Question

我有一个查询，该查询可一次插入两个表（由于链接的外键约束而需要），并且想返回同一级别的两次插入的结果，但是我不确定该怎么做，或者如果可能的话。

插入物看起来像这样，并且工作正常：

with org_b as (
    insert into public.base (
        id,
        organisation_id,
        last_modified_by
    )
    values ('<organisation UUID/>', '<organisation UUID/>', '<user UUID/>')
    returning *
), org_eb as (
    insert into public.entity_base (
        id
    )
    values ('<organisation UUID/>')
    returning *
), org as (
    insert into public.organisation_data (
        id, 
        armicus_name
    )
    values ('<organisation UUID/>', 'Stack Overflow')
    returning *
), use_b as (
    insert into public.base (
        id,
        organisation_id,
        last_modified_by
    )
    values ('<user UUID/>', '<organisation UUID/>', '<user UUID/>')
    returning *
), use_eb as (
    insert into public.entity_base (
        id
    )
    values ('<user UUID/>')
    returning *
), use as (
    insert into public.user_account_data (
        id,
        email
    )
    values ('<user UUID/>', 'stack@overflow.com')
    returning *
)

我想要返回的内容如下：

select to_json(
    select org_b.id, org_b.organisation_id, org_b.created_date, org_b.last_modified_date, org_b.last_modified_by, org_b.armicus_data, org.armicus_name
    from org_b
    join org on org_b.id = org.id
) as organisation, to_json(
    select use_b.id, use_b.organisation_id, use_b.created_date, use_b.last_modified_date, use_b.last_modified_by, use_b.armicus_data, use.email
    from use_b
    join use on use_b.id = use.id
) as user
from org_b

但是这会引发语法错误，有没有办法实现？如果有帮助，我还可以上传SQL来创建表。

编辑以添加SQL以创建表和约束：

创建基本表：

CREATE TABLE base (
    id uuid PRIMARY KEY,
    organisation_id uuid NOT NULL,
    created_date timestamptz NOT NULL DEFAULT now(),
    last_modified_date timestamptz NOT NUll,
    last_modified_by uuid NOT NULL,
    armicus_data jsonb
);

创建实体库：

CREATE TABLE entity_base (
    id uuid PRIMARY KEY REFERENCES base(id) ON DELETE CASCADE,
    template_id uuid
);

创建组织数据：

CREATE TABLE organisation_data (
    id uuid PRIMARY KEY REFERENCES entity_base(id) ON DELETE CASCADE,
    armicus_name TEXT
);

创建用户帐户数据：

CREATE TABLE user_account_data (
    id uuid PRIMARY KEY REFERENCES entity_base(id) ON DELETE CASCADE,
    email text NOT NULL
);

添加组织ID外键：

ALTER TABLE base ADD CONSTRAINT base_organisation_id_fkey FOREIGN KEY (organisation_id) REFERENCES organisation_data(id);

添加最后由外键修改的

ALTER TABLE base ADD CONSTRAINT base_last_modified_by_fkey FOREIGN KEY (last_modified_by) REFERENCES user_account_data(id);

Answer 1

您可以这样表达：

select to_json( (select b.id, b.organisation_id, b.created_date, b.last_modified_date, b.last_modified_by, b.armicus_data, org.armicus_name
                 from org_b b join
                      org o
                      on b.id = o.id
                )
              ) as organisation,
       to_json( (select b.id, b.organisation_id, b.created_date, b.last_modified_date, b.last_modified_by, b.armicus_data, u.email
                 from use_b b join
                      use u
                      on b.id = u.id
                )
              ) as user
from org_b ;

或者我可能会这样写，将json转换移动到子查询：

select (select to_json( (b.id, b.organisation_id, b.created_date, b.last_modified_date, b.last_modified_by, b.armicus_data, org.armicus_name) )
        from org_b b join
             org o
             on b.id = o.id
       ) as organisation,
      (select to_json( (b.id, b.organisation_id, b.created_date, b.last_modified_date, b.last_modified_by, b.armicus_data, u.email) )
       from use_b b join
            use u
            on b.id = u.id
      ) ) as user
from org_b

Answer 2

解决了缺少列名的问题：

SELECT (
    SELECT to_json(o)
    FROM (
        SELECT org_b.id, org_b.organisation_id, org_b.created_date, org_b.last_modified_date, org_b.last_modified_by, org_b.armicus_data, org.armicus_name
        FROM org_b
        JOIN org on org_b.id = org.id
    ) o
) AS organisation, (
    SELECT to_json(u)
    FROM (
        SELECT use_b.id, use_b.organisation_id, use_b.created_date, use_b.last_modified_date, use_b.last_modified_by, use_b.armicus_data, use.email
        FROM use_b
        JOIN use on use_b.id = use.id
    ) u     
) AS user
FROM org_b

在单个查询中从两个表返回

2 个答案: