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(共8篇)
题解 | #计算总和#
from OrderItems group by order_num having total_price>=1000 order by order_num; 解题思路:1⃣️聚合函数+having筛选条件。2⃣️group by order_num,按照order_num分组。
Mysql
2022-06-23
0
145
题解 | #检索产品名称和描述(一)#
from Products where prod_desc LIKE'%toy%'; 题目中要求包含toy,就是用通配符LIKE。内容包含的,需要在字符串两侧都加上%。
Mysql
2022-06-22
1
272
题解 | #浙大不同难度题目的正确率#
sum(if(q1.result='right',1,0))/count(q1.question_id) as correct_rate from question_detail q left join question_practice_detail q1 on q.question_id=q1....
Mysql
2022-06-20
0
126
题解 | #统计每种性别的人数#
SELECT SUBSTRING(profile FROM 15 FOR 6) AS gender, COUNT(blog_url) AS number FROM user_submit GROUP BY gender;
Mysql
2022-01-18
0
207
题解 | #计算用户8月每天的练题数量#
SELECT EXTRACT(DAY FROM date) AS day, COUNT(question_id) AS question_cnt FROM question_practice_detail WHERE date BETWEEN '2021-08-01' AND '2021-08-31...
Mysql
2022-01-18
0
155
题解 | #查看不同年龄段的用户明细#
SELECT device_id,gender, CASE WHEN age<20 THEN '20岁以下' WHEN age BETWEEN 20 AND 24 THEN '20-24岁' WHEN age>=25 THEN '25岁及以上' ELSE '其他' END AS age_...
Mysql
2022-01-18
0
188
题解 | #计算25岁以上和以下的用户数量#
SELECT CASE WHEN age<25 OR age is NULL THEN '25岁以下' WHEN age>=25 THEN '25岁及以上' ELSE NULL END AS age_cnt, count(*) AS number FROM user_profile...
Mysql
2022-01-18
0
156
题解 | #分组过滤练习题#
SELECT university,AVG(question_cnt) AS avg_question_cnt,AVG(answer_cnt) AS avg_answer_cnt FROM user_profile GROUP BY university HAVING avg_question_cn...
Mysql
2021-12-07
0
152