postgresql - 如何返回分组 user_id 的最新值？

Question

我怎样才能使结果只返回每个分组 user_id 的最后一个条目。示例如下。

此外，我们将不胜感激任何改进查询的方法。 properties 是一个 hstore 列。

SELECT user_id, json_agg(to_json(properties -> 'following')
ORDER BY id DESC) AS followings,
json_agg(to_json(properties -> 'assumed_gender') ORDER BY id DESC) AS assumed_genders,
json_agg(to_json(properties -> 'email') ORDER BY id DESC) AS emails,
json_agg(to_json(properties -> 'first_name') ORDER BY id DESC) AS first_names,
json_agg(to_json(properties -> 'last_name') ORDER BY id DESC) AS last_names,
json_agg(to_json(properties -> 'country_name') ORDER BY id DESC) AS country_names,
json_agg(to_json(properties -> 'city_name') ORDER BY id DESC) AS city_names,
json_agg(to_json(properties -> 'mobile_number') ORDER BY id DESC) AS mobile_numbers,
json_agg(to_json(properties -> 'submission_url') ORDER BY id DESC) AS submission_urls
FROM "daily_statistics" WHERE "daily_statistics"."campaign_id" = 72 AND "daily_statistics"."metric" = 'participation'
GROUP BY user_id ORDER BY max(id) DESC;

结果:

 user_id |    followings     | assumed_genders |                emails                 |    first_names     |     last_names     |  country_names   |     city_names     |        mobile_number
s        |                         submission_urls
---------+-------------------+-----------------+---------------------------------------+--------------------+--------------------+------------------+--------------------+---------------------
---------+------------------------------------------------------------------
      16 | ["false", "true"] | ["", ""]        | ["lorem@lorem.com", "lorem@amet.com"] | ["dolor", "ipsum"] | ["lorem", "ipsum"] | ["amet", "amet"] | ["dolor", "ipsum"] | ["9707759365", "2572
943441"] | ["http://www.dolor.com/hgtsjcbn", "http://www.sit.com/qlnogrzd"]
(1 row)

期望的结果:

 user_id |    followings     | assumed_genders |                emails                 |    first_names     |     last_names     |  country_names   |     city_names     |        mobile_number
s        |                         submission_urls
---------+-------------------+-----------------+---------------------------------------+--------------------+--------------------+------------------+--------------------+---------------------
---------+------------------------------------------------------------------
      16 | "true"            | ""              | "lorem@amet.com"                      | "ipsum"            | "ipsum"            | "amet"           | "ipsum"            |         "2572
943441"  | "http://www.sit.com/qlnogrzd"
(1 row)

Answer 1

使用distinct on

select distinct on (user_id)
    user_id,
    to_json(properties -> 'following') as followings,
    to_json(properties -> 'assumed_gender') as assumed_genders,
    to_json(properties -> 'email') as emails,
    to_json(properties -> 'first_name') as first_names,
    to_json(properties -> 'last_name') as last_names,
    to_json(properties -> 'country_name') as country_names,
    to_json(properties -> 'city_name') as city_names,
    to_json(properties -> 'mobile_number') as mobile_numbers,
    to_json(properties -> 'submission_url') as submission_urls
from "daily_statistics"
where
    "daily_statistics"."campaign_id" = 72
    and "daily_statistics"."metric" = 'participation'
order by user_id, id desc

http://www.postgresql.org/docs/current/static/sql-select.html

SELECT DISTINCT ON ( expression [, ...] ) keeps only the first row of each set of rows where the given expressions evaluate to equal. The DISTINCT ON expressions are interpreted using the same rules as for ORDER BY (see above). Note that the "first row" of each set is unpredictable unless ORDER BY is used to ensure that the desired row appears first