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题解 | 返回购买 prod_id 为 BR01 的产品的所有顾客的电子邮件（二）

select C.cust_email from (select order_num from OrderItems where prod_id='BR01') OI join Orders O on OI.order_num = O.order_num join Customers C on O....

2025-07-21

0 12

题解 | 确定哪些订单购买了 prod_id 为 BR01 的产品（二）

select O.cust_id,O.order_date from (select order_num from OrderItems where prod_id = 'BR01') OI join Orders O on OI.order_num = O.order_num order by...

2025-07-19

1 21

题解 | 返回顾客名称和相关订单号以及每个订单的总价

select C.cust_name,O.order_num,sum(OI.quantity*OI.item_price) OrderTotal from Customers C join Orders O on C.cust_id = O.cust_id join OrderItems OI on...

2025-07-19

1 15

题解 | 返回顾客名称和相关订单号

select C.cust_name,O.order_num from Customers C join Orders O on C.cust_id = O.cust_id order by C.cust_name

2025-07-19

1 12

题解 | 从 Products 表中检索所有的产品名称以及对应的销售总数

select P.prod_name,sum(OI.quantity) quant_sold from Products P join OrderItems OI on P.prod_id = OI.prod_id group by P.prod_name

2025-07-19

1 13

题解 | 返回每个顾客不同订单的总金额

select O.cust_id,sum(O_n.item_price*O_n.quantity) total_ordered from OrderItems O_n join Orders O on O_n.order_num = O.order_num group by O.cust_id or...

2025-07-19

1 18

题解 | 返回购买 prod_id 为 BR01 的产品的所有顾客的电子邮件（一）

SELECT DISTINCT C.cust_email FROM Customers C JOIN Orders O ON C.cust_id = O.cust_id JOIN OrderItems OI ON O.order_num = OI.order_num WHERE OI.prod_id...

2025-07-19

1 19

题解 | 分组排序练习题

select university,avg(question_cnt) avg_question_cnt from user_profile group by university order by avg_question_cnt

2025-07-18

2 29

题解 | 分组过滤练习题

select university,avg(question_cnt) avg_question_cnt,avg(answer_cnt) avg_answer_cnt from user_profile group by university having avg_question_cnt <...

2025-07-18

1 20

题解 | 分组计算练习题

select gender,university,count(*) user_num,round(avg(active_days_within_30),1) avg_active_day,round(avg(question_cnt),1) avg_question_cnt from user_pr...

2025-07-18

1 21