OCP-1Z0-051-V9.02-57题
2013-10-17 11:46
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57. The CUSTOMERS table has the following structure:
name Null Type
CUST_ID NOT NULL NUMBER
CUST_FIRST_NAME NOT NULL VARCHAR2(20)
CUST_LAST_NAME NOT NULL VARCHAR2(30)
CUST_INCOME_LEVEL VARCHAR2(30)
CUST_CREDIT_LIMIT NUMBER
You need to write a query that does the following tasks:
1. Display the first name and tax amount of the customers. Tax is 5% of their credit limit.
2. Only those customers whose income level has a value should be considered.
3. Customers whose tax amount is null should not be considered.
Which statement accomplishes all the required tasks?
A. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT
FROM customers
WHERE cust_income_level IS NOT NULL AND
tax_amount IS NOT NULL;
B. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT
FROM customers
WHERE cust_income_level IS NOT NULL AND
cust_credit_limit IS NOT NULL;
C. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT
FROM customers
WHERE cust_income_level <> NULL AND
tax_amount <> NULL;
D. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT
FROM customers
WHERE (cust_income_level,tax_amount) IS NOT NULL;
Answer: B
答案解析:
A,WHERE子句不能跟别名
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level IS NOT NULL AND TAX_AMOUNT IS NOT NULL
3 /
WHERE cust_income_level IS NOT NULL AND TAX_AMOUNT IS NOT NULL
*
ERROR at line 2:
ORA-00904: "TAX_AMOUNT": invalid identifier
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level IS NOT NULL AND cust_credit_limit * .05 IS NOT NULL;
CUST_FIRST_NAME TAX_AMOUNT
-------------------- ----------
Abigail 75
Abigail 350
...
B,正确
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level IS NOT NULL AND cust_credit_limit IS NOT NULL;
CUST_FIRST_NAME TAX_AMOUNT
-------------------- ----------
Abigail 75
Abigail 350
...
C错误,一是where子句不能用别名,二是非null不能用<>来表达
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level <> NULL AND tax_amount <> NULL;
WHERE cust_income_level <> NULL AND tax_amount <> NULL
*
ERROR at line 2:
ORA-00904: "TAX_AMOUNT": invalid identifier
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level <> NULL AND cust_credit_limit<> NULL;
no rows selected
D错误,语法错误
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE (cust_income_level,tax_amount) IS NOT NULL;
WHERE (cust_income_level,tax_amount) IS NOT NULL
*
ERROR at line 2:
ORA-00920: invalid relational operator
name Null Type
CUST_ID NOT NULL NUMBER
CUST_FIRST_NAME NOT NULL VARCHAR2(20)
CUST_LAST_NAME NOT NULL VARCHAR2(30)
CUST_INCOME_LEVEL VARCHAR2(30)
CUST_CREDIT_LIMIT NUMBER
You need to write a query that does the following tasks:
1. Display the first name and tax amount of the customers. Tax is 5% of their credit limit.
2. Only those customers whose income level has a value should be considered.
3. Customers whose tax amount is null should not be considered.
Which statement accomplishes all the required tasks?
A. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT
FROM customers
WHERE cust_income_level IS NOT NULL AND
tax_amount IS NOT NULL;
B. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT
FROM customers
WHERE cust_income_level IS NOT NULL AND
cust_credit_limit IS NOT NULL;
C. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT
FROM customers
WHERE cust_income_level <> NULL AND
tax_amount <> NULL;
D. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT
FROM customers
WHERE (cust_income_level,tax_amount) IS NOT NULL;
Answer: B
答案解析:
A,WHERE子句不能跟别名
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level IS NOT NULL AND TAX_AMOUNT IS NOT NULL
3 /
WHERE cust_income_level IS NOT NULL AND TAX_AMOUNT IS NOT NULL
*
ERROR at line 2:
ORA-00904: "TAX_AMOUNT": invalid identifier
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level IS NOT NULL AND cust_credit_limit * .05 IS NOT NULL;
CUST_FIRST_NAME TAX_AMOUNT
-------------------- ----------
Abigail 75
Abigail 350
...
B,正确
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level IS NOT NULL AND cust_credit_limit IS NOT NULL;
CUST_FIRST_NAME TAX_AMOUNT
-------------------- ----------
Abigail 75
Abigail 350
...
C错误,一是where子句不能用别名,二是非null不能用<>来表达
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level <> NULL AND tax_amount <> NULL;
WHERE cust_income_level <> NULL AND tax_amount <> NULL
*
ERROR at line 2:
ORA-00904: "TAX_AMOUNT": invalid identifier
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE cust_income_level <> NULL AND cust_credit_limit<> NULL;
no rows selected
D错误,语法错误
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
2 WHERE (cust_income_level,tax_amount) IS NOT NULL;
WHERE (cust_income_level,tax_amount) IS NOT NULL
*
ERROR at line 2:
ORA-00920: invalid relational operator
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