PostgreSQL: left outer join syntax

问题

I'm using PostgreSQL 8.4.6 with CentOS 5.5 and have a table of users:

# select * from pref_users where id='DE2';
 id  | first_name | last_name | female |      avatar      |        city         | lat | lng |           login            | last_ip | medals |           logout
-----+------------+-----------+--------+------------------+---------------------+-----+-----+----------------------------+---------+--------+----------------------------
 DE2 | Alex       |           | f      | 2_1280837766.jpg | г. Бохум в Германии |     |     | 2011-01-02 19:26:37.790909 |         |        | 2011-01-02 19:29:30.197062
(1 row)

and another table listing their "virtual money" won in a game each week:

# select * from pref_money where id='DE2';
 id  | money |   yw
-----+-------+---------
 DE2 |    66 | 2010-48
(1 row)

Then I'm trying to display both of these infos for a user, but I'm only interested in the user's money for the current week:

# select u.id,
    u.first_name,
    u.city,
    u.avatar,
    m.money,
    u.login > u.logout as online
from pref_users u, pref_money m where
    m.yw=to_char(current_timestamp, 'YYYY-IW') and
    u.id=m.id and
    u.id='DE2'
order by m.money desc;
 id | first_name | city | avatar | money | online
----+------------+------+--------+-------+--------
(0 rows)

In this case I've got no rows, because the user 'DE2' hasn't earned any "virtual money" this week yet.

I'd like to change my query so that it always returns data for existing users and if they haven't played this week - then the query should return 0.

So I guess I need an outer left join and I'm trying:

 select u.id,
    u.first_name,
    u.city,
    u.avatar,
    m.money,
    u.login > u.logout as online
from pref_users u left outer join pref_money m on (
    m.yw=to_char(current_timestamp, 'YYYY-IW') and
    u.id=m.id and
    u.id='DE2')
order by m.money desc;

but these returns me a lot of rows with different users instead of just one with the id='DE2'.

What am I doing wrong please?

回答1:

select u.id,
       u.first_name,
       u.city,
       u.avatar,
       coalesce(m.money,0),
       u.login > u.logout as online
  from pref_users u left outer join pref_money m on u.id=m.id
       and m.yw=to_char(current_timestamp, 'YYYY-IW') 
 where u.id='DE2'
 order by m.money desc;

回答2:

You want to put the filter condition u.id='DE2' in the WHERE clause, leaving everything else in the ON clause. For every user where the ON doesn't match, it will still output the row with the data from u and nulls for m.

回答3:

Try moving the

u.id='DE2'

into a WHERE clause, like this:

select u.id,
    u.first_name,
    u.city,
    u.avatar,
    m.money,
    u.login > u.logout as online
from pref_users u left outer join pref_money m on u.id=m.id
where m.yw=to_char(current_timestamp, 'YYYY-IW') and
    u.id='DE2'
order by m.money desc;

来源：https://stackoverflow.com/questions/4579957/postgresql-left-outer-join-syntax

标签

postgresql

left-join

outer-join