SQL自定义函数－－隐藏指定长度字符串

环境：SQL2008R2 
去银行ATM取款或查询时，姓名及卡号会隐藏一部分。
今天想试试在SQL中实现，写了个自定义函数。
字符串转星号"*"

if object_id('f_hidenString','fn')>0 
  drop function f_hidenString;
go  
/* f_hidenString():隱藏指定長度字符串， 1) 當字符串長度大於需要隱藏字符長度時 ，隱藏指定長度輸出 
** 2) 當字符串長度小於等於需要隱藏字符長度時，只顯示第一個字符，其他都隱藏   
** 參數：@sourceString: 源字符串 
** @hidenNumber : 隱藏字符串長度  
** 返回： 字符串 
*/
create function f_hidenString(
  @sourceString  nvarchar(4000),
  @hidenNumber  int
)
returns nvarchar(4000)
as 
begin
  declare @strLength  int;
  declare @halfLength int;
  declare @length int;
  declare @startPosition int;
  declare @endPosition int;
  declare @result nvarchar(4000);
  declare @maskString nvarchar(400);  -- 掩码
  
  if coalesce(@sourceString,N'~')<>N'~'
     set @Length = len(@sourceString)
     
  if coalesce(@hidenNumber,0)>0 
  begin
  -- 1) 當字符串長度大於 需要隱藏字符長度時 ，隱藏指定長度輸出  
    if @length > @hidenNumber 
    begin
      set @startPosition =  (@length /2) - (@hidenNumber/2)      
      if @startPosition = 0 
      begin
        set @startPosition = 1
      end;
      set @endPosition = @startPosition + (@hidenNumber);   
    end;
    else if (( @length >= 2 ) and (@length < @hidenNumber ))
  -- 2) 當字符串長度大於等於 需要隱藏字符一半長度時              
    begin
      set @hidenNumber = len(@sourceString)/2
      set @startPosition =  (@length /2) - ceiling(@hidenNumber /2)
      if @startPosition = 0 
      begin
        set @startPosition = 1
      end;
      set @endPosition = @startPosition + (@hidenNumber)      
    end; 
    else if (( @length >= 2 ) and (@length = @hidenNumber ))
  -- 2) 當字符串長度大於等於 需要隱藏字符一半長度時              
    begin
      set @hidenNumber = @hidenNumber-1
      set @startPosition =  (@length /2) - (@hidenNumber /2)      
      if @startPosition = 0 
      begin
        set @startPosition = 1;
        set @endPosition = @length-1;
      end;  
      set @endPosition = @startPosition + (@hidenNumber) ;     
    end;
    else
    begin             
       set @startPosition = 1;
       set @endPosition = 1;
       set @hidenNumber = 0
    end;
      set @maskString = replicate(N'*', @hidenNumber)
      set @result = subString(@sourceString,1,@startPosition) + coalesce(@maskString ,N'')
                  + substring(@sourceString,@endPosition+1,@length) ;                                     
  end    
  else
  begin
      -- 3) 當字符串長度小於等於1 ，原樣輸出   
    set @result = @sourceString;
  end;
    
  return (@result) 

end
go

--測試
select dbo.f_hidenString('13698763747',5) as hidenString
union all 
select dbo.f_hidenString('13612343747',4) as hidenString
union all 
select dbo.f_hidenString(N'東莞市顯示技術有限公司',5) as hidenString
union
select dbo.f_hidenString('1364',3) as hidenString
union all 
select dbo.f_hidenString(N'136',3) as hidenString
union all 
select dbo.f_hidenString(N'13665',3) as hidenString
union all 
select dbo.f_hidenString(N'12',2) as hidenString
union all 
select dbo.f_hidenString(N'1',1) as hidenString