如果没有样本数据，很难回答，但看起来你想在ART_TO_PLACE和ART_TO_UNIT之间使用FULL OUTER JOIN，然后聚合并使用HAVING子句来检查没有行被OUTER JOIN ed（相反，所有行都匹配）：

SELECT b.ID
FROM   PLACE b
       INNER JOIN (
         ART_TO_PLACE a
         FULL OUTER JOIN ART_TO_UNIT t
         ON     t.MATL_TU_IDENT = :AUNITIDENT
            AND a.GRP = t.GRP
       )
       ON b.ALIAS = a.PLACE_ALIAS
GROUP BY b.id
HAVING COUNT(CASE WHEN a.grp IS NULL THEN 1 END) = 0
AND    COUNT(CASE WHEN t.grp IS NULL THEN 1 END) = 0
ORDER BY b.id
FETCH NEXT 1 ROWS ONLY

或者可能是：

SELECT b.ID
FROM   PLACE b
       INNER JOIN (
         SELECT a.place_alias
         FROM   ART_TO_PLACE a
                FULL OUTER JOIN ART_TO_UNIT t
                ON     t.MATL_TU_IDENT = :AUNITIDENT
                   AND a.GRP = t.GRP
         GROUP BY a.place_alias
         HAVING COUNT(CASE WHEN a.grp IS NULL THEN 1 END) = 0
         AND    COUNT(CASE WHEN t.grp IS NULL THEN 1 END) = 0
       ) b
       ON b.ALIAS = a.PLACE_ALIAS
ORDER BY b.id
FETCH NEXT 1 ROWS ONLY

样本数据更新：

选项1：使用集合

定义集合数据类型：

CREATE TYPE string_list IS TABLE OF VARCHAR2(50);

然后，您可以将两个表的值收集到集合中，并在相同的集合值上进行连接：

SELECT workplaceid, unitid
FROM   ( SELECT workplaceid,
                CAST(COLLECT(allowedtype) AS string_list) AS allowedtypes
         FROM   WorkplaceCapabilities
         GROUP BY Workplaceid
       ) w
       INNER JOIN (
         SELECT unitid,
                CAST(COLLECT(containedmaterial) AS string_list) AS containedmaterials
         FROM   TransportUnitContents
         GROUP BY unitid
       ) t
       ON w.allowedtypes = t.containedmaterials;

选项2：比较计数

统计每个表中每个组的类型/材料数量，然后进行连接，并检查连接后的匹配数量是否相同：

SELECT workplaceid, unitid
FROM   ( SELECT wc.*,
                COUNT(allowedtype) OVER (PARTITION BY workplaceid) AS num_allowedtypes
         FROM   WorkplaceCapabilities wc
       ) w
       INNER JOIN (
         SELECT tuc.*,
                COUNT(containedmaterial) OVER (PARTITION BY unitid) AS num_containedmaterials
         FROM   TransportUnitContents tuc
       ) t
       ON (w.allowedtype = t.containedmaterial AND w.num_allowedtypes = t.num_containedmaterials)
GROUP BY
       w.workplaceid, t.unitid
HAVING COUNT(*) = MAX(w.num_allowedtypes)

对于示例数据，两个选项都将输出：
| 工作场所ID|UNITID|
| --------------|--------------|
| 1|一百五十二|
| 第二章|二十五|
| 三|一百二十二|
| 四|十三|
fiddle

如果我对你的问题理解正确的话，你可以像这样使用嵌套的exists条件：

select *
from PLACE b
where
    exists (
        select 1
        from ART_TO_PLACE a
        where b.ALIAS = a.PLACE_ALIAS
        and exists (
                select 1
                from ART_TO_UNIT t
                where a.GRP = t.GRP
                and t.MATL_TU_IDENT = AUNITIDENT
            )
    )