我们按如下方式在sql server中执行某些数据的校验和:
declare @cs int;
select
@cs = CHECKSUM_AGG(CHECKSUM(someid, position))
from
SomeTable
where
userid = @userId
group by
userid;
然后与客户共享此数据。我们希望能够在客户端重复校验和...但是似乎没有关于如何计算上述函数中的校验和的任何信息。谁能开导我?
答案 0 :(得分:6)
答案 1 :(得分:3)
CHECKSUM函数不能提供非常好的质量校验和,IMO在大多数情况下都没用。据我所知,该算法尚未发布。如果你想要一个可以自己重现的检查,那么使用HashBytes函数和一个标准的已发布算法,如MD5或SHA。
答案 2 :(得分:1)
// SQL和C#镜像乌克兰的快速哈希值
private Int64 HASH_ZKCRC64(byte[] Data)
{
Int64 Result = 0x5555555555555555;
if (Data == null || Data.Length <= 0) return 0;
int SizeGlobalBufer = 8000;
int Ost = Data.Length % SizeGlobalBufer;
int LeftLimit = (Data.Length / SizeGlobalBufer) * SizeGlobalBufer;
for (int i = 0; i < LeftLimit; i += 64)
{
Result = Result
^ BitConverter.ToInt64(Data, i)
^ BitConverter.ToInt64(Data, i + 8)
^ BitConverter.ToInt64(Data, i + 16)
^ BitConverter.ToInt64(Data, i + 24)
^ BitConverter.ToInt64(Data, i + 32)
^ BitConverter.ToInt64(Data, i + 40)
^ BitConverter.ToInt64(Data, i + 48)
^ BitConverter.ToInt64(Data, i + 56);
if ((Result & 0x0000000000000080) != 0)
Result = Result ^ BitConverter.ToInt64(Data, i + 28);
}
if (Ost > 0)
{
byte[] Bufer = new byte[SizeGlobalBufer];
Array.Copy(Data, LeftLimit, Bufer, 0, Ost);
for (int i = 0; i < SizeGlobalBufer; i += 64)
{
Result = Result
^ BitConverter.ToInt64(Bufer, i)
^ BitConverter.ToInt64(Bufer, i + 8)
^ BitConverter.ToInt64(Bufer, i + 16)
^ BitConverter.ToInt64(Bufer, i + 24)
^ BitConverter.ToInt64(Bufer, i + 32)
^ BitConverter.ToInt64(Bufer, i + 40)
^ BitConverter.ToInt64(Bufer, i + 48)
^ BitConverter.ToInt64(Bufer, i + 56);
if ((Result & 0x0000000000000080)!=0)
Result = Result ^ BitConverter.ToInt64(Bufer, i + 28);
}
}
byte[] MiniBufer = BitConverter.GetBytes(Result);
Array.Reverse(MiniBufer);
return BitConverter.ToInt64(MiniBufer, 0);
#region SQL_FUNCTION
/* CREATE FUNCTION [dbo].[HASH_ZKCRC64] (@data as varbinary(MAX)) Returns bigint
AS
BEGIN
Declare @I64 as bigint Set @I64=0x5555555555555555
Declare @Bufer as binary(8000)
Declare @i as int Set @i=1
Declare @j as int
Declare @Len as int Set @Len=Len(@data)
if ((@data is null) Or (@Len<=0)) Return 0
While @i<=@Len
Begin
Set @Bufer=Substring(@data,@i,8000)
Set @j=1
While @j<=8000
Begin
Set @I64=@I64
^ CAST(Substring(@Bufer,@j, 8) as bigint)
^ CAST(Substring(@Bufer,@j+8, 8) as bigint)
^ CAST(Substring(@Bufer,@j+16,8) as bigint)
^ CAST(Substring(@Bufer,@j+24,8) as bigint)
^ CAST(Substring(@Bufer,@j+32,8) as bigint)
^ CAST(Substring(@Bufer,@j+40,8) as bigint)
^ CAST(Substring(@Bufer,@j+48,8) as bigint)
^ CAST(Substring(@Bufer,@j+56,8) as bigint)
if @I64<0 Set @I64=@I64 ^ CAST(Substring(@Bufer,@j+28,8) as bigint)
Set @j=@j+64
End;
Set @i=@i+8000
End
Return @I64
END
*/
#endregion
}
答案 3 :(得分:0)
我想出了rotating_file_handler3
算法,至少对于ASCII字符而言。我用JavaScript创建了一个证明(请参见https://stackoverflow.com/a/59014293/9642)。
query_search_info.log
主要要注意的是对字母数字的偏爱(它们的代码相似且升序)。英文字母使用相同的代码,不区分大小写。
我没有测试过高码(128+)或Unicode。