从这里的文档来看，配置单元查询语言似乎支持公共表表达式。
因此，类似于以下内容的工作应该是：

declare @str nvarchar(4000) = ?;        -- String to split.

with n(n) as (select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1)

     -- Select the same number of rows as characters in @str as incremental row numbers.
     -- Cross joins increase exponentially to a max possible 10,000 rows to cover largest @str length.
  ,t(t)   as (select top (select len(isnull(@str,'')) a) row_number() over (order by (select null)) from n n1,n n2,n n3,n n4)

     -- Return the position of every value that follows the specified delimiter.
  ,s(s)   as (select 1 union all select t+1 from t where substring(isnull(@str,''),t,1) = ',')

     -- Return the start and length of every value, to use in the SUBSTRING function.
     -- ISNULL/NULLIF combo handles the last value where there is no delimiter at the end of the string.
  ,l(s,l) as (select s,isnull(nullif(charindex(',',isnull(@str,''),s),0)-s,4000) from s)

     -- Return each individual value in the delimited string along with it's position.
  ,v      as (select row_number() over(order by s) as rn
                    ,substring(@str,s,l) as item
              from l
             )
select *
from v
    join xyz.test_table1 as t
        on v.v = t.f1

如果您不想在所有的数据集中都使用这个僵化的规则，那么您需要将这个逻辑封装到任何与SQLServer表值参数等价的配置单元中，也许是udtf？
在sql server中，函数定义如下：

create function [dbo].[fn_StringSplit4k]
(
  @str nvarchar(4000) = ' '        -- String to split.
 ,@delimiter as nvarchar(1) = ','  -- Delimiting value to split on.
 ,@num as int = null               -- Which value to return.
)
returns table
as
return
     -- Start tally table with 10 rows.
 with n(n)   as (select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1)

     -- Select the same number of rows as characters in @str as incremental row numbers.
     -- Cross joins increase exponentially to a max possible 10,000 rows to cover largest @str length.
  ,t(t)   as (select top (select len(isnull(@str,'')) a) row_number() over (order by (select null)) from n n1,n n2,n n3,n n4)

     -- Return the position of every value that follows the specified delimiter.
  ,s(s)   as (select 1 union all select t+1 from t where substring(isnull(@str,''),t,1) = @delimiter)

     -- Return the start and length of every value, to use in the SUBSTRING function.
     -- ISNULL/NULLIF combo handles the last value where there is no delimiter at the end of the string.
  ,l(s,l) as (select s,isnull(nullif(charindex(@delimiter,isnull(@str,''),s),0)-s,4000) from s)

 select rn
          ,item
 from(select row_number() over(order by s) as rn
    ,substring(@str,s,l) as item
  from l
  ) a
 where rn = @num
  or @num is null;

我想出来了！为其他用户发布答案。
以如下表达式提供查询（在ssrs中的query下）：
=“selectfrom xyz.test\u table1 where f1 in（'”&join（参数）！参数值，“'，'”&“'）”
上面的字符串操作转换为：
从xyz.test\u table1中选择其中f1在（'value1'，'value2'）
注：value1，value2这里是用户选择的多值参数的值