客户端 - 服务器应用程序Java

Question

我写了一个服务器类，我的客户端是我的浏览器。当我在浏览器中输入localhost:8082时，打开了硬编码网站www.mmix.cs.hm.edu。到现在为止还挺好。 一个网站通常有多个页面。无论我点击其他链接，我的服务器只能检索主页www.mmix.cs.hm.edu/index.html。我希望能够导航到这些其他页面。任何人都可以看看我的代码，并告诉我如何继续？

public static void main(String args[]) { 
    String fromClient = "www.mmix.cs.hm.edu"; 

    try(ServerSocket welcomeSocket = new ServerSocket(8082)){ 
     System.out.println("Server started, waiting for clients..."); 
     while(true){ 
      StringBuilder htmlCode = new StringBuilder(); 
      try(Socket connectionSocket = welcomeSocket.accept(); 
        DataOutputStream toClient = new DataOutputStream(connectionSocket.getOutputStream()); 
        BufferedReader fromBrowser = new BufferedReader(new InputStreamReader(connectionSocket.getInputStream()))){ 
        try(InputStream url = new URL("http://"+fromClient+"/index.html").openStream(); 
          BufferedReader getUrl = new BufferedReader(new InputStreamReader(url))){ 
          for(String line = getUrl.readLine(); line != null; line = getUrl.readLine()){ 
           htmlCode.append(line); 
          } 
          String str = htmlCode.toString(); 
          toClient.writeBytes(str); 
           //toClient.write("\r\n"); 
        } 
      } 
     } 
    } 
    catch(IOException io){ 
     io.printStackTrace(); 
    } 
} 

Answer 1

@ ObiWanKenobi-更改了您的代码以提取URL部分。尝试下面的代码片段。请通过代码片段中的评论。运行并确认字符串操作是否有效。谢谢。

public static void main(String args[]) { 
     String fromClient = "www.mmix.cs.hm.edu"; 

     try(ServerSocket welcomeSocket = new ServerSocket(8082)){ 
      System.out.println("Server started, waiting for clients..."); 
      while(true){ 
       StringBuilder htmlCode = new StringBuilder(); 
       try(Socket connectionSocket = welcomeSocket.accept(); 
         DataOutputStream toClient = new DataOutputStream(connectionSocket.getOutputStream()); 
         BufferedReader fromBrowser = new BufferedReader(new InputStreamReader(connectionSocket.getInputStream()))){ 

         String line1 = fromBrowser.readLine(); //Line 1 is of format: GET /index.html HTTP/1.1 
         String dynUrl = line1.substring(line1.indexOf(32)+1,line1.lastIndexOf(32)); //dynUrl is of format:/index.html 
         //Please note that the query string parameters not taken into account and the code may fail if the query string contains space character. 
         //Construct a new URL based on dynUrl 
         try(InputStream url = new URL("http://"+fromClient+dynUrl).openStream(); 
           BufferedReader getUrl = new BufferedReader(new InputStreamReader(url))){ 
           for(String line = getUrl.readLine(); line != null; line = getUrl.readLine()){ 
            htmlCode.append(line); 
           } 
           String str = htmlCode.toString(); 
           toClient.writeBytes(str); 
            //toClient.write("\r\n"); 
         } 
       } 
      } 
     } 
     catch(IOException io){ 
      io.printStackTrace(); 
     } 
    }