3

我在 Postgres 中有一个包含此数据的表,我很难将其转换为 JSON 对象。

node_id    parent_node    name
-------    -----------    ----
1                         node1
2          1              node2
3          1              node3
4          2              node4
5          2              node5
6          2              node6
7          3              node7
8          3              node8

我该如何转换它?

{
    name: 'node1'
    childs: [
        { 
            name: 'node2',
            childs: [
                {
                    name: 'node4',
                    childs: []
                },
                {
                    name: 'node5',
                    childs: []
                },
                {
                    name: 'node6',
                    childs: []
                }
            ]
        },
        ...    
    ]
}

任何建议都会有所帮助。谢谢

4

1 回答 1

5

使用WITH RECURSIVEhttps://www.postgresql.org/docs/current/static/queries-with.html)和 JSON 函数(https://www.postgresql.org/docs/current/static/functions-json.html)我构建了这个解决方案:

db<>小提琴

核心功能:

    WITH RECURSIVE tree(node_id, ancestor, child, path, json) AS  (
      SELECT 
          t1.node_id, 
          NULL::int, 
          t2.node_id,
          '{children}'::text[] || 
             (row_number() OVER (PARTITION BY t1.node_id ORDER BY t2.node_id) - 1)::text,-- C
          jsonb_build_object('name', t2.name, 'children', array_to_json(ARRAY[]::int[])) -- B
      FROM test t1
      LEFT JOIN test t2 ON t1.node_id = t2.parent_node                                   -- A
      WHERE t1.parent_node IS NULL

      UNION

      SELECT
          t1.node_id, 
          t1.parent_node, 
          t2.node_id,
          tree.path || '{children}' || (row_number() OVER (PARTITION BY t1.node_id ORDER BY t2.node_id) - 1)::text, 
          jsonb_build_object('name', t2.name, 'children', array_to_json(ARRAY[]::int[]))
      FROM test t1
      LEFT JOIN test t2 ON t1.node_id = t2.parent_node
      INNER JOIN tree ON (t1.node_id = tree.child)
      WHERE t1.parent_node = tree.node_id                                                -- D
    )
    SELECT                                                                               -- E
        child as node_id, path, json 
    FROM tree 
    WHERE child IS NOT NULL ORDER BY path

EveryWITH RECURSIVE包含一个 startSELECT和一个递归部分(第二个SELECT),由 a 组合而成UNION

A:为了找到 a 的孩子而加入表对自己不利node_id

B:为可以插入其父级的子级构建 json 对象

C:构建必须插入子对象的路径(从根)。窗口函数row_number()https://www.postgresql.org/docs/current/static/tutorial-window.html)在父项的子项数组中生成子项的索引。

D:递归部分作为初始部分工作,但有一个区别:它不是搜索根元素,而是搜索具有最后一次递归的父节点的元素。

E:执行递归并过滤没有任何子元素的所有元素会得到以下结果:

node_id   path                      json
2         children,0                {"name": "node2", "children": []}
4         children,0,children,0     {"name": "node4", "children": []}
5         children,0,children,1     {"name": "node5", "children": []}
6         children,0,children,2     {"name": "node6", "children": []}
3         children,1                {"name": "node3", "children": []}
7         children,1,children,0     {"name": "node7", "children": []}
8         children,1,children,1     {"name": "node8", "children": []}

虽然我发现无法在递归中添加所有子元素(原始 json 不是全局变量;所以它总是知道直接祖先的变化,而不是它们的兄弟姐妹),我不得不在几秒钟内迭代行。

这就是我构建函数的原因。在那里我可以对全局变量进行迭代。使用该函数jsonb_insert,我将所有计算的元素插入到根 json 对象中 - 使用计算的路径。

CREATE OR REPLACE FUNCTION json_tree() RETURNS jsonb AS $$
DECLARE
    _json_output jsonb;
    _temprow record;
BEGIN
    SELECT 
        jsonb_build_object('name', name, 'children', array_to_json(ARRAY[]::int[])) 
    INTO _json_output 
    FROM test 
    WHERE parent_node IS NULL;

    FOR _temprow IN
        /* Query above */
    LOOP
        SELECT jsonb_insert(_json_output, _temprow.path, _temprow.json) INTO _json_output;
    END LOOP;

    RETURN _json_output;
END;
$$ LANGUAGE plpgsql;

最后一步是调用函数并使 JSON 更具可读性 ( jsonb_pretty())

{
    "name": "node1",
    "children": [{
        "name": "node2",
        "children": [{
            "name": "node4",
            "children": []
        },
        {
            "name": "node5",
            "children": []
        },
        {
            "name": "node6",
            "children": []
        }]
    },
    {
        "name": "node3",
        "children": [{
            "name": "node7",
            "children": []
        },
        {
            "name": "node8",
            "children": []
        }]
    }]
}

我确信可以优化查询,但对于草图它可以工作。

于 2018-08-29T11:19:04.893 回答