sql - 选择两个 IP 范围之间的记录

Question

我有一个表，其中存储 ID、Name、Code、IPLow、IPHigh 如:

1, Lucas, 804645, 192.130.1.1, 192.130.1.254
2, Maria, 222255, 192.168.2.1, 192.168.2.254
3, Julia, 123456, 192.150.3.1, 192.150.3.254

现在，如果我有一个 IP 地址 192.168.2.50，如何检索匹配的记录？

编辑

根据戈登的回答(我收到编译错误)，这就是我所拥有的:

select PersonnelPC.*
from (select PersonnelPC.*,
             (
              cast(parsename(iplow, 4)*1000000000 as decimal(12, 0)) +
              cast(parsename(iplow, 3)*1000000 as decimal(12, 0)) +
              cast(parsename(iplow, 2)*1000 as decimal(12, 0)) +
              (parsename(iplow, 1))
             ) as iplow_decimal,
            (
              cast(parsename(iphigh, 4)*1000000000 as decimal(12, 0)) +
              cast(parsename(iphigh, 3)*1000000 as decimal(12, 0)) +
              cast(parsename(iphigh, 2)*1000 as decimal(12, 0)) +
              (parsename(iphigh, 1))
             ) as iphigh_decimal
      from PersonnelPC
     ) PersonnelPC
where 192168002050 between iplow_decimal and iphigh_decimal;

但这给了我一个错误:

Msg 8115, Level 16, State 2, Line 1
Arithmetic overflow error converting expression to data type int.

有什么想法吗？

Answer 1

痛苦。 SQL Server 的字符串操作函数很糟糕。然而，它确实提供了parsename()。此方法将 IP 地址转换为一个大的十进制值以进行比较:

select t.*
from (select t.*,
             (cast(parsename(iplow, 4)*1000000000.0 as decimal(12, 0)) +
              cast(parsename(iplow, 3)*1000000.0 as decimal(12, 0)) +
              cast(parsename(iplow, 2)*1000.0 as decimal(12, 0)) +
              cast(parsename(iplow, 1) as decimal(12, 0))
             ) as iplow_decimal,
             (cast(parsename(iphigh, 4)*1000000000.0 as decimal(12, 0)) +
              cast(parsename(iphigh, 3)*1000000.0 as decimal(12, 0)) +
              cast(parsename(iphigh, 2)*1000.0 as decimal(12, 0)) +
              cast(parsename(iphigh, 1) as decimal(12, 0))
             ) as iphigh_decimal
      from t
     ) t
where 192168002050 between iplow_decimal and iphigh_decimal;

我应该注意到，IP 地址通常以 4 字节无符号整数的形式存储在数据库中。这使得比较变得更加容易。 。 。尽管您需要复杂的逻辑(通常包含在函数中)将值转换为可读格式。