Question

我需要在sql中拆分ip地址。

我已经做了很多发现但找不到任何内置方法来完成任务。

我该如何完成这项任务？

我正在使用sql server

Answer 1

您可以使用以下PARSENAME功能：

with address as(
select '192.168.1.1' as IpAddress
Union
select '192.168.1.2' as IpAddress
Union
select '192.168.1.3' as IpAddress
)
SELECT PARSENAME(IpAddress,4) as first, 
   PARSENAME(IpAddress,3) as second,
   PARSENAME(IpAddress,2) as third,
   PARSENAME(IpAddress,1) as fourth,
FROM address

PARSENAME函数返回对象名称的指定部分。

Answer 2

我发现将IPV4地址从“点字符串”符号转换为（大）整数是有用的，这样它们就可以存储，比较，.... （请注意，以下函数不执行输入验证。）

create function [dbo].[IntegerIPV4Address]( @IPV4Address VarChar(16) )
  returns BigInt
  with SchemaBinding -- Deterministic function.
  begin
  -- NB: ParseName   is non-deterministic.
  declare @Dot1 as Int = CharIndex( '.', @IPV4Address );
  declare @Dot2 as Int = CharIndex( '.', @IPV4Address, @Dot1 + 1 );
  declare @Dot3 as Int = CharIndex( '.', @IPV4Address, @Dot2 + 1 );
  return Cast( Substring( @IPV4Address, 0, @Dot1 ) as BigInt ) * 0x1000000 +
    Cast( Substring( @IPV4Address, @Dot1 + 1, @Dot2 - @Dot1 - 1 ) as BigInt ) * 0x10000 +  
    Cast( Substring( @IPV4Address, @Dot2 + 1, @Dot3 - @Dot2 - 1 ) as BigInt ) * 0x100 +
    Cast( Substring( @IPV4Address, @Dot3 + 1, Len( @IPV4Address ) * 1 ) as BigInt );
  end

转换回用零填充的虚线字符串（以便alpha排序表现良好）：

create function [dbo].[NormalizedIPV4Address]( @IntegerIPV4Address as BigInt )
  returns VarChar(16)
  with SchemaBinding -- Deterministic function.
  begin
  declare @BinaryAddress as VarBinary(4) = Cast( @IntegerIPV4Address as VarBinary(4) );
  return Right( '00' + Cast( Cast( Substring( @BinaryAddress, 1, 1 ) as Int ) as VarChar(3) ), 3 ) +
    '.' + Right( '00' + Cast( Cast( Substring( @BinaryAddress, 2, 1 ) as Int ) as VarChar(3) ), 3 ) +
    '.' + Right( '00' + Cast( Cast( Substring( @BinaryAddress, 3, 1 ) as Int ) as VarChar(3) ), 3 ) +
    '.' + Right( '00' + Cast( Cast( Substring( @BinaryAddress, 4, 1 ) as Int ) as VarChar(3) ), 3 )
  end

使用sql语句拆分字符串（ip地址）

2 个答案: