Oracle SQL 从多个自连接中删除不相同的重复项

Question

我有一个 table 多人在同一个套房我需要在套房上进行匹配，然后将每个人与他们的室友列表一起显示为第一个 ID。

因为他们不止两个，我运行遇到多个结果的情况，其中第一个 ID 被多次列出，室友的顺序不同，或者我只能带回一行4个ID

示例：

我有一个 table 这样的：

Table Name: ROOM_LIST

ID | BUILDING | SUITE | ROOM
01 | BU_1     | SU_1  | RO_1
02 | BU_1     | SU_1  | RO_2
03 | BU_1     | SU_1  | RO_3
04 | BU_1     | SU_1  | RO_4
05 | BU_1     | SU_2  | RO_1
06 | BU_1     | SU_2  | RO_2
07 | BU_2     | SU_1  | RO_1
08 | BU_2     | SU_1  | RO_2

我试过这样的查询：

select A.ID as Primary,
       B.ID as Roomate_1,
       C.ID as Roomate_2,
       D.ID as Roomate_3,
       A.BUILDING as Building,
       A.SUITE As Suite,
       A.ROOM As Room
from ROOM_LIST A
Left Join ROOM_LIST B on A.BUILDING = B.BUILDING and A.SUITE = B.SUITE
Left Join ROOM_LIST C on A.BUILDING = C.BUILDING and A.SUITE = C.SUITE
Left Join ROOM_LIST D on A.BUILDING = D.BUILDING and A.SUITE = D.SUITE
where A.ID > B.ID
and   A.ID > C.ID
and   A.ID > D.ID
and   B.ID > C.ID
and   B.ID > D.ID
and   C.ID > D.ID
order by Primary,Roomate_1,Roomate_2,Roomate_3,Building,Suite,Room;

这消除了额外的重复项，但我每个套件只得到一行，而不是将每个 ID 作为主 ID 的一行。

也尝试了一个类似的，但使用 <> 或 != 而不是 > 然后我得到多个重复项，其中一个 ID 作为第一个，但第二个、第三个和第四个是互换的，所以它们在技术上不是重复项。

这就是为什么我说 "non identical" 重复 :)

我的最终结果如下所示：

| Primary | Roomate_1 | Roomate_2 | Roomate_3 | Building | Suite | Room
| 01      | 02        | 03        | 04        | BU_1     | SU_1  | RO_1
| 02      | 03        | 04        | 01        | BU_1     | SU_1  | RO_2
| 03      | 04        | 01        | 02        | BU_1     | SU_1  | RO_3
| 04      | 01        | 02        | 03        | BU_1     | SU_1  | RO_4
| 05      | 06        | Null      | Null      | BU_1     | SU_2  | RO_1
| 06      | 05        | Null      | Null      | BU_1     | SU_2  | RO_2

我在 select 语句或 from 中尝试了使用不同子 select 的多个查询，但我似乎无法将每个 ID 作为主 ID 归结为一个结果。我已经研究过做一个数据透视表，但是（我的理解）只有当我有相同的 id 有多个结果并且我想将多个结果转换成列时才有效。

想过联合，但不知道如何跨联合查询？如果那是一件事

如有任何帮助，我们将不胜感激

编辑：下面的解决方案仅适用于 12C，但我需要一个 11G 解决方案：

  with dt as (
    select 01 id , 'BU_1' building,  'SU_1' suite ,'RO_1' room from dual union all 
    select 02 id , 'BU_1' building,  'SU_1' suite ,'RO_2' room from dual union all 
    select 03 id , 'BU_1' building,  'SU_1' suite ,'RO_3' room from dual union all 
    select 04 id , 'BU_1' building,  'SU_1' suite ,'RO_4' room from dual union all 
    select 05 id , 'BU_1' building,  'SU_2' suite ,'RO_1' room from dual union all 
    select 06 id , 'BU_1' building,  'SU_2' suite ,'RO_2' room from dual union all 
    select 07 id , 'BU_2' building,  'SU_1' suite ,'RO_1' room from dual union all 
    select 08 id , 'BU_2' building,  'SU_1' suite ,'RO_2' room from dual )
    SELECT  
    A.ID as Primary,
            ( select id from (select id,rownum rn from dt  b where  a.building = b.building AND a.suite = b.suite and b.ID != a.ID  order by id ) where rn=1)  Roomate_1,
            ( select id from (select id,rownum rn from dt  b where  a.building = b.building AND a.suite = b.suite and b.ID != a.ID  order by id ) where rn=2)  Roomate_2,
            ( select id from (select id,rownum rn from dt  b where  a.building = b.building AND a.suite = b.suite and b.ID != a.ID  order by id ) where rn=3)  Roomate_3,
           a.BUILDING as Building,
           A.SUITE As Suite,
           A.ROOM As Room
    FROM
        dt a 
    order by Primary,Roomate_1,Roomate_2,Roomate_3,Building,Suite,Room

我在给定的答案之一中添加了以下内容：and b.ID != a.ID 并更改了 rn=2 to rn=1 以开始计数，因为 0

Answer 1

不确定性能影响。将需要进行分析，但给出的结果符合预期。

12c 答案。

    with dt as (
    select 01 id , 'BU_1' building,  'SU_1' suite ,'RO_1' room from dual union all 
    select 02 , 'BU_1' building,  'SU_1' suite ,'RO_2' room from dual union all 
    select 03 , 'BU_1' building,  'SU_1' suite ,'RO_3' room from dual union all 
    select 04, 'BU_1' building,  'SU_1' suite ,'RO_4' room from dual union all 
    select 05 , 'BU_1' building,  'SU_2' suite ,'RO_1' room from dual union all 
    select 06 , 'BU_1' building,  'SU_2' suite ,'RO_2' room from dual union all 
    select 07 , 'BU_2' building,  'SU_1' suite ,'RO_1' room from dual union all 
    select 08 , 'BU_2' building,  'SU_1' suite ,'RO_2' room from dual )
    SELECT  
    A.ID as Primary,
            ( select id from (select id,rownum rn from dt  b where  a.building = b.building AND a.suite = b.suite  order by id ) where rn=2)  Roomate_1,
            ( select id from (select id,rownum rn from dt  b where  a.building = b.building AND a.suite = b.suite  order by id ) where rn=3)  Roomate_2,
            ( select id from (select id,rownum rn from dt  b where  a.building = b.building AND a.suite = b.suite  order by id ) where rn=4)  Roomate_3,
           a.BUILDING as Building,
           A.SUITE As Suite,
           A.ROOM As Room
    FROM
        dt a 
    order by Primary,Roomate_1,Roomate_2,Roomate_3,Building,Suite,Room

11g 答案。我不确定在 rooommate1 到 roomate3 列中需要多少数据排序。

WITH  dt as (
    select 01 id , 'BU_1' building,  'SU_1' suite ,'RO_1' room from dual union all 
    select 02 id , 'BU_1' building,  'SU_1' suite ,'RO_2' room from dual union all 
    select 03 id , 'BU_1' building,  'SU_1' suite ,'RO_3' room from dual union all 
    select 04 id , 'BU_1' building,  'SU_1' suite ,'RO_4' room from dual union all 
    select 05 id , 'BU_1' building,  'SU_2' suite ,'RO_1' room from dual union all 
    select 06 id , 'BU_1' building,  'SU_2' suite ,'RO_2' room from dual union all 
    select 07 id , 'BU_2' building,  'SU_1' suite ,'RO_1' room from dual union all 
    select 08 id , 'BU_2' building,  'SU_1' suite ,'RO_2' room from dual ),
joindrslt AS (
    SELECT a.*, b.id roommate,
        ROW_NUMBER() OVER(PARTITION BY a.suite, a.building, a.room ORDER BY b.id ) AS ri
    FROM 
    dt a 
    JOIN dt b ON a.building = b.building AND a.suite = b.suite AND b.id != a.id
    ORDER BY b.id
)
SELECT ID Primary,
            roomate_1,
            roomate_2,
            roomate_3,
            Building,Suite,
            Room  FROM
    (
        SELECT
            * 
        FROM
            joindrslt PIVOT (
                MAX ( roommate )
                FOR ri
                IN ( 1 AS roomate_1, 2 AS roomate_2, 3 AS roomate_3 )
            )
    )
ORDER BY
 Primary,Roomate_1,Roomate_2,Roomate_3,Building,Suite,Room

Oracle SQL 从多个自连接中删除不相同的重复项

Oracl SQL remove non-identical duplicates from multiple self join

oracle

sqlplus

oracle11g

我的最终结果如下所示：