青岛网站美工鹤壁做网站优化

当前位置： 首页 > news >正文

青岛网站美工,鹤壁做网站优化,百度竞价培训,装修设计网站有哪些引言 场景介绍#xff1a; 许多互联网平台为了提高用户的参与度和忠诚度#xff0c;会推出各种连续登录奖励机制。例如#xff0c;游戏平台会给连续登录的玩家发放游戏道具、金币等奖励#xff1b;学习类 APP 会为连续登录学习的用户提供积分#xff0c;积分可兑换课程或…引言 场景介绍 许多互联网平台为了提高用户的参与度和忠诚度会推出各种连续登录奖励机制。例如游戏平台会给连续登录的玩家发放游戏道具、金币等奖励学习类 APP 会为连续登录学习的用户提供积分积分可兑换课程或其他福利。通过这些激励措施平台希望用户能够养成持续使用产品的习惯从而提升产品的活跃度和留存率。同时对于平台运营者来说分析用户的连续登录数据可以了解用户的使用习惯和忠诚度进而优化产品功能和运营策略。 题目描述 假设我们有一个记录用户登录信息的表表名为 login_table其中包含两个字段uid用户 ID和 dt登录日期。现在需要完成以下三个任务 查询连续登录超过三天的用户找出在一段时间内连续登录天数大于三天的用户列表。这有助于平台识别出那些高度活跃且对产品有较高忠诚度的用户以便进一步进行精细化运营和奖励。查询每个用户连续登录的最大天数对于每个用户统计其在所有登录记录中连续登录的最长时间段这能帮助我们了解不同用户的活跃程度差异为个性化运营提供数据基础。查询一个用户连续登录的最大天数可隔一天在计算用户连续登录天数时允许中间间隔一天只要整体登录天数最多就是我们要找的结果。比如用户在 1、3、5、6 日登录那么其连续登录的最大天数为 6 天。这种统计方式可以更灵活地评估用户的活跃程度考虑到了用户可能因为某些特殊情况中断一天登录但整体仍保持较高的使用频率。 数据准备与代码实现 数据准备 1 2025-01-01 1 2025-01-02 1 2025-01-03 2 2025-01-07 2 2025-01-08 3 2025-01-09 3 2025-01-10 3 2025-01-12 3 2025-01-131. 查询连续登录超过三天的用户 思路 用户登录记录编号利用row_number()函数按uid分区并依dt升序排序生成序号rn实现对各用户登录时间进行排序编号计算连续登录首日利用date_add函数将dt减去rn计算每行对应的连续登录起始日期first_day。选出连续登录超过三天大用户利用group by按uid和first_day分组结合having筛选出分组行数大于等于3的记录实现找出连续登录超三天的用户uid。 with data as (select 1 as uid,2025-01-01 as dt union allselect 1 as uid,2025-01-02 as dt union allselect 1 as uid,2025-01-03 as dt union allselect 2 as uid,2025-01-07 as dt union allselect 2 as uid,2025-01-08 as dt union allselect 3 as uid,2025-01-09 as dt union allselect 3 as uid,2025-01-10 as dt union allselect 3 as uid,2025-01-12 as dt union allselect 3 as uid,2025-01-13 as dt ), data2 as (select uid,dt,row_number() over (partition by uid order by dt) rn from data ), data3 as (select uid,dt,rn,date_add(dt,-rn) as first_day from data2 ) select uid from data3 group by uid,first_day having count(1) 3;2. 查询每个用户连续登录的最大天数 思路 用户登录记录编号利用窗口函数row_number()按uid分区并依dt升序排序生成序号rn实现对各用户登录时间进行排序编号。计算连续登录首日利用date_add函数将dt减去rn计算每行对应的连续登录起始日期first_day。统计分组登录天数利用group by按uid和first_day分组通过count()统计同一组合的天数login_day以此统计出每个用户每段连续登录的天数。获取用户最大连续登录天数再次使用group by对uid进行分组通过max(login_day)从每个用户的多段连续登录天数中选出最大值最终得到每个用户连续登录的最大天数。 with data as (select 1 as uid,2025-01-01 as dt union allselect 1 as uid,2025-01-02 as dt union allselect 1 as uid,2025-01-03 as dt union allselect 2 as uid,2025-01-07 as dt union allselect 2 as uid,2025-01-08 as dt union allselect 3 as uid,2025-01-09 as dt union allselect 3 as uid,2025-01-10 as dt union allselect 3 as uid,2025-01-12 as dt union allselect 3 as uid,2025-01-13 as dt ), data2 as (select uid,dt,row_number() over (partition by uid order by dt) rn from data ), data3 as (select uid,dt,rn,date_add(dt,-rn) as first_day from data2 ), data4 as (select uid,first_day,count() as login_day from data3 group by uid,first_day) select uid,max(login_day) from data4 group by uid;3. 查询一个用户连续登录的最大天数可以隔一天。解释1、3、5、6登录则最大登录天数为6天。 思路 查找上次登录时间利用lag函数按uid分区并依dt升序排序实现获取每行记录的上一次登录时间prev_dt。打标判断连续登录利用datediff函数计算dt与prev_dt的时间差根据差值情况打标flag实现区分是否连续登录如果差值小于2天或者null表示第一天标记为0都则标记为1。计算连续登录标识和利用sum函数按uid分组并依dt升序对flag求和生成sum_flag实现标识连续登录段。计算每组时间差值利用datediff函数对uid和sum_flag聚类分组后计算max(dt)与min(dt)的差值实现获取每个分组的时间跨度。获取最大连续登录天数利用分组和max函数选出每个用户的最大时间差值max(diff)1实现得到每个用户连续登录的最大天数max_login。 核心点将相差值小于等于2的分到同一组里然后采用分段思想计算每个分组分段的天数即为连续登录的天数。 with data as (select 1 as uid,2025-01-01 as dt union allselect 1 as uid,2025-01-02 as dt union allselect 1 as uid,2025-01-04 as dt union allselect 2 as uid,2025-01-07 as dt union allselect 2 as uid,2025-01-08 as dt union allselect 2 as uid,2025-01-11 as dt union allselect 2 as uid,2025-01-13 as dt union allselect 2 as uid,2025-01-15 as dt union allselect 3 as uid,2025-01-09 as dt union allselect 3 as uid,2025-01-10 as dt union allselect 3 as uid,2025-01-12 as dt union allselect 3 as uid,2025-01-15 as dt ), data2 as (select uid,dt,lag(dt, 1) over (partition by uid order by dt) prev_dt from data ), data3 as (select uid,dt,prev_dt,if(datediff(dt, prev_dt) 2 or datediff(dt, prev_dt) is null, 0 ,1) flag from data2 ), data4 as (select uid,dt,prev_dt,flag,sum(flag) over(partition by uid order by dt) as sum_flag from data3 ), data5 as (select uid,datediff(max(dt),min(dt)) diff from data4 group by uid,sum_flag ) select uid,max(diff)1 as max_login from data5 group by uid;知识点总结 1.窗口函数lag、row_number https://blog.csdn.net/Ahuuua/article/details/127136611 基本语法函数名(参数) OVER (PARTITION BY 子句 ORDER BY 子句 ROWS/RANGE子句) 函数名如sum、max、min、count、avg等聚合函数以及lead、lag行比较函数等over 关键字表示前面的函数是分析函数不是普通的集合函数分组子句over关键字后面挂号内的内容 lag()比较窗口函数 lag/lead(arg1,arg2,arg3)其中arg1为列名arg2为偏移值不能为负默认为1arg3超出记录窗口时的默认值当不指定默认值时则为null。lag:向前取n行 lead:向后取n行
row_number()排序窗口函数 排序窗口函数的主要作用是为查询结果中的每一行数据生成一个唯一的行号。这个行号是基于特定的排序规则生成的并且可以根据不同的分组条件进行独立编号。 rankrow_numberdense_rank10011110012190332

1. 日期计算函数 日期的三种形式 DATEYYYY-MM-DDCURRENT_DATE()DATETIMEYYYY-MM-DD HH:MM:SS、CURRENT_TIMESTAMP()TIMESTAMP时间戳1973-12-30 15:30:00为19731230153000UNIX_TIMESTAMP() 常见计算函数DATEDIFF(end,start)计算end-start单位天数TIMESTAMPDIFF(unit,start,end)计算end-start单位unit unitsecond、minute、hour、day、week、month、quarter季度、year DATE_ADD(date, num)计算datenum后的时间num参数表示要增加的时间间隔数量正数表示增加时间负数表示减少时间。 select CURRENT_DATE(),CURRENT_TIMESTAMP(),UNIX_TIMESTAMP();