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);
原文地址：https://www.cnblogs.com/cnblogs5359/p/10509748.html