[单项选择] Examine the data from the ORDERS and CUSTOMERS tables. ORDERS ORD_ID ORD_DATE CUST_ID 12-JAN- 100 15 10000 09-MAR- 101 40 8000 09-MAR- 102 35 12500 15-MAR- 103 15 12000 25-JUN- 104 15 6000 18-JUL- 105 20 5000 106 18-JUL- 35 7000 21-JUL- 107 20 6500 04-AUG- 109 10 8000 CUSTOMERS CUST_ID CUST_NAME CITY 10 Smith Los Angeles 15 Bob San Francisco 20 Martin Chicago 25 Mary New York 30 Rina Chicago 35 Smith New York 40 Lind New York Evaluate the SQL statement: SELECT * FROM orders WHERE cust_id = (SELECT cust_id FROM customers WHERE cust_name = 'Smith'); What is the result when the query is executed?（）
A. ORD_ID ORD_DATE CUST_ID ORD_TOTAL 09-MAR- 102 35 12500 18-JUL- 106 35 7000 04-AUG- 108 10 8000
B. ORD_ID ORD_DATE CUST_ID ORD_TOTAL 09-MAR- 102 35 12500 18-JUL- 106 35 7000
C. ORD_ID ORD_DATE CUST_ID ORD_TOTAL 04-AUG- 108 10 8000
D. The query fails because the subquery returns more than one row.
E. The query fails because the outer query and the inner query are using different tables.

[单项选择] Examine the data from the ORDERS and CUSTOMERS table. ORDERS ORD_ID ORD_DATE CUST_ID ORD_TOTAL 100 12-JAN-2000 15 10000 09-MAR- 101 40 8000 09-MAR- 102 35 12500 15-MAR- 103 15 12000 104 25-JUN-2000 15 6000 105 18-JUL-2000 20 5000 106 18-JUL-2000 35 7000 107 21-JUL-2000 20 6500 04-AUG- 108 10 8000 CUSTOMERS CUST_ID CUST_NAME CITY 10 Smith Los Angeles 15 Bob San Francisco 20 Martin Chicago 25 Mary New York 30 Rina Chicago 35 Smith New York 40 Linda New York Which SQL statement retrieves the order ID, customer ID, and order total for the orders that are placed on the same day that Martin places his orders?（）
A. SELECT ord_id, cust_id, ord_total FROM orders, customers WHERE cust_name='Mating' AND ord _ date IN ('18-JUL-2000','21-JUL-2000');
B. SELECT ord_id, cust_id, ord_total FROM orders Where ord_date IN (SELECT ord_date FROM orders WHERE cust_id = (SELECT cust_id FROM customers WHERE cust_name = 'MARTIN'));
C. SELECT ord_id, cust_id, ord_total FROM orders Where ord_date IN (SELECT ord_date FROM orders, customers Where cust _ name = 'Martin');
D. SELECT ord_id, cust_id, ord_total FROM orders WHERE cust_id IN (SELECT cust_id FROM customers WHERE cust name = 'Martin');

[多项选择] Examine the SQL statements that creates ORDERS table: CREATE TABLE orders (SER_NO NUMBER UNIQUE, ORDER_ID NUMBER, ORDER_DATE DATE NOT NULL STATUS VARCHARD2(10) CHECK (status IN ('CREDIT','CASH')), PROD_ID_NUMBER REFERENCES PRODUCTS(PRODUCT_ID), ORD_TOTAL NUMBER, PRIMARY KEY (order id, order date)); For which columns would an index be automatically created when you execute the aboveSQL statement? （）
A. SER_NO
B. ORDER_ID
C. STATUS
D. PROD_ID
E. ORD_TOTAL
F. Composite index on ORDER_ID and ORDER_DATE

[单项选择]在下列查询语句中，与SELECT TAB1．* FROM TAB1 WHERE InStr([简历]，"篮球")<>0 功能相同的语句是
A. SELECT TAB1．* FROM TAB1 WHERE TAB1．简历Like "篮球"
B. SELECT TAB1．* FROM TAB1 WHERE TAB1．简历Like "*篮球"
C. SELECT TAB1．* FROM TAB1 WHERE TAB1．简历Like "*篮球*"
D. SELECT TAB1．* FROM TAB1 WHERE TAB1．简历Like "篮球*"

[单项选择] 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');

[单项选择]Data from Voyager II have presented astronomers with a {{U}}puzzle{{/U}} about why our outermost planet exists.
A. problem
B. mystery
C. question
D. point

[单项选择]Data from Voyager Ⅱ have presented astronomers with a {{U}}puzzle{{/U}} about why our outermost planet exists.
A. problem
B. mystery
C. question
D. point

[单项选择]Input data from keyboard are stored in （） for processing.
A. modem
B. bus
C. memory
D. printer

[单项选择]Selected financial data from Krandall, Inc. ' s balance sheet for the year ended December 31 was as follows (in $ ) :
A. Cash
B. $1100000
C. Accounts Payable
D. $ 400000
E. Accounts Receivable
F. 300000
G. Deferred Tax Liability
H. 700000
I. Inventory
J. 2400000
K. Long-term Debt
L.

8200000

M. Property, Plant & Eq.
N. 8000000

[单项选择] You need to dump the data from the master server and import it into a new slave server. Which mysqldump option can be used when dumping data from the master server in order to include the master server's binary log information?（）
A. Include-master-info
B. Master-binlog
C. Include-log-file
D. Master-data

[单项选择]Input data from keyboard are stored in （72） for processing.
A. modem B
B. bus
C. memory
D. printer

[多项选择] Examine the data from the EMP table: EMP_ID DEPT_ID COMMISSION 1 10 500 2 20 1000 3 10 4 10 600 5 30 800 6 30 200 7 10 8 20 300 The COMMISSION column shows the monthly commission earned by the employee. Which three tasks would require subqueries or joins in order to perform in a single step? （）
A. Deleting the records of employees who do not earn commission.
B. Increasing the commission of employee 3 by the average commission earned in department 20.
C. Finding the number of employees who do NOT earn commission and are working for department 20.
D. Inserting into the table a new employee 10 who works for department 20 and earns a commission that is equal to the commission earned by employee 3.
E. Creating a table called COMMISSION that has the same structure and data as the columns EMP_ID and COMMISSIONS of the EMP table.
F. Decreasing the commission by 150 for the employees who are working in department 30 and earning a commission of more then 800.

[单项选择] Examine the data in the EMPLOYEES and EMP_HIST tables: EMPLOYEES NAME DEPT_ID MGR_ID JOB_ID SALARY EMPLOYEE_ID 101 Smith 20 120 SA_REP 4000 102 Martin 10 105 CLERK 2500 103 Chris 20 120 IT_ADMIN 4200 104 John 30 108 HR_CLERK 2500 105 Diana 30 108 IT_ADMIN 5000 106 Smith 40 110 AD_ASST 3000 108 Jennifer 30 110 HR_DIR 6500 110 Bob 40 EX_DIR 8000 120 Ravi 20 110 SA_DIR 6500 EMP HIST EMPLOYEE_ID NAME JOB_ID SALARY 101 Smith SA_CLERK 2000 103 Chris IT_CLERK 2200 104 John HR_CLERK 2000 106 Smith AD_ASST 3000 108 Jennifer HR_MGR 4500 The EMP_HIST table is updated at the end of every year. The employee ID, name, job ID, and salary of each existing employee are modified with the latest data. New employee details are added to the table. Which statement accomplishes this task?（）
A. UPDATE emp_hist SET employee_id, name, job_id, salary = (SELECT employee_id, name, job_id, salary FROM employees) WHERE employee_id IN (SELECT employee_id FROM employees);
B. MERGE INTO emp_hist eh USING employees e ON (eh.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE SET eh.name = e.name, eh.job_id = e.job_id, eh.salary = e.salary WHEN NOT MATCHED THEN INSERT VALUES (e.employee id, e.name, job id, e.salary);
C. MERGE INTO emp_hist eh USING employees e ON (eh.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE emp hist SET eh.name = e.name, eh.job_id = e.job_id, eh.salary = e.salary WHEN NOT MATCHED THEN INSERT INTO emp_hist VALUES (e.employees_id, e.name, e.job_id, e.salary);
D. MERGE INTO emp_hist eh USING employees e WHEN MATCHED THEN UPDATE emp_hist SET eh.name = e.name, eh.job_id = e.job_id, eh.salary = e.salary WHEN NOT MATCHED THEN INSERT INTO emp_hist VALUES (e.employees_id, e.name, e.job_id, e.salary);

[判断题]出现“Resuming from disk，Press TAB to show POST screen”，某些主板的BIOS提供了Suspend to disk（挂起到硬盘）的功能，当使用者以Suspend to disk的方式来关机时，那么在下次开机时就会显示此提示消息。（）

[判断题]myTable=New System.Data.DataTable（），myTable的名称是Table1。