ID #65744

在SQL SERVER中实现RSA加密算法

/*本次修改增加了unicode的支持,但是加密后依然显示为16进制数据,因为进行RSA加密后所得到的unicode编码是无法显示的,所以密文依然采用16进制数据显示。

需要特别注意:如果要对中文进行加密,那么所选取的两个素数要比较大,两个素数的成绩最好要大于65536,即大于unicode的最大编码值

*/

在SQL SERVER中实现RSA加密算法(第二版)

--判断是否为素数

if object_id('f_PRimeNumTest') is not null

drop function f_primeNumTest

go

create function [dbo].[f_primeNumTest]

(@p int)

returns bit

begin

declare @flg bit,@i int

select @flg=1, @i=2

while @i<sqrt(@p)

begin

     if(@p%@i=0 )

     begin

        set @flg=0

       break

     end

     set @[email protected]+1

end

return @flg

end

go

--判断两个数是否互素



if object_id('f_isNumsPrime') is not null

drop function f_isNumsPrime

go

create function f_isNumsPrime

(@num1 int,@num2 int)

returns bit

begin

declare @tmp int,@flg bit

set @flg=1

while (@num2%@num1<>0)

begin

    select @[email protected],@[email protected]%@num1,@[email protected]

end

if @num1=1

     set @flg=0

return @flg

end

go

--产生密钥对

if object_id('p_createKey') is not null

drop proc p_createKey

go

create proc p_createKey

@p int,@q int

as

begin

    declare @n bigint,@t bigint,@flag int,@d int

    if dbo.f_primeNumTest(@p)=0

    begin

       print cast(@p as varchar)+'不是素数,请重新选择数据'

       return

    end

    if dbo.f_primeNumTest(@q)=0

    begin

       print cast(@q as varchar)+'不是素数,请重新选择数据'

       return

end

print '请从下列数据中选择其中一对,作为密钥'

    select @[email protected]*@q,@t=(@p-1)*(@q-1)

    declare @e int

    set @e=2

    while @e<@t

    begin

       if dbo.f_isNumsPrime(@e,@t)=0

       begin

          set @d=2

       while @d<@n

            begin

              if(@e*@d%@t=1)

                 print cast(@e as varchar)+space(5)+cast(@d as varchar)

              set @[email protected]+1

            end

    end

       set @[email protected]+1        

    end

end



/*加密函数说明,@key 为上一个存储过程中选择的密码中的一个 ,@p ,@q 产生密钥对时选择的两个数。获取每一个字符的unicode值,然后进行加密,产生3个字节的16位数据*/



if object_id('f_RSAEncry') is not null

drop function f_RSAEncry

go

create function f_RSAEncry

(@s varchar(100),@key int ,@p int ,@q int)

returns nvarchar(4000)

as

begin

   declare @crypt varchar(8000)

     set @crypt=''

   while len(@s)>0

   begin

              declare @i bigint,@tmp varchar(10),@k2 int,@leftchar int

              select @leftchar=unicode(left(@s,1)),@[email protected]/2,@i=1

              while @k2>0

              begin

                     set @i=(cast(power(@leftchar,2) as bigint)*@i)%(@p*@q)

                     set @[email protected]

              end

              set @i=(@leftchar*@i)%(@p*@q)    

              set @tmp=''

              select @tmp=case when @i%16 between 10 and 15 then char( @i%16+55) else cast(@i%16 as varchar) end [email protected],@[email protected]/16

              from (select number from master.dbo.spt_values where type='p' and number<10 )K

              order by number desc



              set @[email protected]+right(@tmp,6)

    

              set @s=stuff(@s,1,1,'')

end

return @crypt

end



--解密:@key 为一个存储过程中选择的密码对中另一个数字 ,@p ,@q 产生密钥对时选择的两个数

if object_id('f_RSADecry') is not null

drop function f_RSADecry

go

create function f_RSADecry

(@s nvarchar(4000),@key int ,@p int ,@q int)

returns nvarchar(4000)

as

begin

declare @crypt varchar(8000)

    set @crypt=''

while len(@s)>0

    begin

       declare @leftchar bigint

       select @leftchar=sum(data1)

       from (   select case upper(substring(left(@s,6), number, 1)) when 'A' then 10

                                                    when 'B' then 11

                                                    when 'C' then 12

                                                    when 'D' then 13

                                                    when 'E' then 14

                                                    when 'F' then 15

                else substring(left(@s,6), number, 1)

                end* power(16, len(left(@s,6)) - number) data1

         from (select number from master.dbo.spt_values where type='p')K

         where number <= len(left(@s,6))

     ) L

    declare @k2 int,@j bigint

       select @[email protected]/2,@j=1

    while @k2>0

       begin

      set @j=(cast(power(@leftchar,2)as bigint)*@j)%(@p*@q)

      set @[email protected]

    end

      set @j=(@leftchar*@j)%(@p*@q)

    set @[email protected]+nchar(@j)

    set @s=stuff(@s,1,6,'')

    end

return @crypt

end

【测试】

if object_id('tb') is not null

   drop table tb

go

create table tb(id int identity(1,1),col varchar(100))

go

insert into tb values(dbo.f_RSAEncry('中国人',779,1163,59))

insert into tb values(dbo.f_RSAEncry('Chinese',779,1163,59))

select * from tb



id col

1   00359B00E6E000EAF5

2   01075300931B0010A4007EDC004B340074A6004B34



select * ,解密后=dbo.f_RSADecry(col,35039,1163,59)

from tb



id     col                                                                                解密后

1     00359B00E6E000EAF5                                中国人

2     01075300931B0010A4007EDC004B340074A6004B34       Chinese

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2011-08-28 10:53
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