Postgresql 如何返回分组用户id的最近值？

如何使结果只返回每个分组用户id的最后一个条目。下面的示例

此外，任何改进查询的方法都将受到高度赞赏<代码>属性是一个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;

结果是：

 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)

预期结果：

 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)

---

在

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

SELECT DISTINCT ON（表达式[，…]）仅保留给定表达式计算结果相等的每组行的第一行。DISTINCT ON表达式使用与ORDER BY相同的规则进行解释（见上文）。请注意，每个集合的“第一行”是不可预测的，除非使用ORDER BY来确保所需的行首先出现

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这是可行的，但如果我想将其中一列（即

submissions\u url

）保留为数组而不是唯一的不同值，该怎么办？这可能吗？