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题解 | 统计每个用户的平均刷题数

# select university, difficult_level, # round(count(answer_cnt)/count(distinct u.device_id),4) as avg_answer_cnt # from user_profile u, question_pract...

2025-05-17

0 27

题解 | 统计每个学校各难度的用户平均刷题数

# select university, difficult_level, # round(count(q.question_id)/count(distinct q.device_id), 4) as avg_answer_cnt # from user_profile u, question_...

2025-05-17

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题解 | 统计每个学校的答过题的用户的平均答题数

# select u.university, # round(count(q.question_id)/count(distinct(q.device_id)),4) as avg_answer_cnt # from user_profile as u, question_practice_deta...

2025-05-17

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题解 | 浙江大学用户题目回答情况

# 方法一 # select device_id, question_id, result # from question_practice_detail # where device_id in ( # select device_id # from user_profile # ...

2025-05-17

0 26

题解 | 分组排序练习题

select university, round(avg(question_cnt), 4) as avg_question_cnt from user_profile group by university order by avg_question_cnt ASC;

2025-05-16

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题解 | 分组过滤练习题

select university, round(avg(question_cnt), 3) as avg_question_cnt, round(avg(answer_cnt), 3) as avg_answer_cnt from user_profile group by university ...

2025-05-16

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题解 | 分组计算练习题

select gender, university, count(*) as user_num, round(avg(active_days_within_30), 1) as avg_active_day, round(avg(question_cnt), 1) as avg_question_c...

2025-05-16

0 27

题解 | 计算男生人数以及平均GPA

select count(gender) as male_num, avg(gpa) as avg_gpa from user_profile where gender = 'male';

2025-05-16

0 35

题解 | 查找后降序排列

SELECT device_id, gpa, age from user_profile order by gpa DESC, age DESC;

2025-05-16

0 26

题解 | 基本数学函数

select *, abs(value) as absolute_value, ceil(value) as ceiling_value, floor(value) as floor_value, round(value,1) as rounded_value from numbers order ...

2025-05-16

0 31