22 - 游戏玩法分析 IV

• 考点： 聚合函数
  
  

# 日期相加 date_add(min(event_date),INTERVAL 1 DAY) 
select round(count(distinct player_id)/(select count(distinct player_id) from Activity),2) fraction  
fromActivity 
where-- 如果日期加一天的数据能在表中查到，说明连续登录了两天(player_id,event_date) in (select player_id,date_add(min(event_date),INTERVAL 1 DAY) from Activity GROUP BYplayer_id);