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 VARC

第1题：

“球员”表包含以下列： 球员标识NUMBER（9） 姓氏VARCHAR2（20） 名字VARCHAR2（20） 球队标识NUMBER（4） 经理标识NUMBER（9） 位置标识NUMBER（4） 如果要显示“球队标识”列和“经理标识”列的唯一组合，应使用哪条SELECT语句？（）

• A、SELECT*FROM球员；
• B、SELECT球队标识，经理标识FROM球员；
• C、SELECTDISTINCT球队标识，经理标识FROM球员；（*）
• D、SELECT球队标识，DISTINCT经理标识FROM球员；
• E、SELECT球队标识，经理标识DISTINCTFROM球员；

第2题：

DEPARTMENT 表包含以下列： DEPT_ID NUMBER, Primary Key DEPT_ABBR VARCHAR2(4) DEPT_NAME VARCHAR2(30) MGR_ID NUMBER EMPLOYEE 表包含以下列： EMPLOYEE_ID NUMBER EMP_LNAME VARCHAR2(25) EMP_FNAME VARCHAR2(25) DEPT_ID NUMBER JOB_ID NUMBER MGR_ID NUMBER SALARY NUMBER(9,2) HIRE_DATE DATE 请评估以下语句： ALTER TABLE employee ADD CONSTRAINT REFERENTIAL (mgr_id) TO department(mgr_id); 是以下哪个说法（）

• A、ALTER TABLE语句创建从EMPLOYEE表到DEPARTMENT表的引用约束条件
• B、ALTER TABLE语句创建从DEPARTMENT表到EMPLOYEE表的引用约束条件
• C、ADD CONSTRAINT子句存在语法错误，因此ALTER TABLE语句将会失败
• D、ALTER TABLE语句执行成功，但不重新创建引用约束条件

第3题：

TEAMS表包含以下各列： TEAM_ID NUMBER（4） Primary Key TEAM_NAMEVARCHAR2（20） MGR_ID NUMBER（9） TEAMS表当前是空表。您需要允许用户在经理标识值中包含文本字符。应使用以下哪条语句来执行此任务（）

• A、ALTER teams MODIFY（mgr_id VARCHAR2（15））
• B、ALTERTABLE teams MODIFY（mgr_id VARCHAR2（15））
• C、ALTERTABLE teams REPLACE（mgr_id VARCHAR2（15））
• D、ALTER teams TABLEMODIFYCOLUMN（mgr_id VARCHAR2（15））

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）；

第5题：

Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) Which three statements inserts a row into the table? （）

• A、INSERT INTO employees VALUES (NULL, 'JOHN','Smith');
• B、INSERT INTO employees( first_name, last_name) VALUES ('JOHN','Smith');
• C、INSERT INTO employees VALUES ('1000','JOHN','NULL');
• D、INSERT INTO employees(first_name,last_name, employee_id) VALUES ('1000, 'john','Smith');
• E、INSERT INTO employees (employee_id) VALUES (1000);
• F、INSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'john',");

第6题：

Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) Which three statements insert a row into the table? （）

• A、INSERT INTO employees VALUES (NULL, 'John', 'smith');
• B、INSERT INTO employees (first_name, last_name) VALUES ('John', 'smith');
• C、INSERT INTO employees VALUES ('1000, 'John', 'smith');
• D、INSERT INTO employees (first_name, last_name, employee_id) VALUES (1000, 'John', 'smith');
• E、INSERT INTO employees (employee_id) VALUES (1000);
• F、INSERT INTO employees ( employee_id, first_name, last_name, ) VALUES (1000, 'John','');

第7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');

第8题：

Which SQL statement defines a FOREIGN KEY constraint on the DEPTNO column of the EMP table?（）

• A、CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk FOREIGN KEY deptno REFERENCES dept deptno);
• B、CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));
• C、CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk REFERENCES dept (deptno) FOREIGN KEY (deptno));
• D、CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) FOREIGN KEY CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));

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第13题：

您的主管让您修改ORDERS表中的AMOUNT列。他要求将该列配置为接受默认值250。该表包含您需要保留的数据。应执行以下哪条语句来完成此任务（）

• A、ALTER TABLE orders CHANGE DATATYPE amount TO DEFAULT 250
• B、ALTER TABLE orders MODIFY（amount DEFAULT 250）
• C、DROP TABLE orders CREATE TABLE orders（orderno varchar2（5）CONSTRAINT pk_orders_01 PRIMARY KEY，customerid varchar2（5）REFERENCES customers（customerid），orderdate date，amount DEFAULT 250）
• D、DELETE TABLE orders CREATE TABLE orders（orderno varchar2（5）CONSTRAINT pk_orders_01 PRIMARY KEY，customerid varchar2（5）REFERENCES customers（customerid），orderdate date，amount DEFAULT 250）

第14题：

“雇员”表包含以下列： EMPLOYEE_ID NOT NULL, Primary Key SSNUM NOT NULL, Unique LAST_NAME VARCHAR2(25) FIRST_NAME VARCHAR2(25) DEPARTMENT_ID NUMBER 部门表中 DEPARTMENT_ID 列的外键 SALARY NUMBER(8,2) 如果执行以下语句： CREATE INDEX emp_name_idx ON employees(last_name, first_name); 以下哪个说法是的（）

• A、此语句会创建一个基于函数的索引
• B、因为语法错误，此语句将失败
• C、该语句将创建一个组合唯一索引
• D、该语句将创建一个组合非唯一索引

第15题：

请观察PRODUCT表和SUPPLIER表的结构: PRODUCT PRODUCT_ID NUMBER NOT NULL, Primary Key PRODUCT_NAME VARCHAR2 (25) SUPPLIER_ID NUMBER （ SUPPLIER表的SUPPLIER_ID的外键） LIST_PRICE NUMBER (7,2) COST NUMBER (7,2) QTY_IN_STOCK NUMBER QTY_ON_ORDER NUMBER REORDER_LEVEL NUMBER REORDER_QTY NUMBER SUPPLIER SUPPLIER_ID NUMBER NOT NULL, Primary Key SUPPLIER_NAME VARCHAR2 (25) ADDRESS VARCHAR2 (30) CITY VARCHAR2 (25) REGION VARCHAR2 (10) POSTAL_CODE VARCHAR2 (11) 请评估以下语句： ALTER TABLE suppliers DISABLE CONSTRAINT supplier_id_pk CASCADE; 该语句用于执行什么任务（）

• A、删除引用SUPPLIERS表的所有约束条件
• B、删除对FOREIGNKEY表的FOREIGNKEY约束条件
• C、删除引用PRODUCTS表的所有约束条件
• D、禁用所有依赖于PRODUCTS表的SUPPLIER_ID列的完整性约束条件
• E、禁用所有依赖于SUPPLIERS表的SUPPLIER_ID列的完整性约束条件

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;

第17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%

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) 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;

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 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)))

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