SQLite代码：

SELECT ROUND(COUNT(DISTINCT(a.user_id)) *1.0 / COUNT(DISTINCT(c.user_id)),3)
FROM login AS a,login AS c
WHERE EXISTS(
SELECT *
FROM login AS b
WHERE a.user_id=b.user_id AND DATE(a.date, '+1 day')=b.date
)

MYSQL代码：

SELECT ROUND(COUNT(DISTINCT(a.user_id)) *1.0 / COUNT(DISTINCT(c.user_id)),3)
FROM login AS a,login AS c
WHERE EXISTS(
SELECT *
FROM login AS b
WHERE a.user_id=b.user_id AND DATE_ADD(a.date,INTERVAL 1 DAY)=b.date
)