Finding Records that Have Same Options

Question

Trying to figure out how find out which user records have at least 1 similar option enabled and then display these.

Tables:

user (id, username)
thing (id,title)
uthing (id, uid, tid)

Uthing table has a record of what options each user have selected.

uid = user.id
tid = thing.id

I need to compare the options (uthing) from user id 1 with all other users.

How can find users that have 1 or more options(uthing) of USER ID 1 and return these users with the options found?

This query will give me all options for user id 1.

select thing.title,thing.id as thingid 
from thing join uthing on uthing.tid = thing.id where uthing.uid = 1;

Answer 1

You achieve this by following this query

select uthing.id, uthing.uid, thing.title
from uthing left join thing on uthing.tid = thing.id
where uthing.tid in (
        select thing.id
        from thing join uthing on uthing.tid = thing.id
        where uthing.uid = 1
     );

Answer 2

Following your schema I've generated next example:

create table user (id int, username varchar(100));
create table thing (id int, title varchar(100));
create table uthing (id int, uid int, tid int);

insert into user values (1, 'user 1'),(2, 'user 2'),(3, 'user 3'),(4, 'user 4');
insert into thing values (1, 'thing 1'),(2, 'thing 2'),(3, 'thing 3'),(4, 'thing 4'),(5, 'thing 5');
insert into uthing values
(1, 1, 1),(2, 1, 3),
(3, 2, 2),(4, 2, 3),(5, 2, 5),
(6, 3, 4),
(7, 4, 1),(8, 4, 5);

This query select users where id <> 1 that has (exists) at least one thing in common with user id = 1.

select *
from   user u
join   uthing ut
on     ut.uid = u.id
join   thing t
on     t.id = ut.tid
where  u.id <> 1
and    exists (select 1
               from   uthing
               where  uid = 1
               and    tid = t.id);

id | username | id | uid | tid | id | title  
-: | :------- | -: | --: | --: | -: | :------
 2 | user 2   |  4 |   2 |   3 |  3 | thing 3
 4 | user 4   |  7 |   4 |   1 |  1 | thing 1

dbfiddle here