress testing. Which of the following RMAN commands will make that backup usable in the context of the rest of the contents of the recovery catalog? resynccatalog (正确答案)reportlist２３在装入数据后想使主键有效，发现错误ORA-02299: cannot enable(SYS.UX_EMP_01) - duplicate keys found.，DBA哪里找违反主键的行?DBA_CONS_COLUMNSDBA_CONSTRAINTSDBA_CLU_COLUMNSEXCEPTIONS (正确答案)２４数据库处于归档模式时，哪一项不是a direct component of Oracle’s redo mechanism？DBW0 (正确答案)Redo log bufferLGWROnline redo logCKPTArchive redo logs２５使用锁的进程和等待锁的进程可以在哪个工具的输出结果中找到?UTLBSTAT/UTLESTATEXPORTDBMS_APPLICATION_INFOUTLLOCKT (正确答案)２６用户抱怨等待重做日志空间时间太长，DBA可以通过以下方法找到问题，除了?查询视图V$WAITSTAT (正确答案)查询视图V$SYSSTAT查询视图 V$SYSTEM_EVENT查询视图 V$LATCH２７系统平均有35个用户同时往一个表中增加数据导致竞争，为解决问题，可以更改下列哪个区域?LatchesFreelists (正确答案)Redo log bufferLOG_SMALL_ENTRY_MAX_SIZE２８DBA给应用程序开发者哪两个建议来避免死锁?给所有数据更改操作使用显式锁给不同的后台进程分配不同的优先级给所有进程以相同次序获得锁 (正确答案)给长事务分配大的回滚段给所有进程使用较低级别的必需锁 (正确答案)２９从哪个备份资源恢复需要使用RESTORE命令来处理备份内容?Image copyBackup set (正确答案)OS copyImport３０如果数据库运行在NOARCHIVELOG模式且备份策略为完全备份，那么在重建所有数据库时需要哪种恢复类型?RECOVER DATAFILENone (正确答案)RECOVER DATABASERECOVER TABLESPACE３１主数据库和备份数据库中哪些init.ora参数是不一致的?CONTROL_FILES (正确答案)BACKUP SETOS COPYIMPORT３２备份数据库激活为主产品数据库时最容易受到攻击的是?数据文件崩溃任何需要恢复的崩溃 (正确答案)用户访问长时间运行的查询３３哪种类型的恢复自动将归档日志转换到备份数据库?手工恢复自动恢复管理恢复 (正确答案)机械恢复３４当备份数据库处于只读模式下归档日志如何应用到备份数据库? 归档日志连续或依次应用没有归档日志能应用 (正确答案)归档日志不加修改的应用归档日志并行应用３５以下哪项是backup piece 的缺省惯例命名变量?%N%D%U (正确答案)%B３６设计表时为增强列的唯一性，哪些是合适的? (选三)UNIQUE 约束 (正确答案)Bitmap 索引Primary key (正确答案)Foreign keyNOT NULL 约束Partitioned索引Unique索引 (正确答案)CHECK 约束３７能改善备份时文件拥挤程度的init.ora参数是?LARGE_POOL_SIZE (正确答案)BACKUP_ASYNC_IOBACKUP_TAPE_IO_SLAVES (正确答案)BACKUP_IO_SYNCH３８DBA发现数据库的性能在降低，因为Oracle花费大量时间在空间管理上，数据块不足而经常访问数据空闲链,下列哪个空间设置组合能有效解决问题?pctfree = 20, pctused = 75pctfree = 5, pctused = 90pctfree = 5, pctused = 60pctfree = 20, pctused = 30３９Bitmap索引提高下面哪种情况的性能?在查询一个有50000条记录表的表列，该表列值只有四个不同的值时 (正确答案) 当被索引的列值更改时当每次仅删除一或两条记录时当一次插入上百条记录时４０为提高数据库的查询性能，可以使用下列方法，除了哪一项？给经常出现在where子句中的列定义索引给经常更改的列定义bitmap索引 (正确答案)将索引放在单独的表中间，且与表所在的表空间不在同一磁盘上设计以查询语句有关的索引组织表开始答题第23题:用ALTER的MODIFY命令修改将EMP表中已存在的字段ENAME改为长度为12的VARCHAR2型如果EMP表不存在请先用下列语句添加表 CREATE TABLE EMP(EMPNO NUMBER(4) NOT NULL,ENAME VARCHAR2(10),JOB VARCHAR2(10),MGR NUMBER(4),HIREDATE DATE,SAL NUMBER(7,2),COMM NUMBER(7,2),DEPTNO NUMBER(2) NOT NULL);。