there is such a log table , which stores the records of user clicks, and there may be multiple records in one day. Now I want to count out that users who have only one day click records, users who have two days click records, users who have three days click records, , and so on, 4 days, 5 days. How should I write this sql?
related fields in the table:
Let's talk about the results first. After
SELECT TO_DAYS (MAX (time))-TO_DAYS (MIN (time)), USER FROM test GROUP BY USER
is executed, the result is as follows:
0 Wang
1 Li
2
query is grouped according to the user, and then the difference between the maximum time and the minimum time of the user is calculated. For example, the records of 0 and Wang indicate that all the records of the user Wang are on the same day.
-Segmentation Line-
after the hint of @ Xing Aiming, I realized that I was wrong in the first place.
after rethinking, you only need to write a sentence of code.
create a new table by testing yourself. The structure of the test, table is
sql:
only one day is easier to write.
use the maximum click time of the user-the minimum click time of the user. If not more than a day. There's only one day.
other cases
take the minimum time, add it by the number of days, if the maximum time is exceeded. Calculate the number of cycles
come on, logically, first set the click time to a certain range and then re-user_id (distinct), or group group by. This is the first step, and then use the count function to count the total number of entries, that is, how many days there are, and then group them according to this result. There is an one-day, two-day, three-day arrangement. I hope if you don't know the specific sentences, you have to learn them first. I'll write it down below for you.
select a.datenum,group_concat(a.user_id,',') from (select count(*) as datenum,user_id from table group by to_days(click_time),user_id ) as a group by a.datenum; ps:, because there is no specific data test, whether it can be used or not, you still have to test it (if you want to write it with me). That's the logic.
the description of the problem is rather vague. @ Mr. Chen er has given a general idea, @ liuxy to see if it is the result you want.
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