斎虅飛鳥

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题解 | #确定最佳顾客的另一种方式（二）#

select cust_name,sum(item_pricequantity) as total_price from Orders o inner join Customers c on c.cust_id = o.cust_id inner join OrderItems r on o.or...

Mysql

2022-05-09

2 307

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 cust_email from Customers where cust_id in(select cust_id from Orders where order_num in( select order_num from OrderItems where prod_id = "BR0...

Mysql

2022-05-09

0 241

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 cust_email from Orders t2 left join OrderItems t1 on t1.order_num = t2.order_num left join Customers t3 on t2.cust_id = t3.cust_id where prod_i...

Mysql

2022-05-08

0 212

题解 | #返回购买价格为 10 美元或以上产品的顾客列表#

select cust_id from Orders t2 join OrderItems t1 on t1.order_num = t2.order_num group by cust_id having sum(item_price) >= 10 order by cust_id

Mysql

2022-05-08

5 288

题解 | #检索并列出已订购产品的清单#

select order_num,prod_id,quantity from OrderItems where quantity >= 100 and prod_id = 'BR01' union select order_num,prod_id,quantity from OrderItem...

Mysql

2022-05-07

0 210

题解 | #各个视频的平均完播率#

SELECT a.video_id, round(count(if(end_time - start_time >= duration, 1, null))/count(start_time ),3) as avg_comp_play_rate FROM tb_user_video_log a...

Mysql

2022-05-06

0 222

题解 | #计算25岁以上和以下的用户数量#

select user_profile.device_id, university, count(question_practice_detail.device_id) as question...

Mysql

2022-05-06

0 335

题解 | #计算25岁以上和以下的用户数量#

select user_profile.device_id, university, count(question_practice_detail.device_id) as question...

Mysql

2022-04-23

9 347

题解 | #统计每个学校的答过题的用户的平均答题数#

SELECT university, COUNT(question_id)/COUNT(DISTINCT user_profile.device_id) AS avg_answer_cnt FROM user_profile,question_practice_detail WHERE ques...

Mysql

2022-04-22

3 269