Question

-----------------------原始问题------------------- ----

假设我有IP 1.0.88.12。我想查询IPv4数据库并返回正确的国家/地区。

数据库（geodata）表（IP）按以下格式列出IP：

network       country
---------------------
1.0.0.0/24     1212
1.0.1.0/24     1212
...            ...
1.0.88.0/22    3000

在此，我假设1.0.88.0/22代表1.0.88.0，1.0.88.1，1.0.88.2代表1.0.88.22。

如何查询此表，以便了解每个network实际上是一个范围？

-----------------------进度更新------------------- ----

所以我的假设是错的，感谢所有的输入！还有一个问题很突出。

采取以下示例（我现在拥有的）：

/**
 * Convert IP 4-octed set into integer
 */
create function [dbo].[IPAddressToInteger] (@IP as varchar(15))
returns bigint
as
begin
 return (convert(bigint, parsename(@IP,1)) +
         convert(bigint, parsename(@IP,2)) * 256 +
         convert(bigint, parsename(@IP,3)) * 256 * 256 +
         convert(bigint, parsename(@IP,4)) * 256 * 256 * 256)
end


/**
 * Calculate CIDR from mask
 */
create function [dbo].[CIDRFromMask] (@mask as bigint)
returns varchar(5)
as
begin
     declare @maskCalc bigint
     select  @maskCalc = dbo.IPAddressToInteger('255.255.255.255') - @mask + 1

     declare @logCalc int
     select  @logCalc = (32 - log(@maskCalc, 2))

     return '/' + cast(@logCalc as varchar(5))
end

select dbo.IPAddressToInteger('195.65.254.11'); -- 3275881995
select select dbo.CIDRFromMask('3275881995');   -- /2

select '195.65.254.11/2' from geodata.IP        -- 0 results
select '195.65.254.0/2'  from geodata.IP        -- 1 result (seems correct)

重点是：如果我连接IP + CIDR，我在地理数据库中找不到相应的IP（afaik是一个完整的数据库）。如果我手动将195.65.254.11/2转换为195.65.254.0/2，它确实会找到结果，但我不认为这是可行的方法。因此：我还缺少哪一步？

Answer 1

您可以尝试将IP范围转换为Int32。这样可以更容易地查询这些间隔。您可以参考以下链接获得一般概念：
https://www.periscopedata.com/blog/comparing-ip-addresses-in-sql.html

Answer 2

如果它实际上是CIDR表示法，您可以转换为int作为建议@DoğukanTunç应用掩码并与CIDR的地址部分进行比较。制作函数可能是个好主意，取2个参数 - IP和CIDR返回位值（0或1）这里只是插图代码，而不是功能。不处理错误，假设IP地址和CIDR有效

declare @IP nvarchar(20)='192.168.1.111';
declare @CIDRADDR nvarchar(20)='192.168.0.0/22';


--convert to int
declare @IP_as_int bigint;
select @IP_as_int=convert(bigint, parsename(@ip, 1)) +
    convert(bigint, parsename(@ip, 2)) * 256 +
    convert(bigint, parsename(@ip, 3)) * 65536 +
    convert(bigint, parsename(@ip, 4)) * 16777216;

--calculate mask from CIDR: can specify table, or use CASE or use cycle
--mst performance will have CASE, but am to lazy to write CASE with 32 cases  
--the result should be N filled bits from left side of 32-bit integer, where N is number after / in CIDR notation
declare @mask bigint=0;
declare @m1 bigint
declare @m2 bigint=0x80000000;
select @m1=CHARINDEX('/',@CIDRADDR);
select @m1=cast(SUBSTRING(@CIDRADDR,@m1+1,2)  as int);

while(@m1>0)
begin
    set @mask = @mask | @m2;
    set @m2=@m2 /2;
    set @m1=@m1-1;
end



--calculate back
declare @IP_as_bin binary(4)=cast(@IP_as_int & @mask as binary(4));
declare @maskedIP nvarchar(20)= cast(cast((substring(@IP_as_bin,1,1)) as int) as nvarchar)+'.'  +
    cast(cast((substring(@IP_as_bin,2,1)) as int) as nvarchar)+'.'  +
    cast(cast((substring(@IP_as_bin,3,1)) as int) as nvarchar)+'.' + 
    cast(cast((substring(@IP_as_bin,4,1)) as int) as nvarchar);

--compare with CIDR address part
select @m1=CHARINDEX('/',@CIDRADDR);
if(SUBSTRING(@CIDRADDR,1,@m1-1)=@maskedIP)
    print @IP + ' belongs to ' +@CIDRADDR;
else
    print @IP + ' doesn`t belong to ' +@CIDRADDR;

Answer 3

通过@Dogucan扩展这个想法

首先，我将IP字符串转换为整数（忽略＆＃34; / 24＆＃34;）。我使用＆＃34; /＆＃34;用于确定网络路由使用了多少位的部分，在本例中为24.我用它来查找IP地址使用了多少位。在这种情况下，它是32-24 = 8.8位等于2 ^ 8 = 256的范围（因为IP基于零，我取1关，给我的IP范围为0..255）。

然后我将其添加到我的原始IP地址以获取结束IP并将其转换回字符串。

declare @ip varchar(50) = '1.0.0.0/24'


;WITH CTE
AS
(
    SELECT
        convert(bigint, SUBSTRING(parsename(@ip, 1),0, CHARINDEX('/',parsename(@ip, 1)))) +
      convert(bigint, parsename(@ip, 2)) * 256 +
      convert(bigint, parsename(@ip, 3)) * 65536 +
      convert(bigint, parsename(@ip, 4)) * 16777216 as StartIPInt,
      convert(bigint, SUBSTRING(parsename(@ip, 1),0, CHARINDEX('/',parsename(@ip, 1)))) +
      convert(bigint, parsename(@ip, 2)) * 256 +
      convert(bigint, parsename(@ip, 3)) * 65536 +
      convert(bigint, parsename(@ip, 4)) * 16777216 + 
      POWER(2,32- convert(bigint, SUBSTRING(parsename(@ip, 1),CHARINDEX('/',parsename(@ip, 1))+1,20)))-1 AS EndIpInt
)
,
IPRange
As
(
    SELECT 
        StartIPInt / 16777216 As StartOctet1, 
        (StartIPInt - (StartIPInt / 16777216) * 16777216) /65536 As StartOctet2,
        (StartIPInt - ((StartIPInt / 65536) * 65536))/256 As StartOctet3,
        (StartIPInt - ((StartIPInt / 256) * 256)) As StartOctet4,
        EndIpInt / 16777216 As EndOctet1, 
        (EndIpInt - (EndIpInt / 16777216) * 16777216) /65536 As EndOctet2,
        (EndIpInt - ((EndIpInt / 65536) * 65536))/256 As EndOctet3,
        (EndIpInt - ((EndIpInt / 256) * 256)) As EndOctet4
    FROM CTE
)
SELECT  CAST(StartOctet1 AS VARCHAR(3)) + '.' + CAST(StartOctet2 AS VARCHAR(3)) + '.' + CAST(StartOctet3 AS VARCHAR(3)) + '.' + CAST(StartOctet4 AS VARCHAR(3))  StartIPAddress,
        CAST(EndOctet1 AS VARCHAR(3)) + '.' + CAST(EndOctet2 AS VARCHAR(3)) + '.' + CAST(EndOctet3 AS VARCHAR(3)) + '.' + CAST(EndOctet4 AS VARCHAR(3))  EndIPAddress
FROM IPRange

这给出了结果：

StartIPAddress  EndIPAddress
1.0.0.0         1.0.0.255

要获得更清晰的代码，请将IPRange CTE替换为：

IPRange
As
(
-- 1.0.0.0 0x0000000001000000
-- 0.1.0.0 0x0000000000010000
-- 0.0.1.0 0x0000000000000100
-- 0.0.0.255 0x00000000000000FF
    SELECT 
        CAST(SUBSTRING(CAST(StartIPInt AS varbinary),0, 6) AS tinyInt) AS StartOctet1,
        CAST(SUBSTRING(CAST(StartIPInt AS varbinary),5, 2) AS tinyInt) AS StartOctet2,
        CAST(SUBSTRING(CAST(StartIPInt AS varbinary),6, 2) AS tinyInt) AS StartOctet3,
        CAST(SUBSTRING(CAST(StartIPInt AS varbinary),7, 2) AS tinyInt) AS StartOctet4,
        CAST(SUBSTRING(CAST(EndIPInt AS varbinary),0, 6) AS tinyInt) AS EndOctet1,
        CAST(SUBSTRING(CAST(EndIPInt AS varbinary),5, 2) AS tinyInt) AS EndOctet2,
        CAST(SUBSTRING(CAST(EndIPInt AS varbinary),6, 2) AS tinyInt) AS EndOctet3,
        CAST(SUBSTRING(CAST(EndIPInt AS varbinary),7, 2) AS tinyInt) AS EndOctet4
    FROM CTE
)

如何在SQL中读取1.0.0.0/24作为范围（1.0.0.0到1.0.0.24）？

3 个答案: