hasura中的递归/嵌套/树数据获取答案

【问题标题】：Recursive/Nested/Tree Data fetching in hasurahasura中的递归/嵌套/树数据获取
【发布时间】：2020-10-05 07:07:11
【问题描述】：

我的用户表具有与组织单位表 (1-1) 关系的外键，即 1 个用户属于 1 个组织单位。

我有 org-unit 表，其中包含 parent-id(self-reference) 列，即 org-unit 可以嵌套。例如orgUnit_1 的父 ID 为 null，orgUnit_2 的父 ID 为 1，orgUnit_3 的父 ID 为 2。

orgUnit_1
    - orgUnit_2
        -orgUnit_3

user1 belongs to orgUnit_1
user2 belongs to orgUnit_2
user3 belongs to orgUnit_3

现在我希望，当我从 orgUnit_1 获取用户时，它也应该带来它的子 orgUnit 用户......即 user1、user2 和 user3。

这可能吗，在哈苏拉？

【问题讨论】：

标签： postgresql graphql hasura

【解决方案1】：

GraphQL 不支持递归查询，但可以在 PostgreSQL 中使用递归公用表表达式。这可以定义为一个函数并用作表计算域：

drop function if exists organization_descendant_users(organization);
drop table if exists user_info;
drop table if exists organization;

create table organization (
    id serial primary key,
    parent int references organization(id),
    organization text not null
);

insert into organization(id, parent, organization)
values
    (1, null, '/'),
    (2, 1, '/a'),
    (3, 2, '/a/a'),
    (4, 1, '/b'),
    (5, 4, '/b/a'),
    (6, 4, '/b/b');

alter sequence organization_id_seq restart with 7;

create table user_info (
    id serial primary key,
    email text unique,
    organization int not null references organization(id)
);

insert into user_info (id, email, organization)
values
    (1, 'someone@a-a', 3),
    (2, 'someone@b-b', 6),
    (3, 'someone@a', 2),
    (4, 'someone@top', 1);

alter sequence user_info_id_seq restart with 5;

create function organization_descendant_users(organization_row organization)
returns setof user_info as $$
with recursive organization_descendant as (
    select
        id,
        id as descendant
    from organization
    union all
    select
        o.parent as id,
        n.descendant
    from organization o
    join organization_descendant n on n.id = o.id
)
select user_info.*
from organization_descendant
join user_info on user_info.organization = organization_descendant.descendant
where organization_descendant.id is not null and organization_descendant.id = organization_row.id
order by email;
$$ language sql stable;

create index on organization(parent);
create index on user_info(organization);

跟踪user_info 和organization 表并修改表organization 并将organization_descendant_users 作为计算字段添加到表中。

查询示例：

query {
  organization {
    organization
    organization_descendant_users {
      email
    }
  }
}

结果：

{
  "data": {
    "organization": [
      {
        "organization": "/",
        "organization_descendant_users": [
          {
            "email": "someone@a"
          },
          {
            "email": "someone@a-a"
          },
          {
            "email": "someone@b-b"
          },
          {
            "email": "someone@top"
          }
        ]
      },
      {
        "organization": "/a",
        "organization_descendant_users": [
          {
            "email": "someone@a"
          },
          {
            "email": "someone@a-a"
          }
        ]
      },
      {
        "organization": "/a/a",
        "organization_descendant_users": [
          {
            "email": "someone@a-a"
          }
        ]
      },
      {
        "organization": "/b",
        "organization_descendant_users": [
          {
            "email": "someone@b-b"
          }
        ]
      },
      {
        "organization": "/b/a",
        "organization_descendant_users": []
      },
      {
        "organization": "/b/b",
        "organization_descendant_users": [
          {
            "email": "someone@b-b"
          }
        ]
      }
    ]
  }
}

【讨论】：

【解决方案2】：

不，这是不可能的，因为 GraphQL 不支持递归查询请求。

https://github.com/graphql/graphql-spec/issues/91

你能做的最好的事情是：

org_unit {
  id
  name
  users {
    id
    display_name
  }
  child_org_unit {
    id
    name
    users {
      id
      display_name
    }
    child_org_unit {
      id
      name
      users {
        id
        display_name
      }
      child_org_unit {
        id
        name
        users {
          id
          display_name
        }
      }
    }
  }
}

【讨论】：