Question

有没有人知道检查IP是否包含在一组范围内的干净方法？

我收到的回复如下：

"ipRange": "10.12.0.0/16,10.80.0.0/15,10.83.0.0/16,10.101.0.0-10.103.255.255,10.108.0.0/16,10.121.0.0/16,10.123.0.0/16,10.127.0.0/16,10.129.0.0/16,10.131.0.0/16,10.133.0.0/16,10.135.0.0-10.139.255.255,10.208.0.0/14,10.215.0.0/16,10.218.0.0/15,10.233.0.0/16,172.17.128.0-172.19.255.255,172.24.0.0/15,192.168.0.0/16"

我需要查看是否有任何IP工作。带有“斜线”符号的那些应该是直截了当的，但它们有“ - ”的范围怎么样？

Answer 1

我评论了下面的代码。基本上，我拆分了＆＃34; ipRanges＆＃34;通过查找＆＃34;，＆＃34;到不同的ipAddresses或ipAddressRanges中。然后测试每个给定的ip以查看它是否良好。

String ipRange = "10.12.0.0/16,10.80.0.0/15,10.83.0.0/16,10.101.0.0-10.103.255.255,10.108.0.0/16,10.121.0.0/16,10.123.0.0/16,10.127.0.0/16,10.129.0.0/16,10.131.0.0/16,10.133.0.0/16,10.135.0.0-10.139.255.255,10.208.0.0/14,10.215.0.0/16,10.218.0.0/15,10.233.0.0/16,172.17.128.0-172.19.255.255,172.24.0.0/15,192.168.0.0/16";


public IPAddressTest() {
    String ipToTest = "10.101.0.0";
    System.out.println(String.format("[%s] is in range (%b)", ipToTest, isInRange(ipToTest)));
}

private Boolean isInRange(String ipToTest) {
    String[] ranges = ipRange.split(",");

    for (String _range : ranges) {
        String range = _range.indexOf("/") > -1 ? _range.substring(0, _range.indexOf("/")) : _range; // cut off the /xx if it exists
        boolean test = test(ipToTest, range);
        if (test) {
            return true;
        }
    }

    return false;
}

private Boolean test(String ipToTest, String ipRange) {
    String[] ranges = ipRange.split("-"); // Do we have a range or just a single ip
    String loRange = ranges[0]; // grab the first
    String hiRange = ranges.length > 1 ? ranges[1] : null; // If there a range grab the top end, else nothing

    if (hiRange == null) { // if a single ip, just do a string compare
        return ipToTest.equals(loRange);
    } else { // if a range then make sure each part is between the min and max
        String[] partsToTest = ipToTest.split("\\."); // get each part of the ip
        String[] partsLoRange = loRange.split("\\."); // get each part of the ip
        String[] partsHiRange = hiRange.split("\\."); // get each part of the ip
        int test = 0; // this will be our litmus test. if it = 15 when we are done, then each part is good, 63 of ipv6
        for (int i = 0; i < 4; i++) { // change 4 to 6 for ipv6
            int value = Integer.parseInt(partsToTest[i]); //get the int
            int lo = Integer.parseInt(partsLoRange[i]); //get the int
            int hi = Integer.parseInt(partsHiRange[i]); //get the int
            if (value >= lo && value <= hi) { // if we have a good value, set the bit on 'test'
                test += Math.pow(2, i);
            }
        }
        if (test == 15) {
            return true;
        }
    }

    return false;
}

public static void main(String[] args) {
    new IPAddressTest();
}

Answer 2

IPv4地址只不过是无符号的int32。检查具有网络掩码X的IP地址A是否在具有网络掩码Y的子网B中的简单方法：

将A转换为uint32
将B转换为uint32
将Y转换为uint32
如果（A＆amp; Y）==（B＆amp; Y）则是

检查一组范围中是否包含IP？

2 个答案: