【问题标题】：How do I combine multiple one-to-many results with Postgres json_agg?如何将多个一对多结果与 Postgres json_agg 结合起来？
【发布时间】：2018-04-16 22:49:40
【问题描述】：

函数中有多个 json_aggregates

我的 Postgres 数据库的主表与其他表有一对多的关系，每一个都由一个[*]_references 表连接。我需要一个以 JSON 形式从其他表返回的结果。我可以加入以获得我想要的数据：

SELECT ft.id, json_agg(genres) AS genres 
FROM ft_references ft 
INNER JOIN genre_references gr ON gr.reference_id = ft.id 
INNER JOIN genres_catalog genres ON gr.genre_id = genres.id 
WHERE ft.id = 2 GROUP BY ft.id;

 id |                      genres                      
----+--------------------------------------------------
  2 | [{"id":1,"name":"Action","other_info":null},    +
    |  {"id":2,"name":"Adventure","other_info":null}, +
    |  {"id":5,"name":"Comedy","other_info":null},    +
    |  {"id":8,"name":"Drama","other_info":null},     +
    |  {"id":10,"name":"Fantasy","other_info":null},  +
    |  {"id":13,"name":"Horror","other_info":null},   +
    |  {"id":25,"name":"War","other_info":null}]
(1 row)

同样是我的第二组结果：

SELECT ft.id, json_agg(tales) AS tales 
FROM ft_references ft 
INNER JOIN tale_references tr ON tr.reference_id = ft.id 
INNER JOIN tale_catalog tales ON tales.id = tr.tale_catalog_id 
WHERE ft.id = 2 GROUP BY ft.id;

 id |                                      tales                                      
----+---------------------------------------------------------------------------------
  2 | [{"id":8,"tale_id":"124","tale_type":"The Three Brothers","other_info":null},  +
    |  {"id":39,"tale_id":"327A","tale_type":"Hansel and Gretel","other_info":null}, +
    |  {"id":82,"tale_id":"570","tale_type":"Pied Piper","other_info":null}]
(1 row)

这两个结果都是正确的。然而，当我想把它们放在一起时，突然间我得到了我的类型和故事结果的产物（即：三倍结果！）：

SELECT ft.id, json_agg(genres) AS genres, json_agg(tale) AS tales 
FROM ft_references ft 
INNER JOIN genre_references gr ON gr.reference_id = ft.id 
INNER JOIN genres_catalog genres ON gr.genre_id = genres.id 
INNER JOIN tale_references tr ON tr.reference_id = ft.id 
INNER JOIN tale_catalog tale ON tale.id = tr.tale_catalog_id 
WHERE ft.id = 2 GROUP BY ft.id, gr.reference_id, tr.reference_id;
 id |                      genres                      |                                      tales                                      
----+--------------------------------------------------+---------------------------------------------------------------------------------
  2 | [{"id":1,"name":"Action","other_info":null},    +| [{"id":82,"tale_id":"570","tale_type":"Pied Piper","other_info":null},         +
    |  {"id":1,"name":"Action","other_info":null},    +|  {"id":39,"tale_id":"327A","tale_type":"Hansel and Gretel","other_info":null}, +
    |  {"id":1,"name":"Action","other_info":null},    +|  {"id":8,"tale_id":"124","tale_type":"The Three Brothers","other_info":null},  +
    |  {"id":2,"name":"Adventure","other_info":null}, +|  {"id":82,"tale_id":"570","tale_type":"Pied Piper","other_info":null},         +
    |  {"id":2,"name":"Adventure","other_info":null}, +|  {"id":39,"tale_id":"327A","tale_type":"Hansel and Gretel","other_info":null}, +
    |  {"id":2,"name":"Adventure","other_info":null}, +|  {"id":8,"tale_id":"124","tale_type":"The Three Brothers","other_info":null},  +
    |  {"id":5,"name":"Comedy","other_info":null},    +|  {"id":82,"tale_id":"570","tale_type":"Pied Piper","other_info":null},         +
    |  {"id":5,"name":"Comedy","other_info":null},    +|  {"id":39,"tale_id":"327A","tale_type":"Hansel and Gretel","other_info":null}, +
    |  {"id":5,"name":"Comedy","other_info":null},    +|  {"id":8,"tale_id":"124","tale_type":"The Three Brothers","other_info":null},  +
    |  {"id":8,"name":"Drama","other_info":null},     +|  {"id":82,"tale_id":"570","tale_type":"Pied Piper","other_info":null},         +
    |  {"id":8,"name":"Drama","other_info":null},     +|  {"id":39,"tale_id":"327A","tale_type":"Hansel and Gretel","other_info":null}, +
    |  {"id":8,"name":"Drama","other_info":null},     +|  {"id":8,"tale_id":"124","tale_type":"The Three Brothers","other_info":null},  +
    |  {"id":10,"name":"Fantasy","other_info":null},  +|  {"id":82,"tale_id":"570","tale_type":"Pied Piper","other_info":null},         +
    |  {"id":10,"name":"Fantasy","other_info":null},  +|  {"id":39,"tale_id":"327A","tale_type":"Hansel and Gretel","other_info":null}, +
    |  {"id":10,"name":"Fantasy","other_info":null},  +|  {"id":8,"tale_id":"124","tale_type":"The Three Brothers","other_info":null},  +
    |  {"id":13,"name":"Horror","other_info":null},   +|  {"id":82,"tale_id":"570","tale_type":"Pied Piper","other_info":null},         +
    |  {"id":13,"name":"Horror","other_info":null},   +|  {"id":39,"tale_id":"327A","tale_type":"Hansel and Gretel","other_info":null}, +
    |  {"id":13,"name":"Horror","other_info":null},   +|  {"id":8,"tale_id":"124","tale_type":"The Three Brothers","other_info":null},  +
    |  {"id":25,"name":"War","other_info":null},      +|  {"id":82,"tale_id":"570","tale_type":"Pied Piper","other_info":null},         +
    |  {"id":25,"name":"War","other_info":null},      +|  {"id":39,"tale_id":"327A","tale_type":"Hansel and Gretel","other_info":null}, +
    |  {"id":25,"name":"War","other_info":null}]       |  {"id":8,"tale_id":"124","tale_type":"The Three Brothers","other_info":null}]
(1 row)

如何重组我的查询，以便只获得第一个示例中正确的“流派”结果以及第二个示例中正确的“故事”结果，而不是像我的实际（第三个）例子？

【问题讨论】：

你为什么突然按这么多东西分组？像以前一样按 ft.id 分组，这行不通？
否；我尝试了各种分组，结果总是一样的，即使只有一个分组。
我相信你可以通过像json_agg(DISTINCT genres) 这样的事情来解决这个问题，这实际上可能是连接的问题，它会让你重复但在我半睡半醒的状态下我不能告诉你什么。

标签： sql json postgresql aggregate-functions

【解决方案1】：

答案如下：

SELECT ft.id, 
        json_agg(DISTINCT genres.*) AS genres, 
        json_agg(DISTINCT tale.*) AS tales
  FROM ft_references ft 
  JOIN genre_references gr ON gr.reference_id = ft.id
  JOIN genres_catalog genres ON genres.id = gr.genre_id
  JOIN tale_references tr ON tr.reference_id = ft.id
  JOIN tale_catalog tale ON tale.id = tr.tale_catalog_id
  WHERE ft.id = 2 GROUP BY ft.id;

【讨论】：