Sql 如何使用同一表的两个实例

我想知道如何在下面的例子中使用同一个表的两个实例，我知道如何做，但我就是不能让它为我的任务工作

我有以下表格：

• 机构（id_机构，名称）
• 空间（id\U空间、地址）
• 报价（身份证代理、身份证空间）

任务：找出两个不同机构提供的空间的地址、机构名称1、机构名称2（两个机构的组合是唯一的）

我尝试的是：

(一)

(二)

空间

提议

预期产出：

Address     Name1    Name2
----------- -------- -------
address1    Agency1  Agency2

---

要在一行中获得结果，请按照您的要求，每行各对代理机构：

select S.address as address, A1.name as agency_1, A2.name as agency_2
from offer O1
    join offer O2
        on O2.id_space = O1.id_space
        and O2.id_agency != O1.id_agency
    join space S
        on S.id_space = O1.id_space
    join agency A1
        on A1.id_agency = O1.id_agency
    join agency A2
        on A2.id_agency = O2.id_agency
;

这里的“核心功能”是将

offer

no.1（

O1

alias）与

offer

no.2（

O2

alias）在

id\u空间

相等但

id\u机构

不同的情况下连接起来

一个有趣的练习：要将结果分为多行，每行一个代理：

select S.address, A.name as agency, X.number_of_agencies_per_space
from (
        select id_space, id_agency, count(1) over (partition by id_space) as number_of_agencies_per_space
        from offer
    ) X
    join space S
        on S.id_space = X.id_space
    join agency A
        on A.id_agency = X.id_agency
where X.number_of_agencies_per_space > 1
;

---

完成了，谢谢你的帮助。非常感谢你，该死的，我觉得自己太蠢了，我知道我在写代码的时候漏掉了一些东西，我的想法不正确。

id_space  address
--------- --------
1         address1
2         address2
3         address3

id_agency   id_space
----------- --------
1           1
2           1
3           2

Address     Name1    Name2
----------- -------- -------
address1    Agency1  Agency2

select S.address as address, A1.name as agency_1, A2.name as agency_2
from offer O1
    join offer O2
        on O2.id_space = O1.id_space
        and O2.id_agency != O1.id_agency
    join space S
        on S.id_space = O1.id_space
    join agency A1
        on A1.id_agency = O1.id_agency
    join agency A2
        on A2.id_agency = O2.id_agency
;

select S.address, A.name as agency, X.number_of_agencies_per_space
from (
        select id_space, id_agency, count(1) over (partition by id_space) as number_of_agencies_per_space
        from offer
    ) X
    join space S
        on S.id_space = X.id_space
    join agency A
        on A.id_agency = X.id_agency
where X.number_of_agencies_per_space > 1
;