Question

我刚开始使用postgreSQL，现在我正在尝试获取相关表条目的树结构。

假设我有以下表格：

街道，房产，建筑物，楼层，房间

streets (columns: id, name)
properties (columns: id, name, street_id)
buildings (columns: id, name, property_id)
floors (columns: id, name, building_id)
rooms (columns: id, name, floor_id)

现在我尝试获得如下输出：

[{
"id":1,
"name":"Main Street",
"properties": [{
     "id":"1",
     "name":"First Property",
     "buildings": [{
          "id":"1",
          "name":"Main Building",
          "floors": [{...},{...}]
        },{
          "id":"2",
          "name":"Shed",
          "floors": [{...},{...}]
        }]
   },{
     "id":"2",
     "name":"Another Property",
     "buildings": [{...},{...}]
   }]
}]

现在查询是这样的：

SELECT s.*, json_agg(p.*) as properties
FROM streets AS s 
INNER JOIN properties AS p ON s.id = p.street_id
GROUP BY s.id;

它为我提供了上面两层街道的输出和街道内的属性结果。但我无法理解如何将建筑物嵌入房屋，然后将建筑物嵌入建筑物......，

我可以为它们添加另一个INNER JOIN和json_agg（），但是它们不会在属性中。

我应该如何处理这个问题？

编辑：

感谢德米特里，你的回答非常好。但现在让我说我想将所有数据存储在名为“locations”的同一个表中，并通过“parent_id”链接条目

所以我有专栏：

id，name，type，parent_id

使用以下代码我得到了错误的层次结构。

WITH RECURSIVE locationtree AS (
   SELECT
      e.*,
      null::json as parent
   FROM location e
   WHERE parent_id IS NULL

   UNION ALL

   SELECT
      e.*,
      row_to_json(et.*) as parent
   FROM location e
   INNER JOIN locationtree et
      ON et.id = e.parent_id
)
SELECT *
FROM locationtree;

这导致了房间地板建造属性街

但我想在最初的问题和德米特里的答案中遵循以下结构街道属性房屋地板室

有关于此的任何建议吗？

Answer 1

使用 CTE 表达式查询可能如下：

WITH floors_agg AS (
    -- get floors with nested rooms
    SELECT 
      f.*,
      json_agg(r.*) nested_rooms 
    FROM floors f
      LEFT JOIN rooms r ON r.floor_id = f.id
    GROUP BY 1
), buildings_agg AS (
    -- get buildings with nested floors
    SELECT 
      b.*,
      json_agg(fa.*) nested_floors 
    FROM buildings b
      LEFT JOIN floors_agg fa ON fa.building_id = b.id
    GROUP BY 1
), properties_agg AS (
    -- get properties with nested buildings
    SELECT
      p.*,
      json_agg(ba.*)
    FROM properties p
      LEFT JOIN buildings_agg ba ON ba.property_id = p.id
    GROUP BY 1
), streets_agg AS (
  -- get streets with nested properties
    SELECT
      s.*,
      json_agg(pa.*) nested_properties
    FROM streets s
      LEFT JOIN properties_agg pa ON pa.street_id = s.id
    ORDER BY 1
)
SELECT json_agg(sa.*)
FROM streets_agg sa;

如何将多个postgreSQL表嵌套到一个json中？

1 个答案: