OCP-1Z0-051-V9.02-77题
2013-10-18 14:39
513 查看
77. In the CUSTOMERS table, the CUST_CITY column contains the value 'Paris' for the
CUST_FIRST_NAME 'ABIGAIL'.
Evaluate the following query:
SQL> SELECT INITCAP(cust_first_name ' '
UPPER(SUBSTR(cust_city,-LENGTH(cust_city),2)))
FROM customers
WHERE cust_first_name = 'ABIGAIL';
What would be the outcome?
A. Abigail PA
B. Abigail Pa
C. Abigail IS
D. an error message
Answer: B
答案解析:
参考:/article/1628983.html
此为嵌套函数,由里到外可知:
LENGTH(cust_city)=LENGTH('Paris')=5
SUBSTR(cust_city,-5,2)=从右往左第5个,在顺着输出2个字符,即Pa
UPPER(Pa)=PA
INITCAP(cust_first_name ' ' UPPER(SUBSTR(cust_city,-LENGTH(cust_city),2)))=INITCAP('ABIGAIL' '' 'PA')=Abigail Pa
即选B。
CUST_FIRST_NAME 'ABIGAIL'.
Evaluate the following query:
SQL> SELECT INITCAP(cust_first_name ' '
UPPER(SUBSTR(cust_city,-LENGTH(cust_city),2)))
FROM customers
WHERE cust_first_name = 'ABIGAIL';
What would be the outcome?
A. Abigail PA
B. Abigail Pa
C. Abigail IS
D. an error message
Answer: B
答案解析:
参考:/article/1628983.html
此为嵌套函数,由里到外可知:
LENGTH(cust_city)=LENGTH('Paris')=5
SUBSTR(cust_city,-5,2)=从右往左第5个,在顺着输出2个字符,即Pa
UPPER(Pa)=PA
INITCAP(cust_first_name ' ' UPPER(SUBSTR(cust_city,-LENGTH(cust_city),2)))=INITCAP('ABIGAIL' '' 'PA')=Abigail Pa
即选B。
相关文章推荐
- OCP-1Z0-051-V9.02-6题
- OCP-1Z0-051-V9.02-127题
- OCP-1Z0-051-V9.02-60题
- OCP-1Z0-051-V9.02-31题
- OCP-1Z0-051-V9.02-51题
- OCP-1Z0-051-V9.02-61题
- OCP-1Z0-051-V9.02-79题
- OCP-1Z0-051-V9.02-92题
- OCP-1Z0-051-V9.02-129题
- OCP-1Z0-051-V9.02-143题
- OCP-1Z0-051-V9.02-170题
- OCP-1Z0-051-V9.02-73题
- OCP-1Z0-051-V9.02-36题
- OCP-1Z0-051-V9.02-7题
- OCP-1Z0-051-V9.02-128题
- OCP-1Z0-051-V9.02-11题
- OCP-1Z0-051-V9.02-58题
- OCP-1Z0-051-V9.02-30题
- OCP-1Z0-051-V9.02-52题
- OCP-1Z0-051-V9.02-62题