i have a list with peoples id and date, the list say when a person Entered to website (his id and date). how can i show for all the dates how many people enter the site two days in a row?
x
the data ( 30,000 like this in diffrent dates)
01/03/2019 4616
01/03/2019 17584
01/03/2019 7812
01/03/2019 34
01/03/2019 12177
01/03/2019 7129
01/03/2019 11660
01/03/2019 2428
01/03/2019 17514
01/03/2019 10781
01/03/2019 7629
01/03/2019 11119
I succeeded to show the amount of pepole enter the site on the same day but i didnt succeeded to add a column that show the pepole that enter 2 days in row.
date number_of_entrance
2019-03-01 7099
2019-03-02 7021
2019-03-03 7195
2019-03-04 7151
2019-03-05 7260
2019-03-06 7169
2019-03-07 7076
2019-03-08 7081
2019-03-09 6987
2019-03-10 7172
select date,count(*) as number_of_entrance
fROM [finalaa].[dbo].[Daily_Activity]
group by Date
order by date;
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Answer
how can i show for all the dates how many people enter the site two days in a row?
I would just use lag()
:
select count(distinct person)
from (select t.*,
lag(date) over (partition by person order by date) as prev_date
from t
) t
where prev_date = dateadd(day, -1, date);
Your code suggests SQL Server, so I used the date functions in that database.
If you want this per date:
select date, count(distinct person)
from (select t.*,
lag(date) over (partition by person order by date) as prev_date
from t
) t
where prev_date = dateadd(day, -1, date)
group by date;