将 MSSQL 'FOR XML PATH' 转换为 Oracle

F. *_*aum 5 sql sql-server oracle code-conversion for-xml-path

我有一个语句来填充我的 MSSQL 数据库上的一个表。它将一些值连接在一起,用分号分隔。

INSERT INTO XXAArcDocSWSB (ArcDocINr, SWorte)
SELECT A.ArcDocINr, B.SWorte FROM XXAArcDoc A 
LEFT JOIN (
SELECT DISTINCT T2.ArcDocINr,
SUBSTRING(
    (
        SELECT ';' + T1.SWort  AS [text()]
        FROM (SELECT D.ArcDocINr, SW.SWort FROM XXAArcDoc D, XXAArcSW SW WHERE D.ArcDocINr = SW.ArcDocINr) T1
        WHERE T1.ArcDocINr = T2.ArcDocINr
        For XML PATH ('')
    ), 2, 255) [SWorte]
FROM (SELECT D.ArcDocINr, SW.SWort FROM XXAArcDoc D, XXAArcSW SW WHERE D.ArcDocINr = SW.ArcDocINr) T2
) B ON A.ArcDocINr = B.ArcDocINr 
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我没有足够的知识将其转换为 Oracle。它应该给我与 MSSQL 相同的输出。有人能帮我吗?

编辑:

以下是一些示例数据:

XXAArcDoc:

ArcDocINr | ...
----------|----------
1         |
2         |
3         |
.         |
.         |
.         |
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XXAArcSW:

ArcSWINr | ArcDocINr | SWort
---------|-----------|---------
6        | 1         | Müller
7        | 1         | 100
8        | 2         | 111111
9        | 2         | 13579
10       | 2         | 002
11       | 3         | TM-AH
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这是我想要的输出:

ArcDocINr | SWorte
----------|---------
1         | Müller;100
2         | 111111;13579;002
3         | TM-AH
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MT0*_*MT0 4

使用LISTAGG

SELECT ArcDocINr,
       LISTAGG(
          SWort,
          ';'
       ) WITHIN GROUP ( ORDER BY ArcSWINr ) AS SWorte
FROM   XXAArcSW
GROUP BY ArcDocINr;
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更新

XXAArcDoc如果您使用表中的值插入表中XXAArcSW,则类似于:

INSERT INTO XXAArcDoc ( ArcDocINr, SWorte )
SELECT ArcDocINr,
       LISTAGG( SWort, ';' ) WITHIN GROUP ( ORDER BY ArcSWINr )
FROM   XXAArcSW
GROUP BY ArcDocINr
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