连续登录2-7天用户

建表

sql">create table continuous_login(user_id1 integer comment '用户id',date_login date comment '登陆日期'
) comment '用户登录表';insert into continuous_login
values(1,'2022-01-10'),(1,'2022-01-11'),(1,'2022-01-13'),(1,'2022-01-14'),(1,'2022-01-15'),(1,'2022-01-16'),(1,'2022-01-17'),(1,'2022-01-18'),(1,'2022-01-19'),(2,'2022-01-05'),(2,'2022-01-06'),(2,'2022-01-07'),(2,'2022-01-11'),(3,'2022-01-01'),(3,'2022-01-03'),(3,'2022-01-11');

排名找规律

sql">with tmp as(-- 过滤 保证无重复日期select *from continuous_logingroup by user_id1,date_login
),tmp1 as(-- 日期排名 等差数列select*,dense_rank() over(partition by user_id1 order by date_login) drfrom continuous_login ),tmp2 as(-- 等差 - 等差 =  差一样selectuser_id1,count(1) days_login_consecutive, -- 连续登录天数concat(min(date_login),'/',max(date_login)) date_finish_begin -- 拼接 连续登录起止日期from tmp1group by user_id1, adddate(date_login,-dr) -- 连续登录日期 等差数列having count(1) between 2 and 7 -- 连续登录所有情况中过滤出连续登录2-7天)
select -- 输出格式user_id1,group_concat(days_login_consecutive) days_login_consecutive,group_concat(date_finish_begin) date_finish_begin
from tmp2
group by user_id1;

最大连胜次数

建表

sql">create table if not exists match1(player_id integer unsigned not null default 0 comment '玩家ID',match_day date default '1970-01-01' comment '比赛日期',result1 varchar(5) comment '比赛结果'
) comment '玩家比赛表';insert into match1
value('1','2022-01-17','Win'),('1','2022-01-18','Win'),('1','2022-01-25','Win'),('1','2022-01-31','Draw'),('1','2022-02-08','Win'),('2','2022-02-06','lose'),('2','2022-02-08','lose'),('3','2022-03-30','Win');

只输出连胜结果

sql">with tmp as(-- 日期多次排名 找规律selectplayer_id,match_day,result1,if(result1 = 'Win',1,0) result1_if, -- if标记 为了计数dense_rank() over(partition by player_id order by match_day) dr,dense_rank() over(partition by player_id,result1 order by match_day) dr1from match1
),tmp1 as(-- 连续同种结果 差一样selectplayer_id,sum(result1_if) consecutive_countfrom tmpgroup by player_id,dr - dr1)
select -- 输出格式player_id,'Win' result1, -- 比赛结果max(consecutive_count) max_consecutive_count -- 最大连胜次数
from tmp1
group by player_id;

输出所有连续结果

sql">with tmp as(-- 日期多次排名 找规律selectplayer_id,match_day,result1,dense_rank() over(partition by player_id order by match_day) dr,dense_rank() over(partition by player_id,result1 order by match_day) dr1from match1
)
select -- 连续同种结果 差一样player_id,any_value(result1) result1, -- 比赛结果count(result1) consecutive_count -- 同种结果 连续次数
from tmp
group by player_id,dr - dr1;