127. Examine the data in the CUSTOMERS table:
CUSTNO CUSTNAME CITY
1 KING SEATTLE
2 GREEN BOSTON
3 KOCHAR SEATTLE
4 SMITH NEW YORK
You want to list all cities that have more than one customer along with the customer details.
Evaluate the following query:
SQL>SELECT c1.custname, c1.city
FROM Customers c1 __________________ Customers c2
ON (c1.city=c2.city AND c1.custname<>c2.custname);
Which two JOIN options can be used in the blank in the above query to give the correct output? (Choose
two.)
A. JOIN
B. NATURAL JOIN
C. LEFT OUTER JOIN
D. FULL OUTER JOIN
E. RIGHT OUTER JOIN
Answer: AE
答案解析：
scott@TESTDB> SELECT * FROM CUSTOMERS;
CUSTNO CUSTNAME CITY
---------- ---------- ----------
1 KING SEATTLE
2 GREEN BOSTON
3 KOCHAR SEATTLE
4 SMITH NEW YORK
你想要列出所有的城市,有一个以上的客户以及客户的细节
A
scott@TESTDB> select c1.custname,c1.city from customers c1 join customers c2
2 on (c1.city=c2.city and c1.custname<>c2.custname);
CUSTNAME CITY
---------- ----------
KOCHAR SEATTLE
KING SEATTLE
B，select列表中c1.city有限定
scott@TESTDB> select c1.custname,c1.city from customers c1 natural join customers c2
2 on (c1.city=c2.city and c1.custname<>c2.custname);
on (c1.city=c2.city and c1.custname<>c2.custname)
*
ERROR at line 2:
ORA-00933: SQL command not properly ended
C
scott@TESTDB> select c1.custname,c1.city from customers c1 left outer join customers c2
2 on (c1.city=c2.city and c1.custname<>c2.custname);
CUSTNAME CITY
---------- ----------
KOCHAR SEATTLE
KING SEATTLE
SMITH NEW YORK
GREEN BOSTON
D
scott@TESTDB> select c1.custname,c1.city from customers c1 full outer join customers c2
2 on (c1.city=c2.city and c1.custname<>c2.custname);
CUSTNAME CITY
---------- ----------
KOCHAR SEATTLE
KING SEATTLE
SMITH NEW YORK
GREEN BOSTON
E
scott@TESTDB> select c1.custname,c1.city from customers c1 right outer join customers c2
2 on (c1.city=c2.city and c1.custname<>c2.custname);
CUSTNAME CITY
---------- ----------
KING SEATTLE
KOCHAR SEATTLE