Question

我有一个显示列表，列出城市中的每个用户，以及他们在表中按照最新消息排序的最后一条消息：

Users
------
Caleb - Hey what's up?
------
Bill - Is there anything up tonight?
------
Jon - What's up man?
------

任何帮助优化下面的查询和/或帮助确定添加索引的位置都会非常有帮助。

我可能会对last_message_created_at进行非规范化并存储users，但希望避免这种回填。

用户表：

                    Table "public.users"
      Column       |            Type             | Modifiers
-------------------+-----------------------------+-----------
 id                | integer                     | not null
 role              | user_role                   | not null
 last_message_id   | integer                     |
 city_id           | integer                     |

Indexes:
    "users_pkey" PRIMARY KEY, btree (id)
    "ix_users_city_id" btree (city_id)
    "ix_users_last_message_id" btree (last_message_id)
Foreign-key constraints:
    "messages_last_message_id_fkey" FOREIGN KEY (last_message_id) REFERENCES messages(id)
    "users_city_id_fkey" FOREIGN KEY (city_id) REFERENCES cities(id)
Referenced by:
    TABLE "messages" CONSTRAINT "messages_from_user_id_fkey" FOREIGN KEY (from_user_id) REFERENCES users(id)
    TABLE "messages" CONSTRAINT "messages_to_user_id_fkey" FOREIGN KEY (to_user_id) REFERENCES users(id)

消息表：

                                        Table "public.messages"
      Column      |            Type             |                       Modifiers
------------------+-----------------------------+-------------------------------------------------------
 id               | integer                     | not null default nextval('messages_id_seq'::regclass)
 content          | character varying           |
 from_user_id     | integer                     |
 to_user_id       | integer                     |
 created_at       | timestamp without time zone |

Indexes:
    "messages_pkey" PRIMARY KEY, btree (id)
    "ix_messages_from_user_id" btree (from_user_id)
    "ix_messages_to_user_id" btree (to_user_id)
Foreign-key constraints:
    "messages_from_user_id_fkey" FOREIGN KEY (from_user_id) REFERENCES users(id)
    "messages_to_user_id_fkey" FOREIGN KEY (to_user_id) REFERENCES users(id)
Referenced by:
    TABLE "users" CONSTRAINT "messages_last_message_id_fkey" FOREIGN KEY (last_message_id) REFERENCES messages(id)

查询和计划：

EXPLAIN ANALYZE
SELECT users.city_id AS users_city_id, users.id AS users_id, users.last_message_id AS users_last_message_id
FROM users
JOIN messages ON messages.id = users.last_message_id
WHERE users.city_id = 1 ORDER BY users.last_message_id DESC;

QUERY PLAN
---------------------------------------------------------------------------------------------------------------------------
 Sort  (cost=951606.67..951811.07 rows=81760 width=12) (actual time=1934.501..1998.216 rows=79454 loops=1)
   Sort Key: users.last_message_id
   Sort Method: quicksort  Memory: 6797kB
   ->  Nested Loop  (cost=1575.71..944935.41 rows=81760 width=12) (actual time=26.784..1817.478 rows=79454 loops=1)
         ->  Bitmap Heap Scan on users  (cost=1575.71..33209.21 rows=84040 width=12) (actual time=26.656..393.749 rows=85348 loops=1)
               Recheck Cond: (city_id = 1)
               ->  Bitmap Index Scan on ix_users_city_id  (cost=0.00..1554.70 rows=84040 width=0) (actual time=20.679..20.679 rows=85348 loops=1)
                     Index Cond: (city_id = 1)
         ->  Index Only Scan using messages_pkey on messages  (cost=0.00..10.84 rows=1 width=4) (actual time=0.012..0.013 rows=1 loops=85348)
               Index Cond: (id = users.last_message_id)
               Heap Fetches: 79454
 Total runtime: 2058.134 ms
(12 rows)

Answer 1

除了排除没有消息的用户之外，您永远不会将查询中的messages表用于任何其他目的，因此这可能会更快：

select
    u.city_id as users_city_id,
    u.id as users_id,
    u.last_message_id as users_last_message_id
from users u
where
    u.city_id = 1
    and exists (
        select 1
        from messages
        where id = u.last_message_id
    )
order by u.last_message_id desc

从连接列优化ORDER BY

1 个答案: