java - 无论如何从 URL 中删除参数

Question

假设我们有一个如下所示的 URL:

https://www.google.com/search?q=X&b=Y

我喜欢直接获取 https://www.google.com/search。有一些简单的方法可以做到这一点，就像这样String baseUrl = url.split("?")[0]; 有什么更好/安全的方法来做到这一点？有内置的东西吗？

Answer 1

您可以使用 URL 类型及其方法。

public static void main(String [] args){
    try
    {
      URL url = new URL("https://www.google.com/search?q=X&b=Y");


      System.out.println("protocol is " + url.getProtocol());
      // |-> prints: 'https'
      System.out.println("hot is "+ url.getHost()); 
      // url.getAuthority() is valid too, but may include the port 
      // incase it's included in the URL
      // |-> prints: 'www.google.com'
      System.out.println("path is " + url.getPath());
      // |-> prints: '/search'

      //WHat you asked for          
      System.out.println(url.getProtocol()+"://"+ url.getAuthority()+url.getPath());
      // |-> prints: 'https://www.google.com/search'
    }catch(IOException e){
      e.printStackTrace();
    }
} 

输出:

protocol is https

host is www.google.com

path is /search

https://www.google.com/search