用于接受有效主机名、IPv4 或 IPv6 地址的 Java 正则表达式
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Java regex for accepting a valid hostname,IPv4, or IPv6 address
提问by ljbade
Anyone have a good (preferably tested) regex for accpeting only a valid DNS hostname, IPv4 or IPv6 address?
任何人都有一个好的（最好是经过测试的）正则表达式来仅接受有效的 DNS 主机名、IPv4 或 IPv6 地址？
回答by Stephen C
I understand that you may be forcedto use a regex. However, if possible it is better to avoid using regexes for this task and use a Java library class to do the validation instead.
我知道您可能被迫使用正则表达式。但是，如果可能，最好避免在此任务中使用正则表达式，而是使用 Java 库类来进行验证。
If you want to do validation and DNS lookup together, then InetAddress.getByName(String)is a good choice. This will cope with DNS, IPv4 and IPv6 in one go, and it returns you a neatly wrapped InetAddressinstance that contains both the DNS name (if provided) and the IPv4 or IPv6 address.
如果你想一起做验证和DNS查找，那么InetAddress.getByName(String)是一个不错的选择。这将一次性处理 DNS、IPv4 和 IPv6，并返回一个整齐包装的InetAddress实例，其中包含 DNS 名称（如果提供）和 IPv4 或 IPv6 地址。
If you just want to do a syntactic validation, then Apache commons has a couple of classes that should do the job: DomainValidatorand InetAddressValidator.
如果您只想进行语法验证，那么 Apache commons 有几个类可以完成这项工作：DomainValidator和InetAddressValidator.
回答by Kevin Bourrillion
Guava has a new class HostSpecifier. It will even validate that the host name (if it is a host name) ends in a valid "public suffix" (e.g., ".com", ".co.uk", etc.), based on the latest mozilla public suffix list. That's something you would NOT want to attempt with a hand-crafted regex!
Guava 有一个新类HostSpecifier。它甚至会根据最新的 mozilla 公共后缀验证主机名（如果是主机名）是否以有效的“公共后缀”（例如，“.com”、“.co.uk”等）结尾列表。这是您不想使用手工制作的正则表达式尝试的事情！
回答by Sean F
As others have said, doing this with a regex is quite a challenge and not advisable. But it is easy to do with the IPAddress Java librarywhich can parse host names, IPv4 and IPv6 addresses, without triggering DNS lookup. Disclaimer: I am the project manager of that library.
正如其他人所说，使用正则表达式执行此操作是一项相当大的挑战，不可取。但是使用IPAddress Java 库很容易做到，它可以解析主机名、IPv4 和 IPv6 地址，而不会触发 DNS 查找。免责声明：我是那个图书馆的项目经理。
Sample code:
示例代码：
check("1.2.3.4");
check("::1");
check("a.b.com");
static void check(String hostStr) {
HostName host = new HostName(hostStr);
try {
host.validate(); // triggers exception for invalid
if(host.isAddress()) {
IPAddress address = host.asAddress();
System.out.println(address.getIPVersion() + " address: " + address);
} else {
System.out.println("host name: " + host);
}
} catch(HostNameException e) {
System.out.println(e.getMessage());
}
}
Output:
输出：
IPv4 address: 1.2.3.4
IPv6 address: ::1
host name: a.b.com
回答by Henno Vermeulen
Inspired by the code I found in this post, I created the following validator method that seems to suit simple validation needs quite nicely. By reading the JavaDoc of URI I removed some false positives such as "host:80" and "hostname/page", but I cannot guarantee there are some false positives left.
受我在这篇文章中找到的代码的启发，我创建了以下验证器方法，它似乎非常适合简单的验证需求。通过阅读 URI 的 JavaDoc，我删除了一些误报，例如“主机：80”和“主机名/页面”，但我不能保证还有一些误报。
public static boolean isValidHostNameSyntax(String candidateHost) {
if (candidateHost.contains("/")) {
return false;
}
try {
// WORKAROUND: add any scheme and port to make the resulting URI valid
return new URI("my://userinfo@" + candidateHost + ":80").getHost() != null;
} catch (URISyntaxException e) {
return false;
}
}
