OCP-051-008
发表于:2024-09-22 作者:千家信息网编辑
千家信息网最后更新 2024年09月22日,View the Exhibit and examine the structure of the CUSTOMERS table.Which two tasks would require subq
千家信息网最后更新 2024年09月22日OCP-051-008View the Exhibit and examine the structure of the CUSTOMERS table.
Which two tasks would require subqueries or joins to be executed in a single statement? (Choose two.)
观察下面的表结构, 哪儿个任务需要在一个语句中包含子查询或者joins
A. listing of customers who do not have a credit limit and were born before 1980
B. finding the number of customers, in each city, whose marital status is 'married'
C. finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'
D. listing of those customers whose credit limit is the same as the credit limit of customers residing in the city 'Tokyo'
E. finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers
A. select * from CUSTOMERS where cust_credit_limit=0 and cust_year_of_birth <1980;
B. select cust_city,count(*) from CUSTOMERS where customer_marital='married' group by cust_city;
C. select avg(cust_credit_limit) from CUSTOMERS where customer_city in('Tokyo','Sydney');
D. select * from customers where cust_credit_limit in(select cust_credit_limit from customers where customer_city='Tokyo')
E. select count(*),customer_city from CUSTOMERS where customer_credit_limit>(select avg(customer_credit_limit) from CUSTOMERS ) group by customer_city;
Which two tasks would require subqueries or joins to be executed in a single statement? (Choose two.)
观察下面的表结构, 哪儿个任务需要在一个语句中包含子查询或者joins
A. listing of customers who do not have a credit limit and were born before 1980
B. finding the number of customers, in each city, whose marital status is 'married'
C. finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'
D. listing of those customers whose credit limit is the same as the credit limit of customers residing in the city 'Tokyo'
E. finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers
A. select * from CUSTOMERS where cust_credit_limit=0 and cust_year_of_birth <1980;
B. select cust_city,count(*) from CUSTOMERS where customer_marital='married' group by cust_city;
C. select avg(cust_credit_limit) from CUSTOMERS where customer_city in('Tokyo','Sydney');
D. select * from customers where cust_credit_limit in(select cust_credit_limit from customers where customer_city='Tokyo')
E. select count(*),customer_city from CUSTOMERS where customer_credit_limit>(select avg(customer_credit_limit) from CUSTOMERS ) group by customer_city;
任务
结构
语句
面的
E.
中包
查询
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