在postgres中将关系树转换为嵌套的json文档答案

【问题标题】：Converting relational tree to nested json document in postgres在postgres中将关系树转换为嵌套的json文档
【发布时间】：2022-01-17 23:45:17
【问题描述】：

在 PostgreSQL 中，有一个关系数据模型来表示组织内的层次结构。

create table employee (
    id              integer primary key, 
    name            varchar(40) not null, 
    supervisor_id   integer references employee 
);

只有 CEO 有supervisor_id=NULL。其他所有员工都有一名主管，其中有一些supervisor_id。我想将数据导出为单个嵌套的 json 文档

{ 
  "id": 1, 
  "name": "Name of company's CEO", 
  "supervises": [
    { 
      "id": 2, 
      "name": "Name of 1st EC member",
      "supervises": [ ... nested employees ... ]
    }, 
    {
      "id": 3, 
      "name": "Name of 2nd EC member",
      "supervises": [ ... nested employees ... ]
    }
    ... 
  ]
}

我按照https://www.postgresqltutorial.com/postgresql-recursive-query/ 中的示例进行操作，但它只能帮助我使用 WITH RECURSIVE 子句从上到下沿报告线识别所有员工。我知道我需要从树中深度最高的员工（不仅仅是叶节点）开始聚合，然后自下而上聚合它们，但我没有设法编写完成这项工作的查询。

感谢您的帮助！

【问题讨论】：

标签： postgresql recursion

【解决方案1】：

乍一看你的问题使用递归查询听起来很简单，但这并不是因为可能有同一个主管的员工列表，这需要一个聚合函数来构建相应的json数组，而是递归查询中不允许使用聚合函数...

第一步 - 我们为每个员工构建 json 对象，并在他/她是主管时构建相应的受监管员工数组：

SELECT s.id AS father
     , CASE WHEN array_agg(e.id) = array[NULL :: integer] THEN NULL ELSE array_agg(e.id) END  AS children_list
     , jsonb_build_object
       ( 'id', s.id
       , 'name', s.name
       , 'supervises', array_agg(e.id)
       ) AS json_tree
  FROM employee AS s
  LEFT JOIN employee AS e
    ON s.id = e.supervisor_id
 GROUP BY s.id, s.name

第二步 - 使用递归查询，我们以自上而下的方式遍历员工树，我们为每个员工分配一个等级并构建将在下一步：

WITH RECURSIVE elt AS
(
SELECT s.id AS father
     , CASE WHEN array_agg(e.id) = array[NULL :: integer] THEN NULL ELSE array_agg(e.id) END  AS children_list
     , jsonb_build_object
       ( 'id', s.id
       , 'name', s.name
       , 'supervises', array_agg(e.id)
       ) AS json_tree
  FROM employee AS s
  LEFT JOIN employee AS e
    ON s.id = e.supervisor_id
 GROUP BY s.id, s.name
), list (father, json_tree, children_list, rank, path) AS
(
SELECT c.father, c.json_tree, c.children_list, 1, '{}' :: text[]
  FROM elt AS c
  LEFT JOIN elt AS f
    ON array[c.father] <@ f.children_list
 WHERE f.father IS NULL
UNION ALL
SELECT c.father
     , c.json_tree
     , c.children_list
     , f.rank + 1
     , f.path || array['supervises',(array_position(f.children_list, c.father)-1) :: text]
  FROM list AS f
 INNER JOIN elt AS c
    ON array[c.father] <@ f.children_list
) --, ordered_list AS (
SELECT *
  FROM list
 ORDER BY rank DESC

第三步 - 我们创建 jsonb_set 函数的聚合版本，以便在迭代前一个结果列表时构建最终的 jsonb 数据

CREATE OR REPLACE FUNCTION jsonb_set(x jsonb, y jsonb, p text[], e jsonb, b boolean)
RETURNS jsonb LANGUAGE sql AS $$
SELECT CASE WHEN x IS NULL THEN e ELSE jsonb_set(x, p, e, b) END ; $$ ;

CREATE OR REPLACE AGGREGATE jsonb_set_agg(x jsonb, p text[], e jsonb, b boolean)
( STYPE = jsonb, SFUNC = jsonb_set) ;

最终查询

WITH RECURSIVE elt AS
(
SELECT s.id AS father
     , CASE WHEN array_agg(e.id) = array[NULL :: integer] THEN NULL ELSE array_agg(e.id) END  AS children_list
     , jsonb_build_object
       ( 'id', s.id
       , 'name', s.name
       , 'supervises', array_agg(e.id)
       ) AS json_tree
  FROM employee AS s
  LEFT JOIN employee AS e
    ON s.id = e.supervisor_id
 GROUP BY s.id, s.name
), list (father, json_tree, children_list, rank, path) AS
(
SELECT c.father, c.json_tree, c.children_list, 1, '{}' :: text[]
  FROM elt AS c
  LEFT JOIN elt AS f
    ON array[c.father] <@ f.children_list
 WHERE f.father IS NULL
UNION ALL
SELECT c.father
     , c.json_tree
     , c.children_list
     , f.rank + 1
     , f.path || array['supervises',(array_position(f.children_list, c.father)-1) :: text]
  FROM list AS f
 INNER JOIN elt AS c
    ON array[c.father] <@ f.children_list
)
SELECT jsonb_set_agg(NULL :: jsonb, path, json_tree, true ORDER BY rank ASC)
  FROM list

测试结果在dbfiddle

【讨论】：