我使用IPAddressUtil.isIPv6LiteralAddress (ipAddress)方法验证IPv6，但此方法因IPV6的ipv6-address / prefix-length格式（格式在RFC 4291第2.3节中提到）而失败。
有谁知道任何验证“ipv6-address / prefix-length”格式的验证器？
IPV6的法律陈述
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
}
近一年前我购买了一个名为RegexMagic的非常有用的程序，用于我计划使用的一些复杂的正则表达式。
这被认为是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以匹配您的示例。
您可以使用Guava库，特别是使用com.google.common.net.InetAddresses类，调用isInetAddress()。
严格地说，2.3节没有描述地址表示，而是表示地址前缀（甚至“全长”前缀与地址不同）。
IPv6地址前缀由符号表示：ipv6-address / prefix-length其中ipv6-address是第2.2节中列出的任何符号中的IPv6地址。
这意味着如果您需要验证地址，您可以安全地忽略此格式。
我的想法是将它分为两部分，前缀地址和前缀len。
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.前缀地址只能有'：'小于
prefix len divided by 16的结果 4.其他逻辑也必须考虑，离开TODO，对不起:(
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;
}
我在java中尝试了以下正则表达式，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();
}
}
The IPAddress Java library支持以多态方式解析IPv4和IPv6 CIDR子网（即地址/前缀格式）。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，上述方法的输出是：
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
正如你所看到的，关于FFFF的问题是不正确的：129.144.52.38是有效的，关于2001：db8 :: cd30 / 60和2001：db8 :: cd3 / 60无效。第一个如果是:: FFFF：129.144.52.38则有效