Sql 选项以满足我的要求，因为我有动态类别字符串，其大小不是预定义的类别列表可能在单独的行中。一旦我完成查询，我会更新我的答案。@RaviPatel你只需要知道可能的最高选项数并相应地更新，我认为你可以接受这个答案，看起来很棒！看起来很棒！ Table_Id |

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选项以满足我的要求，因为我有动态类别字符串，其大小不是预定义的类别列表可能在单独的行中。一旦我完成查询，我会更新我的答案。@RaviPatel你只需要知道可能的最高选项数并相应地更新，我认为你可以接受这个答案，看起来很棒！看起来很棒！

Table_Id | Categories         | Value
<ID>     | ABC1; ABC1-1; XYZ  | 10
<ID>     | ABC1; ABC1-2; XYZ  | 15
<ID>     | XYZ                | 5
.....

Table_Id | Categories         | Value
<ID>     | ABC1               | 25
<ID>     | ABC1-1             | 10
<ID>     | ABC1-2             | 15
<ID>     | XYZ                | 30
.....

WITH
values as (
    select distinct category
    from (
            select distinct split_part(categories,';',1) as category from your_table
            union select distinct split_part(categories,';',2) from your_table
            union select distinct split_part(categories,';',3) from your_table
     )
     where nullif(category,'') is not null
)
SELECT
 t2.category
,sum(t1.value)
FROM your_table t1
JOIN values t2
ON split_part(categories,';',1)=t2.category
OR split_part(categories,';',2)=t2.category
OR split_part(categories,';',3)=t2.category

select row_number() over(order by 1) as r1, 
        to_char(timestamptz 'epoch' + date_time * interval '1 second', 'yyyy-mm-dd') AS DAY,
        split_part(categories, ';', numbers.n) as catg,
        value
    from <TABLE>
    join numbers
    on numbers.n <= regexp_count(category_string, ';') + 1 <OTHER_CONDITIONS>