pro*_*oxy 7 java uri http amazon-s3
我得到了一个奇怪的方法:
import java.net.URI
URI url = new URI("https://pmi_artifacts_prod.s3.amazonaws.com");
System.out.println(url.getHost()); /returns NULL
URI url2 = new URI("https://s3.amazonaws.com");
System.out.println(url2.getHost()); //returns s3.amazonaws.com
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我想先url.getHost()成为pmi_artifacts_prod.s3.amazonaws.com,但它给了我NULL.事实证明,问题在于域名中的下划线,这是一个已知的错误,但仍然可以做什么,因为我需要完全使用这个主机?
该错误不在 Java 中,而是在命名主机中,因为下划线在主机名中不是有效字符。尽管被广泛错误地使用,Java 拒绝处理此类主机名。
https://en.wikipedia.org/wiki/Hostname#Restrictions_on_valid_hostnames
一个可能的解决方法:
public static void main(String...a) throws URISyntaxException, NoSuchFieldException, SecurityException, IllegalArgumentException, IllegalAccessException {
URI url = new URI("https://pmi_artifacts_prod.s3.amazonaws.com");
System.out.println(url.getHost()); //NULL
URI uriObj = new URI("https://pmi_artifacts_prod.s3.amazonaws.com");
if (uriObj.getHost() == null) {
final Field hostField = URI.class.getDeclaredField("host");
hostField.setAccessible(true);
hostField.set(uriObj, "pmi_artifacts_prod.s3.amazonaws.com");
}
System.out.println(uriObj.getHost()); //pmi_artifacts_prod.s3.amazonaws.com
URI url2 = new URI("https://s3.amazonaws.com");
System.out.println(url2.getHost()); //s3.amazonaws.com
}
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