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全部文章（共13篇）

题解 | #截取出年龄# 截取两遍是关键

正确题解： select substring_index(substring_index(profile,',',3),',',-1) as age, #重点 count(device_id) as number from user_submit group by age ⚠️Stella的两个迷糊...

2023-01-09

0 228

题解 | #统计每种性别的人数# substring_

正确题解1： select case when substring_index(profile,',',-1) ='male' then 'male' when substring_index(profile,',',-1) ='female' then 'female' end as gend...

2023-01-09

0 306

题解 | #计算25岁以上和以下的用户数量# case语句

正确题解： select case when age<25 or age is null then '25岁以下' when age>=25 then '25岁及以上' end as age_cut, count(device_id) as number from user_profi...

2023-01-09

1 476

题解 | #21年8月份练题总数# 三种限定时间条件的方法

正确题解： select count(distinct device_id) as did_cnt, count(question_id) as question_cnt from question_practice_detail where year(date)=2021 and month(da...

2023-01-08

0 307

题解 | #计算用户8月每天的练题数量# 关于时间的函数

正确题解： select day(date) as day, #day() count(question_id) as question_cnt from question_practice_detail where year(date)=2021 and month(date)=8 #year()...

2023-01-08

0 383

题解 | #查找山东大学或者性别为男生的信息# union

正确题解： select device_id, gender, age, gpa from user_profile where university = '山东大学' union all #本题重点 select device_id, gender, age, gpa from user_pr...

2023-01-08

0 268

题解 | #统计每个用户的平均刷题数# 第一次一次做对中等题

正确解法： select university, difficult_level, count(a.question_id) / count(distinct a.device_id) as avg_answer_cnt from question_practice_detail as a lef...

2023-01-08

0 268

题解 | #统计每个学校各难度的用户平均刷题数# 中等

正确解法： select university, difficult_level, round(count(a.question_id) / count(distinct a.device_id), 4) as avg_answer_cnt from question_practic...

2023-01-08

0 249

题解 | #统计每个学校的答过题的用户的平均答题数# 注意

解法： select university, count(b.question_id)/count(distinct b.device_id) as avg_answer_cnt from user_profile as a join question_practice_detail as b on...

2023-01-08

0 297

题解 | #浙江大学用户题目回答情况# 第一遍没做出来

解法1： select a.device_id, a.question_id, a.result from question_practice_detail as a, user_profile as b where a.device_id=b.device_id and b.university=...

2023-01-08

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