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 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 @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
【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
责任编辑：小草
