java 如何在给定开始和结束 IP 地址的情况下生成 IP 地址范围？
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How do I get generate an IP address range given start and end IP address?
提问by Vishal
How can generate a range of IP addresses from a start and end IP address?
如何从开始和结束 IP 地址生成一系列 IP 地址？
Example for a network "192.168.0.0/24":
网络“192.168.0.0/24”的示例：
String start = "192.168.0.2"
String end = "192.168.0.254"
I want to have:
我希望有：
192.168.0.2
192.168.0.3
192.168.0.4
192.168.0.5
...
192.168.0.254
PS: Network, start and end IP can be dynamic above is just an example.
PS：网络，开始和结束IP可以是动态的，上面只是一个例子。
Thanks...
谢谢...
回答by Eric J.
Recognize that each of the 4 components of an IPv4 address is really a hex number between 00 and FF.
认识到 IPv4 地址的 4 个组成部分中的每一个实际上都是一个介于 00 和 FF 之间的十六进制数。
If you change your start and end IP addresses into 32 bit unsigned integers, you can just loop from the lowest one to the highest one and convert each value you loop through back into the IP address format.
如果将起始 IP 地址和结束 IP 地址更改为 32 位无符号整数，则只需从最低的一个循环到最高的一个，然后将循环的每个值转换回 IP 地址格式。
The range in the example you give is C0A80002 to C0A800FE.
您给出的示例中的范围是 C0A80002 到 C0A800FE。
Here's a link to code that converts between a hex number and an IPv4 address
这是在十六进制数和 IPv4 地址之间转换的代码的链接
http://technojeeves.com/joomla/index.php/free/58-convert-ip-address-to-number
http://technojeeves.com/joomla/index.php/free/58-convert-ip-address-to-number
回答by Bohemian
Here's simple implementation that outputs what you asked for:
这是输出您要求的简单实现：
public static void main(String args[]) {
String start = "192.168.0.2";
String end = "192.168.0.254";
String[] startParts = start.split("(?<=\.)(?!.*\.)");
String[] endParts = end.split("(?<=\.)(?!.*\.)");
int first = Integer.parseInt(startParts[1]);
int last = Integer.parseInt(endParts[1]);
for (int i = first; i <= last; i++) {
System.out.println(startParts[0] + i);
}
}
Note that this will onlywork for ranges involving the lastpart of the IP address.
请注意，这仅适用于涉及IP 地址最后一部分的范围。
回答by Ryan Stewart
Start at 2, count to 254, and put a "192.168.0." in front of it:
从 2 开始，数到 254，然后输入“192.168.0”。在它面前：
for (int i = 2; i <= 254; i++) {
System.out.println("192.168.0." + i);
}
回答by Vikas
void main(String args[])
{
String start = "192.168.0.2";
String end = "192.168.0.254";
String[] startParts = start.split("(?<=\.)(?!.*\.)");
String[] endParts = end.split("(?<=\.)(?!.*\.)");
int first = Integer.parseInt(startParts[1]);
int last = Integer.parseInt(endParts[1]);
for (int i = first; i <= last; i++)
{
System.out.println(startParts[0] + i);
}
}
