How to get common IDs in each group from a group by SQL clause?

Question

I have data of calls for customers. I want to get those customers between two dates that have activity against every date. They did at least one activity every day. I tried following query.

Following is the query:

select date_id , count (distinct customer_id) from usage_analysis 
where usage_direction_type_id = 1
and date_id => 20130608 and date_id <= 20130612
group by date_id 

That returns:

DATE_ID         COUNT
----------------------------
20130608        23451
20130609        9878
20130610        56122   
20130611        7811
20130612        12334

But I want to get those customers that are common in each group. It may happen a person who called on 8 June does not exist on the next day. So I only want those customers that exist in every group.

Any idea who can I do that in SQL?

Answer 1

You can count the distinct dates for each customer. Only customers with five distinct dates would then pass the test. The following provides the list of customers:

select customer_id
from usage_analysis 
where usage_direction_type_id = 1 and
      date_id >= 20130608 and date_id <= 20130612
group by customer_id
having count(distinct date_id) = 5

Answer 2

@Gordon Linoff answer should be working fine for your situation. When you tried with 2 days, did you make sure to change the count value from 5 to 2?