Find the chat both persons are a part of
chat_person --- id chat_id person_id 1 1 20 2 1 19 3 2 19 4 2 3 5 3 19 6 3 2
I am trying to find the chat_id where p1=20 and p2=2 are both in. If there are none, return none.
SELECT DISTINCT "t1".chat_id FROM "chat_person" t1 WHERE EXISTS ( SELECT 1 FROM "chat_person" t2 WHERE "t2".person_id = 20 ) AND "t1".person_id = 2
This query is incorrectly returning
chat_id: 3. There is no common chat_id with both person_id=20 and person_id=2, so it should not return anything.
I think you may have missed to add where condition in exist.
SELECT DISTINCT "t1".chat_id FROM "chat_person" t1 WHERE EXISTS ( SELECT 1 FROM "chat_person" t2 WHERE "t2".person_id = 20 and t2.ChatID = "t1".chat_id ) AND "t1".person_id = 2
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The most simple way is aggregation:
select chat_id from chat_person group by chat_id having bool_or(person_id = 2) and bool_or(person_id = 20);
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you could try like below if you need all the others fields
select t1.* from chat_person t1 where exists ( select 1 from chat_person t2 where t2.chat_id=t1.chat_id and person_id in (2,20) having count(distinct person_id)=2)
or you can simply do like below if you just need
select chat_id from cte t2 where person_id in (2,20) group by chat_id having count(distinct person_id)=2
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Is this what you want?
SELECT chat_id, count(distinct person_id) from table group by chat_id having count(case when person_id in (2,20) then person_id end)=2
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- @Thorston Just wanted to know is there any way like
sum(person_id)=22 and count(*)=2in having to compensate person_id
- @Himanshu Ahuja: Yes. Here are two options: (1)
having sum(when person_id in (2,22)) then 1 else 0 end) = 2. (2)
where person_id in (2,22) group by chat_id having count(*) = 2.
- Perfect but second
wherestatement should not work in all cases as the Op's query says 20,2 can be there with other values as well Thanks mate :)
- @imanshu Ahuja: The second statement works very well. We use
WHEREto only get rows with persons 2 and 20. Then we see whether we get them both (
count(*) = 2). Only if there could be duplicates in the table (which wouldn't make sense), we'd have to
- Yeah it could have been an error if the groups were to not fall in place as i guess groups are falling in place meaning 1 group 1 count, 2nd group 2nd count
- Now it looks fine.