Java 8 获取没有最后一段的完整 URL 路径(有或没有尾部斜杠)

Question

我有如下 URL，需要按如下方式修剪它们，去掉最后一段。 可能有也可能没有尾部斜线。

localhost:8080/myapp -> localhost:8080/

https://myapp-dev.myhost.com/app/ -> https://myapp-dev.myhost.com/

https://myapp-dev.myhost.com/app/app2 -> https://myapp-dev.myhost.com/app//p>

当然我可以尝试像这样的解决方案

String[] tokens = uri.split("/");//然后连接之前的...

或

Path path = Paths.get(uri.getPath());
String secondToLast = path.getName(path.getNameCount() - 2).toString();

但难道就没有更强大的实用程序或方法吗？

Answer 1

尝试将 url 字符串传递到 URL 对象，然后提取所需的段:

URL someURL = new URL("https://myapp-dev.myhost.com/app/");
System.out.println("PATH = " + someURL.getPath());
System.out.println("HOST = " + someURL.getHost());
System.out.println("PROTOCOL = " + someURL.getProtocol());
System.out.println("PORT = " + someURL.getPort());

输出:

路径 =/app/

HOST = myapp-dev.myhost.com

协议(protocol) = https

端口 = 8080