（若oracle数据库不了解，可以用sql）test1ID Varchar2(6) 主键NAME Varchar2(15)Test2Use_ID Varchar2(6) 主键关联TEST1的主键的IDPASSword Varchar2(10)

题目

（若oracle数据库不了解，可以用sql）

test1

ID Varchar2(6) 主键

NAME Varchar2(15)

Test2

Use_ID Varchar2(6) 主键关联TEST1的主键的ID

PASSword Varchar2(10)

相似考题

1.The CUSTOMERS table has these columns:CUSTOMER_ID NUMBER (4) NOT NULLCUSTOMER_NAME VARCHAR2 (100) NOT NULLSTREET_ADDRESS VARCHAR2 (150)CITY_ADDRESS VARHCAR2 (50)STATE_ADDRESS VARCHAR2 (50)PROVINCE_ADDRESS VARCHAR2 (50)COUNTRY_ADDRESS VARCHAR2 (50)POSTAL_CODE VARCHAR2 (12)CUSTOMER_PHONE VARCHAR2 (20)The CUSTOMER_ID column is the primary key for the table.You need to determine how dispersed your customer base is.Which expression finds the number of different countries represented in the CUSTOMERS table?（）A. COUNT(UPPER(country_address))B. COUNT(DIFF(UPPER(country_address)))C. COUNT(UNIQUE(UPPER(country_address)))D. COUNT DISTINTC UPPER(country_address)E. COUNT(DISTINTC (UPPER(country_address)))

2.Scott issues the SQL statements:CREATE TABLE dept(deptno NUMBER(2),dname VARCHAR2(14),loc VARCHAR2(13)};GRANT SELECTON DEPTT0 SUE;If Sue needs to select from Scott‘s DEPT table, which command should she use?（）A.SELECT*FROMDEPT；B.SELECT*FROMSCOTT.DEPT；C.SELECT*FROMDBASCOTTDEPT；D.SELECT*FROMALL_USERSWHEREUSER_NAME=‘；SCOTT‘；ANDTABLENAME=‘；DEPT‘；；

3.The CUSTOMERS table has these columns:CUSTOMER_ID NUMBER(4) NOT NULLCUSTOMER_NAME VARCHAR2(100) NOT NULLSTREET_ADDRESS VARCHAR2(150)CITY_ADDRESS VARCHAR2(50)STATE_ADDRESS VARCHAR2(50)PROVINCE_ADDRESS VARCHAR2(50)COUNTRY_ADDRESS VARCHAR2(50)POSTAL_CODE VARCHAR2(12)CUSTOMER_PHONE VARCHAR2(20)The CUSTOMER_ID column is the primary key for the table.You need to determine how dispersed your customer base is. Which expression finds the number of different countries represented in the CUSTOMERS table?（）A.COUNT(UPPER(country_address))B.COUNT(DIFF(UPPER(country_address)))C.COUNT(UNIQUE(UPPER(country_address)))D.COUNT DISTINCT UPPER(country_address)E.COUNT(DISTINCT (UPPER(country_address)))

4.You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns:CUST_ID NUMBER(4) NOT NULLCUST_NAME VARCHAR2(100) NOT NULLCUST_ADDRESS VARCHAR2(150)CUST_PHONE VARCHAR2(20)Which SELECT statement accomplishes this task?（）A. SELECT* FROM customers;B. SELECT name, address FROM customers;C. SELECT id, name, address, phone FROM customers;D. SELECT cust_name, cust_address FROM customers;E. SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;

更多“（若oracle数据库不了解，可以用sql）test1ID Varchar2(6) 主键NAME Varchar2(15)Test2Use_ID Varchar2(6) 主键关联TEST1的主键的IDPASSword Varchar2(10)”相关问题

  • 第1题：

    talbe :

    pub(pub_code char(2),pub_name varchar2(32))

    book(book_code char(8),book_name varhcar2(128),pub_code char92),author_code

    char(4),book_mem varchar2(128))

    author(author_code char(4),author_name varchar2(10),sex char(1), age

    number(3),dir varchar2(64))

    auth_list(auth_date date,author_code char(4))

    在PL/SQL中有以下两种字义

    v_book_row book%rowtype;

    v_book_code book.bank_code%type;

    请说明v_book_row,v_book_code变量的含义以及数据类型


    正确答案：
     

  • 第2题：

    你要操纵Oracle数据，下列哪个不是SQL命令？（）

    • A、select*from dual；
    • B、set define
    • C、update emp set ename=6543where ename=‘SMITHERS’；
    • D、create table employees（empid varchar2（10）primary key）；

    正确答案:B

  • 第3题：

    数据库中的中的PROFITS表中PRODUCT_NAME与SALE_PERIOD列建立主键。下列哪个语句无法定义这个主键？（）

    • A、create table profits（product_name varchar2（10），sale_perion varchar2（10），profit number， constraint pk_profits_01 primary key（product_name，sale_period））
    • B、alter table profits add constraint pk_profits_01 primary key（product_name，sale_period）deferrabl
    • C、以上两者都是

    正确答案:B

  • 第4题：

    修改表Department添加列Leader数据类型varchar2（10）的语句是：Alter table Department add Leader varchar2（10）；


    正确答案:错误

  • 第5题：

    You accepted the recommended SQL Profile by executing the following code：   DECLARE   sqlprofile_name varchar2（30）；  BEGIN   sqlprofile_name ：= DBMS_SQLTUNE.ACCEPT_SQL_PROFILE（  task_name => ’my_task’，   profile_name => ’my_profile’）；   END；   Which advisor will analyze this profile？（）  

    • A、 SQL Access Advisor
    • B、 Undo Advisor
    • C、 Segment Advisor
    • D、 SQL Tuning Advisor

    正确答案:D

  • 第6题：

    Examine the structure of the EMP_DEPT_VU view: Column Name Type Remarks EMPLOYEE_ID NUMBER From the EMPLOYEES table EMP_NAME VARCHAR2(30) From the EMPLOYEES table JOB_ID VARCHAR2(20) From the EMPLOYEES table SALARY NUMBER From the EMPLOYEES table DEPARTMENT_ID NUMBER From the DEPARTMENTS table DEPT_NAME VARCHAR2(30) From the DEPARTMENTS table Which SQL statement produces an error?（）

    • A、SELECT*FROMemp_dept_vu；
    • B、SELECTdepartment_id，SUM（salary）FROMemp_dept_vuGROUPBYdepartment_id；
    • C、SELECTdepartment_id，job_id，AVG（salary）FROMemp_dept_vuGROUPBYdepartment_id，job_id；
    • D、SELECTjob_id，SUM（salary）FROMemp_dept_vuWHEREdepartment_idIN（10，20）GROUPBYjob_idHAVINGSUM（salary）>；20000；
    • E、Noneofthestatementsproduceanerror；allarevalid

    正确答案:E

  • 第7题：

    Evaluate the following code：   SQL>VARIABLE task_name VARCHAR2（255）； SQL>VARIABLE sql_stmt VARCHAR2（4000）； SQL>BEGIN ：sql_stmt ：= ’SELECT COUNT（*） FROM customers  WHERE cust_state_province =’’CA’’’； ：task_name ：= ’MY_QUICKTUNE_TASK’；  DBMS_ADVISOR.QUICK_TUNE（DBMS_ADVISOR.SQLACCESS_ADVISOR，  ：task_name， ：sql_stmt）；  END；   What is the outcome of this block of code？（）  

    • A、 It creates a task and workload， and executes the task.
    • B、 It creates a task and workload but does not execute the task.
    • C、 It produces an error because a template has not been created.
    • D、 It produces an error because the SQL Tuning Set has not been created.

    正确答案:A

  • 第8题：

    Examine the description of the CUSTOMERS table: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) The CUSTOMER_ID column is the primary key for the table. Which statement returns the city address and the number of customers in the cities Los Angeles or San Francisco?（）

    • A、SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco');
    • B、SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco') GROUP BY city_address;
    • C、SELECT city_address, COUNT(customer_id) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco') GROUP BY city_address, customer_id;
    • D、SELECT city_address, COUNT(customer_id) FROM customers GROUP BY city_address IN ('Los Angeles', 'San Francisco');

    正确答案:B

  • 第9题：

    Scott issues the SQL statements: CREATE TABLE dept (deptno NUMBER(2), dname VARCHAR2(14), loc VARCHAR2(13)}; GRANT SELECT ON DEPT T0 SUE; If Sue needs to select from Scott's DEPT table, which command should she use?（）

    • A、SELECT*FROMDEPT；
    • B、SELECT*FROMSCOTT.DEPT；
    • C、SELECT*FROMDBASCOTTDEPT；
    • D、SELECT*FROMALL_USERSWHEREUSER_NAME='；SCOTT'；ANDTABLENAME='；DEPT'；；

    正确答案:B

  • 第10题：

    单选题
    You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns: CUST_ID NUMBER(4) NOT NULL CUST_NAME VARCHAR2(100) NOT NULL CUST_ADDRESS VARCHAR2(150) CUST_PHONE VARCHAR2(20) Which SELECT statement accomplishes this task?（）
    A

    SELECT* FROM customers;

    B

    SELECT name, address FROM customers;

    C

    SELECT id, name, address, phone FROM customers;

    D

    SELECT cust_name, cust_address FROM customers;

    E

    SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;


    正确答案： C
    解析： 暂无解析

  • 第11题：

    单选题
    评估EMPLOYEE表的结构： EMPLOYEE_IDNUMBER（9） LAST_NAMEVARCHAR2（25） FIRST_NAMEVARCHAR2（25） DEPARTMENT_IDNUMBER（9） MANAGER_IDNUMBER（9） SALARYNUMBER（7，2） 您使用以下哪条语句可将LAST_NAME列（当前包含200条记录）的长度增加到35个字节（）
    A

    ALTER employee TABLEAL TERCOLUMN（last_name VARCHAR2（35））

    B

    ALTER TABLE employee RENAME last_name VARCHAR2（35）

    C

    ALTER TABLE employee MODIFY（last_name VARCHAR2（35））

    D

    不能增大LAST_NAME列的宽度


    正确答案： B
    解析： 暂无解析

  • 第12题：

    单选题
    You executed the following code：   SQL> CREATE TABLE COUNTRY   （COUNTRY_ID CHAR（2） CONSTRAINT COUNTRY_ID_nn NOT NULL，   COUNTRY_NAME VARCHAR2（20），   CURRENCY_NAME VARCHAR2（20），   CONSTRAINT COUNTRY_ID_PK PRIMARY KEY （COUNTRY_ID））   ORGANIZATION INDEX；   In which tablespace will the mapping table be created？（）
    A

     SYSTEM tablespace

    B

     SYSAUX tablespace

    C

     Undo tablespace

    D

     The tablespace of the Index Organized Table （IOT）


    正确答案： D
    解析： 暂无解析

  • 第13题：

    找出"作者"库表中没有出现在"获奖名单"库表中的所有作者信息的SQL语句,(提示:使用

    NOT IN,NOT EXISTS,以及外部连接(+)三种方法),并说明那种方法是不优化的

    出版社: 出版社代码 char(2),出版社名称 varchar2(32)

    图书: 图书编号 char(8),图书名称 varchar2(128),出版社代码 char(2),作者代号

    char(4),图书简介 varchar2(128)

    作者: 作者代号 char(4),作者名称 varchar2(10),性别 char(1),年龄 number(3),文学

    方向 varchar2(64)

    获奖名单: 获奖日期 date,获奖人员 char(4)


    正确答案：
     

  • 第14题：

    You executed the following code：   SQL> CREATE TABLE COUNTRY   （COUNTRY_ID CHAR（2） CONSTRAINT COUNTRY_ID_nn NOT NULL，   COUNTRY_NAME VARCHAR2（20），   CURRENCY_NAME VARCHAR2（20），   CONSTRAINT COUNTRY_ID_PK PRIMARY KEY （COUNTRY_ID））   ORGANIZATION INDEX；   In which tablespace will the mapping table be created？（）

    • A、 SYSTEM tablespace
    • B、 SYSAUX tablespace
    • C、 Undo tablespace
    • D、 The tablespace of the Index Organized Table （IOT）

    正确答案:D

  • 第15题：

    ORACLE中char类型与varchar2类型的区别，描述正确的是（）

    • A、char为定长字符数据类型
    • B、char为不定长字符数据类型
    • C、varchar2为不定长数据类型
    • D、char与varchar2没有区别
    • E、varchar2为定长数据类型

    正确答案:A,C

  • 第16题：

    The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) The CUSTOMER_ID column is the primary key for the table. Which two statements find the number of customers? （）

    • A、SELECT TOTAL(*) FROM customer;
    • B、SELECT COUNT(*) FROM customer;
    • C、SELECT TOTAL(customer_id) FROM customer;
    • D、SELECT COUNT(customer_id) FROM customer;
    • E、SELECT COUNT(customers) FROM customer;
    • F、SELECT TOTAL(customer_name) FROM customer;

    正确答案:B,D

  • 第17题：

    You executed the following code：   SQL> CERATE TABLE COUNTRY   （COUNTRY_ID CHAR（2） CONSTRAINT COUNTRY_ID_nn NOT NULL，   COUNTRY_NAME VARCHAR2（20），   CURRENCY_NAME VARCHAR2（20），   CONSTRAINT COUNTRY_ID_PK PRIMARY KEY （COUNTRY_ID））   ORGANIZATION INDEX；   Which types of tables will be created automatically？（）

    • A、 journal table
    • B、 clustered table
    • C、 mapping table
    • D、 partitioned table

    正确答案:C

  • 第18题：

    The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) A promotional sale is being advertised to the customers in France. Which WHERE clause identifies customers that are located in France?（）

    • A、WHERE lower(country_address) = "france"
    • B、WHERE lower(country_address) = 'france'
    • C、WHERE lower(country_address) IS 'france'
    • D、WHERE lower(country_address) = '%france%'
    • E、WHERE lower(country_address) LIKE %france%

    正确答案:B

  • 第19题：

    The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) Which statement finds the rows in the CUSTOMERS table that do not have a postal code? （）

    • A、SELECT customer_id, customer_name FROM customers WHERE postal_code CONTAINS NULL;
    • B、SELECT customer_id, customer_name FROM customers WHER postal_code = ' ___________';
    • C、SELECT customer_id, customer_name FROM customers WHERE postal _ code IS NULL;
    • D、SELECT customer_id, customer_name FROM customers WHERE postal code IS NVL;
    • E、SELECT customer_id, customer_name FROM customers WHERE postal_code = NULL;

    正确答案:C

  • 第20题：

    The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER (4) NOT NULL CUSTOMER_NAME VARCHAR2 (100) NOT NULL STREET_ADDRESS VARCHAR2 (150) CITY_ADDRESS VARHCAR2 (50) STATE_ADDRESS VARCHAR2 (50) PROVINCE_ADDRESS VARCHAR2 (50) COUNTRY_ADDRESS VARCHAR2 (50) POSTAL_CODE VARCHAR2 (12) CUSTOMER_PHONE VARCHAR2 (20) The CUSTOMER_ID column is the primary key for the table. You need to determine how dispersed your customer base is. Which expression finds the number of different countries represented in the CUSTOMERS table?（）

    • A、COUNT(UPPER(country_address))
    • B、COUNT(DIFF(UPPER(country_address)))
    • C、COUNT(UNIQUE(UPPER(country_address)))
    • D、COUNT DISTINTC UPPER(country_address)
    • E、COUNT(DISTINTC (UPPER(country_address)))

    正确答案:E

  • 第21题：

    单选题
    FACULTY表包含以下各列： FACULTYID VARCHAR2(5) NOT NULL PRIMARY KEY FIRST_NAME VARCHAR2(20) LAST_NAME VARCHAR2(20) ADDRESS VARCHAR2(35) CITY VARCHAR2(15) STATE VARCHAR2(2) ZIP NUMBER(9) TELEPHONE NUMBER(10) STATUS VARCHAR2(2) NOT NULL COURSE 表包含以下各列： COURSEID VARCHAR2(5) NOT NULL PRIMARY KEY SUBJECT VARCHAR2(5) TERM VARCHAR2(6 FACULTYID VARCHAR2(5) NOT NULL FOREIGN KEY 您需要制定一个报表，用于确定在下学期任教的所有副教授。您要创建一个视图来简化报表的创建过程。以下哪条CREATE VIEW语句将完成此任务（）
    A

    CREATE VIEW（SELECT first_name，last_name，status，courseid，subject，term FROM faculty，course WHERE facultyid=facultyid）

    B

    CREATE VIEW pt_view ON（SELEC Tfirst_name，last_name，status，courseid，subject，term FROM faculty f and coursec WHERE f.facultyid=c.facultyid）

    C

    CREATE VIEW pt_view IN（SELECT first_name，last_name，status，courseid，subject，term FROM faculty course）

    D

    CREATE VIEW pt_view AS（SELECT first_name，last_name，status，courseid，subject，term FROM facultyf，coursec WHERE f.facultyid=c.facultyid）


    正确答案： D
    解析： 暂无解析

  • 第22题：

    单选题
    The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) The CUSTOMER_ID column is the primary key for the table. You need to determine how dispersed your customer base is. Which expression finds the number of different countries represented in the CUSTOMERS table?（）
    A

    COUNT(UPPER(country_address))

    B

    COUNT(DIFF(UPPER(country_address)))

    C

    COUNT(UNIQUE(UPPER(country_address)))

    D

    COUNT DISTINCT UPPER(country_address)

    E

    COUNT(DISTINCT (UPPER(country_address)))


    正确答案： A
    解析： 暂无解析

  • 第23题：

    单选题
    Evaluate the following code：   SQL>VARIABLE task_name VARCHAR2（255）； SQL>VARIABLE sql_stmt VARCHAR2（4000）； SQL>BEGIN ：sql_stmt ：= ’SELECT COUNT（*） FROM customers  WHERE cust_state_province =’’CA’’’； ：task_name ：= ’MY_QUICKTUNE_TASK’；  DBMS_ADVISOR.QUICK_TUNE（DBMS_ADVISOR.SQLACCESS_ADVISOR，  ：task_name， ：sql_stmt）；  END；   What is the outcome of this block of code？（）
    A

     It creates a task and workload， and executes the task.

    B

     It creates a task and workload but does not execute the task.

    C

     It produces an error because a template has not been created.

    D

     It produces an error because the SQL Tuning Set has not been created.


    正确答案： D
    解析： 暂无解析

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