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60.View the Exhibit and examine the structure of CUSTOMERS table.

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132.View the Exhibit and examine the structure of the SALES and PRODUCTS tables. In the SALES table, PROD_ID is the foreign key referencing PROD_ID in the PRODUCTS table, You want to list each product ID and the number of times it has been

60.View the Exhibit and examine the structure of CUSTOMERS table.
Evaluate the following query:
SQL>SELECT cust_id, cust_city FROM customers WHERE 
cust_first_name NOT LIKE 'A_%g_%' AND
cust_credit_limit BETWEEN 5000 AND 15000 AND
cust_credit_limit NOT IN (7000, 11000) AND
cust_city NOT BETWEEN 'A' AND 'B';

Which statement is true regarding the above query?


A.It executes successfully.
B.It produces an error because the condition on the CUST_CITY column is not valid.
C.It produces an error because the condition on the CUST_FIRST_NAME column is not valid.
D.It produces an error because conditions on the CUST_CREDIT_LIMIT column are not valid.
答案:A
A:正确
B:错误

cust_city NOT BETWEEN 'A' AND 'B',这种是进行的ascii进行的判断,比如

SQL> select 1 from dual where 'B' between 'A' and 'C';

         1
----------

         1

‘B' between ’A' and ‘C'  相当于  ’B' >= 'A' and  'B' <='C'

在比如

SQL> select 1 from dual where 'ABC' between 'A' and 'C';

         1
----------
         1

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163.View the Exhibit and examine the description for the CUSTOMERS table. You want to update the CUST_CREDIT_LIMIT column to NULL for all the customers, where CUST_INCOME_LEVEL has NULL in the CUSTOMERS table. Which SQL statement will accom

SQL> select 1 from dual where 'CBA' between 'A' and 'C';


no rows selected

第一个有结果而第二个没有,这是因为‘ABC' 首字母为A,和between 的开头A相等,因此开始判断第二个字母,由于between开始的字符 没有第二个字母所以符合

第二个没有结果是因此,第一个为C和and的C相等,然后 判断第二个,第二个为B ,但是and的C没有第二个字符结果给大了

SQL> select 1 from dual where 'CBA' between 'A' and 'CC';


         1
----------
         1

这个就符合了,因此B<C



C:错误

cust_first_name NOT LIKE 'A_%g_%',这种匹配的是A开头,然后任意字符中间有个g,然后是任意字符,比如
SQL> select 1 from dual where 'Awwwwwgwwww' LIKE 'A_%g_%';


         1
----------
         1
D:错误

原文

60.View the Exhibit and examine the structure of CUSTOMERS table. Evaluate the following query: SQL>SELECT cust_id, cust_city FROM customers WHERE  cust_first_name NOT LIKE 'A_%g_%

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