Trying to get a count in 3 tables on MySql - mysql

Question

So I have 3 tables in a database
One is hotel which has hotel_id and status
one is partner which has partner_id and partner_name
and one is partner_hotel which has hotel_id and partner_id
What I am trying to get is the count for each partner that has a hotel with the status = 1
The closest I have gotten is
select p.partner_name,count(hotel_id)
from partner_hotel ph
join partner p on p.partner_id = ph.partner_id
group by ph.partner_id;
The problem is that does not limit to ones with a status of 1 and nothing I seem to be doing seems to work.

Answer 1

you should join hotel too if th hotel contain status and group by p.partner_id
select p.partner_name,count(*)
from partner_hotel ph
inner join partner p on p.partner_id = ph.partner_id
inner join hotel h on ph.hotel_id = h.hotel_id
where h.status = 1
group by p.partner_id;
or ig hotel use id
select p.partner_name,count(*)
from partner_hotel ph
inner join partner p on p.partner_id = ph.partner_id
inner join hotel h on ph.hotel_id = h.id
where h.status = 1
group by p.partner_id;

Answer 2

I guess you're trying to find the number of partners with one or more hotels having a status = 1. If my guess is correct, this gets you the one number.
SELECT COUNT(DISTINCT p.partner_id) partner_count
FROM partner p
JOIN partner_hotel ph ON p.partner_id = ph.partner_id
JOIN hotel h ON ph.hotel_id = h.hotel_id AND h.status = 1
If your requirement is to show one result row for each partner showing the number of hotels with status = 1, that will go like this.
SELECT COUNT(*) hotel_count,
p.partner_id
FROM partner p
JOIN partner_hotel ph ON p.partner_id = ph.partner_id
JOIN hotel h ON ph.hotel_id = h.hotel_id AND h.status = 1
GROUP BY p.partner_id
Pro tip: When preparing to write a query, it's helpful to imagine the result set you want. Ask yourself these questions:
How many rows must the result set have?
It should have one row per ____ (row per what? Hotel, Partner, Guest, Hotel with status = 1?).
Each row should show ____ values?
If you do this work of imagination, writing the query will be much easier.