将 PostgreSQL 关系查询结果嵌套为 JSON 对象答案

【问题标题】：Nesting PostgreSQL relational query results as JSON objects将 PostgreSQL 关系查询结果嵌套为 JSON 对象
【发布时间】：2020-05-02 12:10:07
【问题描述】：

我的数据库中有 3 个表：parent、children 和 grandchildren。由于parent-children和children-grandchildren是多对多相关的，所以也有2个关系表。

我有一个查询，我希望将所有 parent 元素与嵌套的 children 和 grandchildren 行（如果有）作为 JSON 对象：

SELECT
  p.*,
  COALESCE(json_agg(json_build_object(
    'child_id', c.child_id,
    'child_name', c.child_name
    -- * GET ALL grandchildren FOR THIS child HERE!
  )) FILTER (WHERE c.child_id IS NOT NULL), '[]')
  AS children
FROM parent p
LEFT JOIN parent_children pc
ON pc.parent_id = p.parent_id
LEFT JOIN children c
ON c.child_id = pc.child_id
GROUP BY p.parent_id;

这个查询，后来在我的 JS 代码中，最终成为一个不错的 JSON 对象：

[
  {
    "parent_id": 1,
    "parent_name": "whatever",
    "children": [
      {
        "child_id": 1,
        "child_name": "le child"
      },
      {
        "child_id": 2,
        "child_name": "le second child"
      }
    ]
  },
  {
    "parent_id": 2,
    "parent_name": "second",
    "children": []
  }
]

正如我在查询中所评论的，我想为每个 children 元素添加所有 grandchildren 行。我有一个与上面相同的查询，除了它使用不同的表来获取该关系：

SELECT
  c.*,
  COALESCE(json_agg(json_build_object(
    'grandchild_id', gc.grandchild_id,
    'grandchild_name', gc.grandchild_name
  )) FILTER (WHERE gc.grandchild_id IS NOT NULL), '[]')
  AS grandchildren
FROM children c
LEFT JOIN children_grandchildren cg
ON cg.child_id = c.child_id
LEFT JOIN grandchildren g
ON g.grandchild_id = gc.grandchild_id
GROUP BY c.child_id;

我不确定这是否可能，如果可以，如何精确嵌套查询以获得如下结果：

[
  {
    "parent_id": 1,
    "parent_column": "whatever",
    "children": [
      {
        "child_id": 1,
        "child_name": "le child",
        "grandchildren": [
          {
            "grandchild_id": 1,
            "grandchild_name": "foo"
          },
          {
            "grandchild_id": 2,
            "grandchild_name": "bar"
          }
        ]
      },
      {
        "child_id": 2,
        "child_name": "le second child",
        "grandchildren": []
      }
    ]
  },
  {
    "parent_id": 2,
    "parent_name": "second",
    "children": []
  }
]

【问题讨论】：

标签： json postgresql nested aggregate

【解决方案1】：

我设法通过简单地LEFT JOINing grandchildren 表解决了这个问题：

SELECT
  p.*,
  COALESCE(json_agg(json_build_object(
    'child_id', c.child_id,
    'child_name', c.child_name
    -- * GET ALL grandchildren FOR THIS child HERE!
  )) FILTER (WHERE c.child_id IS NOT NULL), '[]')
  AS children
FROM parent p
LEFT JOIN parent_children pc
ON pc.parent_id = p.parent_id
LEFT JOIN children c
ON c.child_id = pc.child_id
LEFT JOIN (
  SELECT
    c.*,
    COALESCE(json_agg(json_build_object(
      'grandchild_id', g.grandchild_id,
      'grandchild_name', g.grandchild_name
    ))
    FILTER (WHERE g.grandchild_id IS NOT NULL), '[]') 
    AS grandchildren
  FROM children c
  LEFT JOIN children_grandchildren cg
  ON c.child_id = cg.child_id
  LEFT JOIN grandchildren g
  ON g.grandchild_id = cg.grandchild_id
  GROUP BY c.child_id
) grandchildren ON grandchildren.child_id = c.child_id
GROUP BY p.parent_id;

这会生成具有我需要的关系的 JSON 对象：

[
  {
    "parent_id": 1,
    "parent_column": "whatever",
    "children": [
      {
        "child_id": 1,
        "child_name": "le child",
        "grandchildren": [
          {
            "grandchild_id": 1,
            "grandchild_name": "foo"
          },
          {
            "grandchild_id": 2,
            "grandchild_name": "bar"
          }
        ]
      },
      {
        "child_id": 2,
        "child_name": "le second child",
        "grandchildren": []
      }
    ]
  },
  {
    "parent_id": 2,
    "parent_name": "second",
    "children": []
  }
]

【讨论】：