OCP-12c2019年CUUG OCP 071考试题库（74题）

Posted 2021-02-12 cnblogs5359

74、View the exhibit and examine the structure of ORDERS and CUSTOMERS tables.

 

ORDERS

Name     Null?       Type

ORDER_ID  NOT NULL  NUMBER(4)

ORDER_DATE  NOT NULL   DATE

ORDER_MODE   VARCHAR2(8)

CUSTOMER_ID NOT NULL  NUMBER(6)

ORDER_TOTAL   NUMBER(8, 2)

 

CUSTOMERS

Name  Null?  Type

CUSTOMER_ID  NOT NULL  NUMBER(6)

CUST_FIRST_NAME NOT NULL  VARCHAR2(20)

CUST_LAST_NAME  NOT NULL  VARCHAR2(20)

CREDIT_LIMIT  NUMBER(9,2)

CUST_ADDRESS  VARCHAR2(40)

Which INSERT statement should be used to add a row into the ORDERStable for the customer whose CUST_LAST_NAMEis Robertsand CREDIT_LIMITis 600?Assume there exists only one row with CUST_LAST_NAME as Roberts and CREDIT_LIMIT as 600.

A. INSERT INTO (SELECT o.order_id, o.order_date, o.order_mode, c.customer_id, o.order_total

FROM orders o, customers c

WHERE o.customer_id = c.customer_id AND c.cust_last_name=‘Roberts‘ AND c.credit_limit=600)

VALUES (1,‘10-mar-2007‘, ‘direct‘, (SELECT customer_id

FROM customers

WHERE cust_last_name=‘Roberts‘ AND credit_limit=600), 1000);

 

B. INSERT INTO orders (order_id, order_date, order_mode,

(SELECT customer id FROM customers

WHERE cust_last_name=‘Roberts‘ AND credit_limit=600), order_total);

VALUES (1,‘10-mar-2007‘, ‘direct‘, &customer_id, 1000);

 

C. INSERT INTO orders

VALUES (1,‘10-mar-2007‘, ‘direct‘,

(SELECT customer_id

FROM customers

WHERE cust_last_name=‘Roberts‘ AND credit_limit=600), 1000);

 

D. INSERT INTO orders (order_id, order_date, order_mode,

(SELECT customer_id FROM customers

WHERE cust_last_name=‘Roberts‘ AND credit_limit=600), order_total);

VALUES (1,‘10-mar-2007‘, ‘direct‘, &customer_id, 1000);

 

Correct Answer: C

Section: (none)

Explanation 相关的语句，注意子查询查出来的是某个列的值：

INSERT INTO emp (empno,ename,job,deptno,sal)

VALUES (1,‘cuug‘, ‘direct‘,

(SELECT deptno

FROM dept

 

WHERE dname=‘RESEARCH‘ ), 1000);