跪求要一个offer
跪求要一个offer
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算法和编程题解
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题解 | #查看不同年龄段的用户明细#
#sql中case when end的用法 select device_id,gender, case when age>=25 then '25岁及以上' when age>=20 then '20-24岁' else '其他' ...
Mysql
2022-01-12
0
183
题解 | #删除记录(二)#
delete from exam_record where submit_time is null #sql中使用timestampdiff函数获取时间差 or timestampdiff(minute,start_time,submit_time)<5 order b...
Mysql
2022-01-10
0
153
题解 | #计算用户8月每天的练题数量#
#使用sql中的substring_index字符串截取函数 SELECT SUBSTRING_INDEX(date,"-",-1) as day, count(question_id) as question_cnt from question_practice_detail where ...
Mysql
2022-01-09
0
179
题解 | #截取出年龄#
select #sql中subString_index的嵌套使用的用法 SUBSTRING_INDEX(SUBSTRING_INDEX(profile,',',3),',',-1) as age, count(device_id) as NUMBER from user_s...
Mysql
2022-01-06
0
185
题解 | #统计每种性别的人数#
#sql中分割字符串的方法 select SUBSTRING_INDEX(profile,',',-1) as gender, COUNT(device_id) as NUMBER from user_submit group by gender;
Mysql
2022-01-05
0
242
题解 | #计算25岁以上和以下的用户数量#
select ( case when age>=25 then '25岁及以上' else '25岁以下' end) as age_cut, count(device_id) as number...
Mysql
2022-01-05
0
272
题解 | #查找山东大学男生的GPA#
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="ma...
Mysql
2021-11-03
0
213
题解 | #更新记录(二)#
UPDATE exam_record set submit_time='2099-01-01 00:00:00', score=0 where start_time<'2021-09-01 00:00:00' and submit_time is null;
Mysql
2021-11-01
0
264
题解 | #浙江大学用户题目回答情况#
select user_profile.device_id,question_practice_detail.question_id, question_practice_detail.result from user_profile,question_practice_detail where u...
Mysql
2021-11-01
1
227
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);
Mysql
2021-11-01
0
245
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