日期范围内每位用户每天的登录次数

Question

I saw many posts around that area, but couldn't find the exact one. I have a table that registers all of users logins to my app and it contains two columns - userID and timeOfLogin. It looks like that:

userID, timeOfLogin
1     , 14-01-10 00:07:38
2     , 14-01-10 01:28:45
3     , 14-01-10 01:28:45
1     , 14-01-09 02:04:08
1     , 14-01-09 06:14:54
etc....

I want to have a table that counts the number of unique user logins per day since a specific "day1" that I define in the query (day2 is the following day and so on). The table should look something like:

userID, numOFLogins day1, numOfLogins Day2, numOfLogins Day3, ...., numOfLogins DayN
1     ,      10         , 12              , 0               , ...., 12
2     ,      3          , 6               , 7               , ...., 15
132   ,      0          , 5               , 9               , ...., 14

Answer 1

You can do this with conditional aggregation:

select userId,
       sum(date(timeOfLogin) = date(@day1)) as NumLogins_0,
       sum(date(timeOfLogin) = date(date(@day1) + 1)) as NumLogins_1,
       sum(date(timeOfLogin) = date(date(@day1) + 2)) as NumLogins_2,
       sum(date(timeOfLogin) = date(date(@day1) + 3)) as NumLogins_3,
       sum(date(timeOfLogin) = date(date(@day1) + 4)) as NumLogins_4
from table t
group by userId;

In MySQL, date(timeOfLogin) = @day1 is treated as a 0 when the expression is false and 1 when it is true.

I'm just using @day1 to represent your variable for the date, whatever that is.

This will work for a fixed number of columns (such as the 5 days shown above). If you want a variable number of columns, then you cannot do this with a simple SQL statement. You will need to construct the SQL in a string, use prepare , and execute it.