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题解 | 统计每个学校各难度的用户平均刷题数

select up.university,qd.difficult_level, round(count(qd.question_id)/count(distinct qpd.device_id),4)avg_answer_cnt from user_profile up join question...

2025-07-21

0 13

题解 | 基本数学函数

select id,value,abs(value) absolute_value,ceil(value) ceiling_value,floor(value) floor_value,round(value,1) rounded_value from numbers

2025-07-21

0 12

题解 | 纠错4

做完了SQL必知必会：）@烤点老白薯(637174235) SELECT cust_name, cust_contact, cust_email FROM Customers WHERE cust_state = 'MI' UNION SELECT cust_name, cust_conta...

2025-07-21

1 21

题解 | 组合 Products 表中的产品名称和 Customers 表中的顾客名称

select prod_name from Products union select cust_name from Customers order by prod_name

2025-07-21

0 13

题解 | 将两个 SELECT 语句结合起来（一）

select prod_id,quantity from OrderItems where quantity = 100 or prod_id like "BNBG%"

2025-07-21

0 12

题解 | 列出供应商及其可供产品的数量

select V.vend_id,count(P.prod_id) prod_id from Vendors V left join Products P using (vend_id) group by V.vend_id order by V.vend_id

2025-07-21

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题解 | 返回产品名称和与之相关的订单号

select prod_name,order_num from Products A left join OrderItems B on A.prod_id=B.prod_id union select prod_name,order_num from Products C right join O...

2025-07-21

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题解 | 检索每个顾客的名称和所有的订单号（二）

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

2025-07-21

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题解 | 检索每个顾客的名称和所有的订单号（一）

这题是中等？ 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

2025-07-21

0 12

题解 | 确定最佳顾客的另一种方式（二）

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

2025-07-21

0 11