IPv6 验证

Question

我使用 IPAddressUtil.isIPv6LiteralAddress (ipAddress) 方法来验证 IPv6，但此方法对于 ipv6-address/prefix-length 格式失败（格式在 RFC 4291 第 2.3 节） IPV6。

有人知道验证“ ipv6-address/prefix-length ”格式的验证器吗？

IPV6的法律表示

1. ABCD:EF01:2345:6789:ABCD:EF01:2345:6789
2. 2001:DB8:0:0:8:800:200C:417A
3. FF01:0:0:0:0 :0:0:101
4. 0:0:0:0:0:0:0:1
5. 0:0:0:0:0:0:0:0
6. 2001:DB8::8:800:200C:417A
7. FF01::101
8. ::1
9. ::
10. 0:0:0:0:0:0:13.1。 68.3
11. 0:0:0:0:0:FFFF:129.144.52.38
12. ::13.1.68.3
13. FFFF:129.144.52.38
14. 2001:0DB8:0000:CD30:0000:0000:0000:0000/60
15. 2001:0DB8::CD30:0:0:0:0/60
16. 2001:0DB8:0:CD30::/60

< strong>不是 IPV6 的法律代表

1. 2001:0DB8:0:CD3/60
2. 2001:0DB8::CD30/60
3. 2001:0DB8::CD3/60

Answer 1

您可以使用 Guava 库，特别是使用 com.google.common.net.InetAddresses 类，调用 isInetAddress()。

---

isInetAddress

public static boolean isInetAddress(String ipString)

如果提供的字符串是有效的 IP 字符串文字，则返回 true，否则返回 false。

参数：ipString - 要计算为 IP 字符串文字的字符串

返回：true（如果参数是有效 IP）字符串文字

Answer 2

IPAddress Java 库 支持以 a 格式解析 IPv4 和 IPv6 CIDR 子网（即地址/前缀格式）多态方式。免责声明：我是项目经理。

以下方法是用于验证的示例代码：

static void parse(String str) {
    IPAddressString addrString = new IPAddressString(str);
    try {
         IPAddress addr = addrString.toAddress();
         IPAddress hostAddr = addrString.toHostAddress();
         Integer prefix = addr.getNetworkPrefixLength();
         if(prefix == null) {
             System.out.println(addr + " has no prefix length"); 
         } else {
             System.out.println(addr + " has host address " + hostAddr + " and prefix length " + prefix);
         }
    } catch(AddressStringException e) {
        System.out.println(addrString + " is invalid: " + e.getMessage());
    }
}

使用问题中提供的示例，上述方法的输出为：

abcd:ef01:2345:6789:abcd:ef01:2345:6789 has no prefix length
2001:db8::8:800:200c:417a has no prefix length
ff01::101 has no prefix length
::1 has no prefix length
:: has no prefix length
2001:db8::8:800:200c:417a has no prefix length
ff01::101 has no prefix length
::1 has no prefix length
:: has no prefix length
::d01:4403 has no prefix length
::ffff:8190:3426 has no prefix length
::d01:4403 has no prefix length
FFFF:129.144.52.38 is invalid: FFFF:129.144.52.38 IP Address error: address has too few segments
2001:db8:0:cd30::/60 has host address 2001:db8:0:cd30:: and prefix length 60
2001:db8:0:cd30::/60 has host address 2001:db8:0:cd30:: and prefix length 60
2001:db8:0:cd30::/60 has host address 2001:db8:0:cd30:: and prefix length 60
2001:0DB8:0:CD3/60 is invalid: 2001:0DB8:0:CD3/60 IP Address error: address has too few segments
2001:db8::cd30/60 has host address 2001:db8::cd30 and prefix length 60
2001:db8::cd3/60 has host address 2001:db8::cd3 and prefix length 60

如您所见，问题关于 FFFF:129.144.52.38 有效且关于 2001:db8::cd30 是不正确的/60 和 2001:db8::cd3/60 无效。如果第一个是 ::FFFF:129.144.52.38 则有效

Answer 3

看看这是否有效：

try {
    if (subjectString.matches(
        "(?ix)\\A(?:                                                  # Anchor address\n" +
        " (?:  # Mixed\n" +
        "  (?:[A-F0-9]{1,4}:){6}                                # Non-compressed\n" +
        " |(?=(?:[A-F0-9]{0,4}:){2,6}                           # Compressed with 2 to 6 colons\n" +
        "     (?:[0-9]{1,3}\\.){3}[0-9]{1,3}                     #    and 4 bytes\n" +
        "     \\z)                                               #    and anchored\n" +
        "  (([0-9A-F]{1,4}:){1,5}|:)((:[0-9A-F]{1,4}){1,5}:|:)  #    and at most 1 double colon\n" +
        " |::(?:[A-F0-9]{1,4}:){5}                              # Compressed with 7 colons and 5 numbers\n" +
        " )\n" +
        " (?:(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\\.){3}  # 255.255.255.\n" +
        " (?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])           # 255\n" +
        "|     # Standard\n" +
        " (?:[A-F0-9]{1,4}:){7}[A-F0-9]{1,4}                    # Standard\n" +
        "|     # Compressed\n" +
        " (?=(?:[A-F0-9]{0,4}:){0,7}[A-F0-9]{0,4}               # Compressed with at most 7 colons\n" +
        "    \\z)                                                #    and anchored\n" +
        " (([0-9A-F]{1,4}:){1,7}|:)((:[0-9A-F]{1,4}){1,7}|:)    #    and at most 1 double colon\n" +
        "|(?:[A-F0-9]{1,4}:){7}:|:(:[A-F0-9]{1,4}){7}           # Compressed with 8 colons\n" +
        ")/[A-F0-9]{0,4}\\z                                                    # Anchor address")) 
        {
        // String matched entirely
    } else {
        // Match attempt failed
    } 
} catch (PatternSyntaxException ex) {
    // Syntax error in the regular expression
}

大约一年前，我购买了一个非常有用的程序，名为 RegexMagic，用于我计划使用的一些复杂的正则表达式。

这应该是Java，所以它应该编译，我假设/60可以在0000和FFFF的范围之间，你可以修改最后一部分。

/[A-F0-9]{0,4} 是我添加到正则表达式中以匹配您的示例的内容。

Answer 4

我的想法是把它分成两部分，前缀地址和前缀len。

<强>1。验证前缀地址使用一些正则表达式来验证 IPv6 地址
2.验证前缀 len 必须是整数
3.前缀地址中的 ':' 只能小于前缀 len 除以 16
4.还必须考虑其他逻辑，将 TODO 留在这里，抱歉:(

  private int validateIPv6AddrWithPrefix(String address) {
        int occurCount = 0;
        for(char c : address) {
            if(c=='/'){
                occurCount++;
            }
        }
        if(occurCount != 1){
         //not good, to much / character
            return -1;
        }
        /* 2nd element should be an integer */
        String[] ss = pool.getAddress().split("/");
        Integer prefixLen = null;
        try{
            prefixLen = Integer.valueOf(ss[1]);
                    // TODO validate the prefix range(1, 128)

        }catch(NumberFormatException e) {
            /* not a Integer */
            return -1;
        }
        /* 1st element should be ipv6 address */
        if(!IPaddrUtilities.isIPv6Address(ss[0])) {
            return -1;
        }
        /* validate ':' character logic */
        occurCount = 0;
        for(char c : ss[0].toCharArray()){
            if(c==':') {
                occurCount++;
            }
        }
        if(occurCount >= prefixLen/16) {
            // to much ':' character
            return -1;
        }
        return 0;
    }

Answer 5

严格来说，2.3节描述的不是地址表示，而是地址前缀的表示（即使是“全长”前缀也与地址不同）。

IPv6 地址前缀是
用符号表示：
ipv6 地址/前缀长度
在哪里
ipv6-address 是采用列出的任何表示法的 IPv6 地址
在第 2.2 节中。

这意味着如果您需要验证地址，您可以安全地忽略此格式。

Answer 6

我在java中尝试过下面的正则表达式，它适用于IPV4和IPV6

public class Utilities {
private static Pattern VALID_IPV4_PATTERN = null;
private static Pattern VALID_IPV6_PATTERN = null;
private static final String ipv4Pattern = "(([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.){3}([01]?\\d\\d?|2[0-4]\\d|25[0-5])";
private static final String ipv6Pattern = "([0-9a-f]{1,4}:){7}([0-9a-f]){1,4}";

static {
try {
  VALID_IPV4_PATTERN = Pattern.compile(ipv4Pattern, Pattern.CASE_INSENSITIVE);
  VALID_IPV6_PATTERN = Pattern.compile(ipv6Pattern, Pattern.CASE_INSENSITIVE);
   } catch (PatternSyntaxException e) {
  //logger.severe("Unable to compile pattern", e);
 }
}

 /**
 * Determine if the given string is a valid IPv4 or IPv6 address.  This method
 * uses pattern matching to see if the given string could be a valid IP address.
 *
 * @param ipAddress A string that is to be examined to verify whether or not
 * it could be a valid IP address.
 * @return <code>true</code> if the string is a value that is a valid IP address,
 *  <code>false</code> otherwise.
 */
 public static boolean isIpAddress(String ipAddress) {

Matcher m1 = Utilities.VALID_IPV4_PATTERN.matcher(ipAddress);
if (m1.matches()) {
  return true;
}
Matcher m2 = Utilities.VALID_IPV6_PATTERN.matcher(ipAddress);
return m2.matches();
  }


}