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) POST

第1题：

“雇员”表包含以下列：姓氏VARCHAR2（25） 名字VARCHAR2（25） 电子邮件VARCHAR2（50）如果要编写以下SELECT语句来检索具有电子邮件地址的雇员的姓名。SELECT姓氏||’，’||名字"雇员姓名"FROM雇员；则应使用哪条WHERE子句来完成此条语句？（）

• A、WHERE电子邮件=NULL；
• B、WHERE电子邮件!=NULL；
• C、WHERE电子邮件ISNULL；
• D、WHERE电子邮件ISNOTNULL；

第2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3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4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)))

第5题：

You need to create a table named ORDERS that contains four columns: 1.an ORDER_ID column of number data type 2.a CUSTOMER_ID column of number data type 3.an ORDER_STATUS column that contains a character data type 4.a DATE_ORDERED column to contain the date the order was placed When a row is inserted into the table, if no value is provided for the status of the order, the value PENDING should be used instead. Which statement accomplishes this?（）

• A、CREATE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status NUMBER(10) DEFAULT 'PENDING', date_ordered DATE );
• B、CREATE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) = 'PENDING', date_ordered DATE );
• C、CREATE OR REPLACE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) DEFAULT 'PENDING', date_ordered DATE );
• D、CREATE OR REPLACE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) = 'PENDING', date_ordered DATE );
• E、CREATE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) DEFAULT 'PENDING', date_ordered DATE );
• F、CREATE TABLE orders ( order_id NUMBER(10), customer_id NUMBER(8), order_status VARCHAR2(10) DEFAULT 'PENDING', date_ordered VARCHAR2 );

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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）

第14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15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16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));

第17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 Fransisco');
• B、SELECT city_address, COUNT (*) FROM customers WHERE city address IN ( 'Los Angeles', 'San Fransisco') GROUP BY city_address;
• C、SELECT city_address, COUNT(customer_id) FROM customers WHERE city_address IN ( 'Los Angeles', 'San Fransisco') GROUP BY city_address, customer_ id;
• D、SELECT city_address, COUNT (customer_id) FROM customers GROUP BY city_ address IN ('Los Angeles','San Fransisco');

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