Question

我试图从 PostgreSQL 查询返回一个带有嵌套数组的对象，我该怎么做。？这是我迄今为止尝试过的方法，有人可以指导我吗？

SELECT d.id as distributor_id,
       d.name,
       json_agg(json_build_object('route_id', r.id, 'route_name', r.name,'shops',
           json_agg((select * from retailers where retailers."distributorId" = d.id and retailers."routeId" = r.id)) )) as route
FROM users U
         INNER JOIN tsi_distributors td ON U.user_id = td."userUserId" AND td.delete_at IS NULL
         INNER JOIN distributors d ON td."distributorId" = d.id AND d.delete_at IS NULL
         LEFT JOIN routes r on td.id = r."distributorId" AND r.delete_at IS NULL
WHERE U.user_id = 'TSI_84826'
GROUP BY d.id

这会引发错误。 subquery must return only one column

绑定返回这样的对象

[
  {
    distributor_id: 1,
    name: 'Distributor_1',
    route: [
      {
        route_id: 1,
        route_name: 'Route 1',
        shops: [
          {
            id: 1,
            name: 'shop 1',
          },
          {
            id: 2,
            name: 'shop 2',
          },
        ],
      },
      {
        route_id: 2,
        route_name: 'Route 2',
        shops: [
          {
            id: 3,
            name: 'shop 3',
          },
        ],
      },
    ],
  },
];

Answer 1

您错误地使用了 json_agg：

您不应将子查询的结果集用作 json_agg（或任何其他聚合函数）的参数
json_agg 接受 json 作为参数，它不会将任何内容转换为 json。

尝试将 json_agg 转移到子选择中，例如：

( select json_agg(to_json(retailers)) 
  from retailers 
  where retailers."distributorId" = d.id 
    and retailers."routeId" = r.id)

子查询必须只返回一列 Postgrsql

1 个答案: