Sie benötigen einen zusätzlichen Schritt:Finden Sie zuerst unterschiedliche Werte und aggregieren Sie sie dann. Zum Beispiel:
SQL> with test (id, col) as
2 (select 1, 'x' from dual union all
3 select 1, 'x' from dual union all
4 --
5 select 2, 'w' from dual union all
6 select 2, 't' from dual union all
7 select 2, 'w' from dual union all
8 --
9 select 3, 'i' from dual
10 ),
11 -- first find distinct values ...
12 temp as
13 (select distinct id, col from test)
14 -- ... then aggregate them
15 select id,
16 listagg(col, ';') within group (order by col) result
17 from temp
18 group by id;
ID RESULT
---------- ----------
1 x
2 t;w
3 i
SQL>
