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题解 | #查看不同年龄段的用户明细#

select device_id, gender, case when age >= 25 then '25岁及以上' when age >= 20 then '20-24岁' when age < 20 ...

Mysql

2022-03-18

0 212

题解 | #统计每个学校各难度的用户平均刷题数#

select case when age >= 25 then '25岁及以上' else '25岁以下' //这里注意用else：包含 null end as age_cut, count(*) number from user_profile group by age_cut;

Mysql

2022-03-18

0 235

题解 | #统计每个学校各难度的用户平均刷题数#

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...

Mysql

2022-03-18

1 317

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(qpd.question_id)/count(distinct qpd.device_id)),4) as avg_answer_cnt //注意：共同列必须指明来自那张...

Mysql

2022-03-18

2 335

题解 | #统计每个学校的答过题的用户的平均答题数#

select university, round((count(question_id)/count(distinct q.device_id)),4) as avg_answer_cnt //保留四位小数；计算平均做题数：每个学校总题数/每个学校用户数（设备数-设备...

Mysql

2022-03-15

1 237

题解 | #浙江大学用户题目回答情况#

select device_id,question_id,result from question_practice_detail where device_id in (select device_id from user_profile where university = '浙江大学');//...

Mysql

2022-03-13

1 221

题解 | #浙江大学用户题目回答情况#

select q.device_id,question_id,result //注意：两个表***有的字段必须：表别名.字段名指明来自那张表？ from question_practice_detail q join user_profile u on q.device_id = u.device...

Mysql

2022-03-13

1 208

题解 | #分组排序练习题#

select university,avg(question_cnt) as avg_question_cnt from user_profile group by university order by avg_question_cnt;//默认升序

Mysql

2022-03-13

0 220

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...

Mysql

2022-03-13

0 182

题解 | #分组过滤练习题#

select university, avg(question_cnt) as avg_question_cnt,//平均 avg(answer_cnt) as avg_answer_cnt//平均 from user_profile group by universit...

Mysql

2022-03-13

1 201