SELECT COUNT(DISTINCT q2.date, q2.device_id) / COUNT(DISTINCT q1.date, q1.device_id) AS avg_ret /*排除用户在同一天多次提交*/ FROM question_practice_detail AS q1 LEFT OUTER JOIN question_practice_detail AS q2 ON q1.device_id = q2.device_id/*设备相同*/ AND DATEDIFF(q2.date, q1.date) = 1;/*相差一天*/ 思...