题解 | 平均活跃天数和月活人数

SELECT DATE_FORMAT(start_time, '%Y%m') AS month, 
       ROUND(COUNT(DISTINCT uid, DATE(start_time)) / COUNT(DISTINCT uid), 2) AS avg_active_days, # 考虑不同用户的登录天数且去除同一用户一天多次登录的情况
       COUNT(DISTINCT uid) AS mau
FROM exam_record
WHERE score IS NOT NULL
AND YEAR(start_time) = 2021
GROUP BY DATE_FORMAT(start_time, '%Y%m');