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sql练习-面试前刷题

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题解 | 统计每种性别的人数

select profile_gender gender,count(*) number from ( select substr(profile,15) profile_gender from user_submit )a group by profile_gender

2025-05-16

0 54

题解 | 计算用户的平均次日留存率

select count(if(datediff(b.date,a.date) = 1,a.device_id,null))/count(if(datediff(b.date,a.date) = 0,a.device_id,null)) avg_ret from ( select ...

2025-05-14

0 53

题解 | 计算用户8月每天的练题数量

select day(date) day, count(question_id) question_cnt from question_practice_detail where month(date) = '08' group by date

2025-05-14

0 56

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 >= 20 and age < 25 then '20-24岁' when...

2025-05-14

0 53

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' else '25岁及以上' end as age_cut, count(*) as number from user_profi...

2025-05-13

0 51

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' else '25岁及以上' end as age_cut, count(*) as number from user_profi...

2025-05-13

0 34

题解 | 查找山东大学或者性别为男生的信息

select device_id,gender,age,gpa from user_profile where university = '山东大学' union all select device_id,gender,age,gpa from user_profile where gender =...

2025-05-13

0 47

题解 | 统计每个用户的平均刷题数

select university,difficult_level,count(answer_cnt)/count(distinct a.device_id) avg_answer_cnt from user_profile a inner join question_practice_detail...

2025-05-13

0 44

题解 | 统计每个学校各难度的用户平均刷题数

select university,difficult_level,count(difficult_level)/count(distinct a.device_id) avg_answer_cnt from user_profile a inner join ( select b.devi...

2025-05-09

0 47

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 a.device_id) avg_answer_cnt from user_profile a inner join question_practice_detail b on a.device_...

2025-05-09

0 47