两种解法

解法一：

# 方法一：用左连接
SELECT COUNT(date2)/ COUNT(date1) AS avg_ret
FROM (
    SELECT DISTINCT 
  		qpd.device_id,
		qpd.date AS date1,
		uniq_id_date.date AS date2
    FROM question_practice_detail AS qpd
    LEFT JOIN (
        SELECT DISTINCT device_id, date
        FROM question_practice_detail 
    ) AS uniq_id_date
    ON qpd.device_id = uniq_id_date.device_id
        AND DATE_ADD(qpd.date, INTERVAL 1 DAY) = uniq_id_date.date
) AS id_last_next_date

解法二：

# 方法二：由内而外展开，理解这个问题，为什么要加distinct
SELECT AVG(IF(DATEDIFF(date2, date1) = 1, 1, 0)) AS avg_ret
FROM (
    SELECT 
        device_id, 
        date AS date1,
        LEAD(date) OVER (PARTITION BY device_id ORDER BY date) AS date2
    FROM (
        SELECT DISTINCT device_id, date
        FROM question_practice_detail
    ) AS uniq_id_date
) AS id_last_next_date

几个注意点

1. 客户次日留存率是什么
2. 所有子查询必须取别名
3. date_add(your_date, INTERVAL expr TYPE)

     expr可以是数字，也可以是表达式，TYPE可以是年、天、月、时、分、秒

4. DISTINCT作用于多个属性，直接在DISTINCT后面加上属性的名称

SELECT DISTINCT device_id, date FROM question_practice_detail