/*本次修改增加了unicode的支持，但是加密後依然顯示為16進制數據，因為進行RSA加密後所得到的unicode編碼是無法顯示的，所以密文依然采用16進制數據顯示。
需要特別注意：如果要對中文進行加密，那麼所選取的兩個素數要比較大，兩個素數的成績最好要大於65536，即大於unicode的最大編碼值
在SQL Server中實現RSA加密算法(第二版)
--判斷是否為素數
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 @i=@i+1
end
return @flg
end
go
--判斷兩個數是否互素
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 @tmp=@num1,@num1=@num2%@num1,@num2=@tmp
end
if @num1=1
set @flg=0
return @flg
end
go
--產生密鑰對
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)+'不是素數，請重新選擇數據'
return
end
if dbo.f_primeNumTest(@q)=0
begin
print cast(@q as varchar)+'不是素數，請重新選擇數據'
return
end
print '請從下列數據中選擇其中一對，作為密鑰'
select @n=@p*@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 @d=@d+1
end
end
set @e=@e+1
end
end
/*加密函數說明，@key 為上一個存儲過程中選擇的密碼中的一個 ,@p ,@q 產生密鑰對時選擇的兩個數。獲取每一個字符的unicode值，然後進行加密，產生3個字節的16位數據*/
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)),@k2=@key/2,@i=1
while @k2>0
begin
set @i=(cast(power(@leftchar,2) as bigint)*@i)%(@p*@q)
set @k2=@k2-1
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 +@tmp,@i=@i/16
from (select number from master.dbo.spt_values where type='p' and number<10 )K
order by number desc
set @crypt=@crypt+right(@tmp,6)
set @s=stuff(@s,1,1,'')
end
return @crypt
end
--解密：@key 為一個存儲過程中選擇的密碼對中另一個數字 ,@p ,@q 產生密鑰對時選擇的兩個數
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 @k2=@key/2,@j=1
while @k2>0
begin
set @j=(cast(power(@leftchar,2)as bigint)*@j)%(@p*@q)
set @k2=@k2-1
end
set @j=(@leftchar*@j)%(@p*@q)
set @crypt=@crypt+nchar(@j)
set @s=stuff(@s,1,6,'')
end
return @crypt
end
【測試】
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('中國人',779,1163,59))
insert into tb values(dbo.f_RSAEncry('Chinese',779,1163,59))
select * from tb
id col
1 00359B00E6E000EAF5
2 01075300931B0010A4007EDC004B340074A6004B34
select * ,解密後=dbo.f_RSADecry(col,35039,1163,59)
from tb
id col 解密後
1 00359B00E6E000EAF5 中國人
2 01075300931B0010A4007EDC004B340074A6004B34 Chinese
