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题解 | #确定最佳顾客的另一种方式（二）#

```select c.cust_name, ois.total_price from Customers c join Orders o on c.cust_id = o.cust_id join (select order_num, sum(item_price * quantity) as t...

Mysql

2022-05-29

0 294

题解 | #返回顾客名称和相关订单号以及每个订单的总价#

select cust_name, t1.order_num, oi.OrderTotal from Customers join Orders t1 using(cust_id) join (select&n...

Mysql

2022-05-28

0 275

题解 | #返回每个顾客不同订单的总金额#

select o.cust_id, oi.total_ordered from Orders o join (select os.order_num, sum(item_price * q...

Mysql

2022-05-28

0 233

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 where order_num in (select distinct order_num ...

Mysql

2022-05-28

0 252

题解 | #检索并列出已订购产品的清单#

select order_num, prod_id, quantity from OrderItems group by order_num, prod_id, quantity having quantity...

Mysql

2022-05-25

1 656

题解 | #返回更多的产品#

select u.order_num from(select order_num, sum(quantity) as num from OrderItems group by order_num having ...

Mysql

2022-05-25

0 290

题解 | #计算用户的平均次日留存率#

COUNT(q2.device_id) / COUNT(q1.device_id) AS avg_ret FROM (SELECT DISTINCT device_id, date FROM question_practice_detail)as q1 LEFT JOIN (SE...

Mysql

2022-05-25

0 331

题解 | #找出每个学校GPA最低的同学#

select device_id,university,gpa from user_profile where (university,gpa) in (select university,min(gpa) from u...

Mysql

2022-05-24

0 300

题解 | #提取博客URL中的用户名#

select device_id, substring_index(blog_url, '/', -1) as user_name from user_submit 和上一题一样用到 su...

Mysql

2022-05-24

0 281

题解 | #截取出年龄#

```select substring_index(substring_index(profile, ',', -2), ',', 1) as age, count(device_id) as number from user_submit group by age substr...

Mysql

2022-05-24

0 329