select cust_name,sum(item_price*quantity) total_price from Customers t1 inner join Orders t2 on t1.cust_id=t2.cust_id inner join OrderItems t3 on t2.order_num=t3.order_num group by cust_name having total_price>=1000 order by total_price

select t1.cust_name,t2.order_num,sum(t3.quantity*t3.item_price) OrderTotal from Customers t1,Orders t2,OrderItems t3 where t1.cust_id=t2.cust_id and t2.order_num=t3.order_num group by t1.cust_name,t2.order_num order by t1.cust_name

select t2.cust_id,sum(t1.quantity*t1.item_price) total_ordered from OrderItems t1,Orders t2 where t1.order_num=t2.order_num group by t2.cust_id order by total_ordered desc

select count(distinct device_id) as did_cnt, count(question_id) as question_cnt from question_practice_detail where month(date)=8

select t3.difficult_level, round(sum(if(t2.result='right',1,0))/count(t2.question_id),4) as correct_rate from user_profile t1,question_practice_detail t2,question_detail t3 where t1.device_id = t2.device_id and t2.question_id = t3.question_id and t1.university='浙江大学' group by t3.difficult_level orde...