amit kohan 提供的解决方案有效，但对我来说太慢了。对于大字符串数据，每个字符的迭代不是最优的。我做了以下函数，虽然不是很紧凑但速度很快。
CREATE FUNCTIONdbo.F_StripLowAscii
(
@Name nvarchar(max)
)
RETURN nvarchar(max) as
BEGIN
DECLARE @Result nvarchar(max)
If @Name IS NULL
RETURN @Name
DECLARE @BlankRange VARCHAR(15)
DECLARE @FoundAt INTEGER
-- ASCII CHAR #0 needs a special treatment
SET @BlankRange = '%[' + CHAR(0) + ']%'
SET @FoundAt = PATINDEX(@BlankRange ,@Name COLLATE SQL_Latin1_General_CP850_Bin)
WHILE @FoundAt > 0
BEGIN
SET @name = left(@name, @FoundAt-1 ) + SUBSTRING(@name, @FoundAt+1, LEN(@Name))
SET @FoundAt = PATINDEX(@BlankRange ,@Name COLLATE SQL_Latin1_General_CP850_Bin)
END
SET @BlankRange = '%[' + CHAR(1)+'-'+CHAR(8) + ']%'
SET @FoundAt = PATINDEX(@BlankRange ,@Name COLLATE SQL_Latin1_General_CP850_Bin)
WHILE @FoundAt > 0
BEGIN
SET @name = Replace(@Name, SUBSTRING(@Name, @FoundAt,1),'')
SET @FoundAt = PATINDEX(@BlankRange ,@Name COLLATE SQL_Latin1_General_CP850_Bin)
END
SET @BlankRange = '%[' + CHAR(11)+'-'+CHAR(12) + ']%'
SET @FoundAt = PATINDEX(@BlankRange ,@Name COLLATE SQL_Latin1_General_CP850_Bin)
WHILE @FoundAt > 0
BEGIN
SET @name = Replace(@Name, SUBSTRING(@Name, @FoundAt,1),'')
SET @FoundAt = PATINDEX(@BlankRange ,@Name COLLATE SQL_Latin1_General_CP850_Bin)
END
SET @BlankRange = '%[' + CHAR(14)+'-'+CHAR(31) + ']%'
SET @FoundAt = PATINDEX(@BlankRange ,@Name COLLATE SQL_Latin1_General_CP850_Bin)
WHILE @FoundAt > 0
BEGIN
SET @name = Replace(@Name, SUBSTRING(@Name, @FoundAt,1),'')
SET @FoundAt = PATINDEX(@BlankRange ,@Name COLLATE SQL_Latin1_General_CP850_Bin)
END
RETURN @Name
END
GO