Return all records of non-given id if just one of those records matches the given id of another field

Question

After searching for a damn long time, I've not found a query to make this happen.

I have an "offers" table with a "listing_id" field and a "user_id" field and I need to get ALL the records for all listing_id's where at least one record matches the given user_id.

In other words, I need a query that determines the listing_id's that the given user is involved in, and then returns all the offer records of those listing_id's regardless of user_id.

That last part is the problem. It's getting all the other user's offer records to return when I'm only providing one user's id and no listing id's

I was thinking of first determining the listing_ids in a separate query and then using a php loop to create a WHERE clause for a second query that would consist of a bunch of "listing_id = $var ||" but then I couldn't bring myself to do it because I figured there must be a better way.

Hopefully this is easy and the only reason it has escaped me is because I've had my head up my ass. Will be happy to get this one behind me.

Thanks for taking the time.

Josh

Answer 1

您可以在MySQL方面进行两个查询，如下所示：

SELECT * FROM offers WHERE listing_id IN (SELECT listing_id FROM offers WHERE user_id = 1)

Answer 2

If I understand what you are after you should join offers on itself on listingid match and userid = given

select * from offers AS t1
inner join offers AS t2 on t1.listingid = t2.listingid and t1.userid = 1;