Postgres: how to find absent users in DB

Question

Regular tasks when need to compare two db. Somene has sent me list of users, that should be in my database. 127 positions.

When I make SELECT * FROM users WHERE id

Answer 1

You can use an outer join to a values list:

SELECT t.id 
FROM (
  values 
     ('089477'),('089485'),('089417'),('089419'),('089416'),('089415'),('082513'),('087201'),('084769'),('083467'),('089498'),
     ('089394'),('085097'),('084818'),('089497'),('085208'),('082924'),('087204'),('087257'),('084708'),('0844187'),('089119'),
     ('088475'),('089448'),('084824'),('089436'),('085200'),('086431'),('089444'),('089479'),('089486'),('089460'),('089442'),
     ('089449'),('089413'),('089420'),('084917'),('084702'),('089433'),('089437'),('089443'),('081804'),('088813'),('089480'),
     ('089441'),('087184'),('081806'),('089435'),('081784'),('089401'),('089434'),('089423'),('089384'),('089422'),('089382'),
     ('089476'),('089473'),('089406'),('089461'),('089404'),('089409'),('089410'),('089412'),('089411'),('089396'),('089006'),
     ('089381'),('089379'),('089378'),('089397'),('089405'),('080006'),('089293'),('089478'),('084846'),('085210'),('089453'),
     ('089400'),('089452'),('089389'),('089383'),('089456'),('089402'),('089394'),('089418'),('089392'),('089387'),('089399'),
     ('089101'),('089117'),('080163'),('086021'),('081059'),('089414'),('089108'),('089288'),('089447'),('089446'),('089388'),
     ('089445'),('089386'),('089430'),('088828'),('088375'),('089407'),('083429'),('088645'),('089377'),('089342'),('089337'),
     ('089332'),('081635'),('089426'),('087197'),('089425'),('087767'),('088395'),('089341'),('089349'),('082114'),('082123'),
     ('084687'),('089333'),('089297'),('087371'),('089331')
) as t(id)
  left join users u on t.id = u.trader_systems_id 
where u.trader_systems_id  is null;

You can also return found and not found users:

select t.id, 
       u.*,
from (
   values ( .... )
) as t(id)
  left join users u on t.id = u.trader_systems_id;

---

If you get a single string with comma separated IDs, you can use unnest() and string_to_array() to turn that in a proper set:

select t.id, 
       u.*,
from unnest(string_to_array('089477,089485,089417,089419,089416,..', ',',  null)) as t(id)
  left join users u on t.id = u.trader_systems_id;

Answer 2

Based on your comment:

select * from [given_list] EXCEPT select * from [already_in_database] 

This will return the users who are in one list but not the other.

Obviously, without more information it is hard to make use of primary keys, and so forth