您可以尝试以下方法：

DECLARE @Data TABLE ( ID INT, [Name] VARCHAR(10), To_Replace VARCHAR(10) );
INSERT INTO @Data ( ID, [Name], To_Replace ) VALUES
( 1, 'Fruits', '1' ),
( 2, 'Apple', '1-2' ),
( 3, 'Citrus', '1-3' ),
( 4, 'Orange', '1-3-4' ),
( 5, 'Empire', '1-2-5' ),
( 6, 'Fuji', '1-2-6' );

SELECT
    *
FROM @Data AS d
OUTER APPLY (

    SELECT STRING_AGG ( [Name], '-' ) AS Replaced FROM @Data WHERE ID IN (
        SELECT CAST ( [value] AS INT ) FROM STRING_SPLIT ( d.To_Replace, '-' )
    )

) List
ORDER BY ID;

退货

+----+--------+------------+----------------------+
| ID |  Name  | To_Replace |       Replaced       |
+----+--------+------------+----------------------+
|  1 | Fruits | 1          | Fruits               |
|  2 | Apple  | 1-2        | Fruits-Apple         |
|  3 | Citrus | 1-3        | Fruits-Citrus        |
|  4 | Orange | 1-3-4      | Fruits-Citrus-Orange |
|  5 | Empire | 1-2-5      | Fruits-Apple-Empire  |
|  6 | Fuji   | 1-2-6      | Fruits-Apple-Fuji    |
+----+--------+------------+----------------------+

更新
确保聚合名称时保持id列表顺序。

DECLARE @Data TABLE ( ID INT, [Name] VARCHAR(10), To_Replace VARCHAR(10) );
INSERT INTO @Data ( ID, [Name], To_Replace ) VALUES
    ( 1, 'Fruits', '1' ),
    ( 2, 'Apple', '1-2' ),
    ( 3, 'Citrus', '1-3' ),
    ( 4, 'Orange', '1-3-4' ),
    ( 5, 'Empire', '1-2-5' ),
    ( 6, 'Fuji', '1-2-6' ),
    ( 7, 'Test', '6-2-7' );

SELECT
    *
FROM @Data AS d
OUTER APPLY (

    SELECT STRING_AGG ( [Name], '-' ) AS Replaced FROM (

        SELECT TOP 100 PERCENT
            Names.[Name]
        FROM ( SELECT CAST ( '<ids><id>' + REPLACE ( d.To_Replace, '-', '</id><id>' ) + '</id></ids>' AS XML ) AS id_list ) AS xIds
        CROSS APPLY (
            SELECT 
                x.f.value('.', 'INT' ) AS name_id, 
                ROW_NUMBER() OVER ( ORDER BY ( SELECT NULL ) ) AS row_id
            FROM xIds.id_list.nodes('//ids/id') x(f)
        ) AS ids
        INNER JOIN @Data AS Names ON Names.ID = ids.name_id
        ORDER BY row_id

    ) AS x

) List
ORDER BY ID;

退货

+----+--------+------------+----------------------+
| ID |  Name  | To_Replace |       Replaced       |
+----+--------+------------+----------------------+
|  1 | Fruits | 1          | Fruits               |
|  2 | Apple  | 1-2        | Fruits-Apple         |
|  3 | Citrus | 1-3        | Fruits-Citrus        |
|  4 | Orange | 1-3-4      | Fruits-Citrus-Orange |
|  5 | Empire | 1-2-5      | Fruits-Apple-Empire  |
|  6 | Fuji   | 1-2-6      | Fruits-Apple-Fuji    |
|  7 | Test   | 6-2-7      | Fuji-Apple-Test      |
+----+--------+------------+----------------------+

我确信这里可以进行优化，但是这个解决方案似乎可以保证保持列表顺序。

虽然您所要求的在技术上是可行的（可能使用递归查询或计数），但我将采取不同的立场，建议您改为修复数据模型。
不应将多个值作为分隔列表存储在单个数据库列中。这违背了关系数据库的目的，使简单的事情变得不必要的复杂和低效。
相反，您应该有一个单独的表来存储这些数据，每个替换都需要存储这些数据 id 在一个单独的行上，也可能在一个列上，该列指示列表中每个元素的顺序。
对于您的示例数据，如下所示：

id    replace_id    seq
1     1             1
2     1             1
2     2             2
3     1             1
3     3             2
4     1             1
4     3             2
4     4             3
5     1             1
5     2             2
5     5             3
6     1             1
6     2             2
6     6             3

现在您可以使用连接、子查询或横向连接高效地生成预期结果。假设您的表被调用 mytable Map表是 mymapping ，横向连接解决方案为：

select t.*, r.*     
from mytable t
outer apply (
    select string_agg(t1.name) within group(order by m.seq) replaced
    from mymapping m
    inner join mytable t1 on t1.id = m.replace_id
    where m.id = t.id
) x