Question

我正在尝试从服务器下载XML文档：

http://api.yr.no/weatherapi/locationforecast/1.9/?lat=60.10;lon=9.58;msl=70

我使用java来执行此操作并且还需要使用套接字。

import java.net.*;
import java.io.*;

public class main 
{

/**
 * @param args
 */
public static void main(String[] args) 
{
    try
    {
        byte[] data = new byte[100000];
        Socket clientSocket = new Socket();
        InetSocketAddress ip = new InetSocketAddress("api.yr.no", 80);
        clientSocket.connect(ip);
        DataInputStream inData = new DataInputStream(clientSocket.getInputStream());
        OutputStream outData = clientSocket.getOutputStream();

        PrintWriter pw = new PrintWriter(outData, false);
        pw.print("GET " + "/weatherapi/locationforecast/1.9/?lat=60.10;lon=9.58;msl=70" + " HTTP/1.0\r\n");
        pw.print("\r\n");
        pw.flush();

        int bytesread = inData.read(data);
        String translateddata = new String(data);
        System.out.print(translateddata);
    } 
    catch (IOException e) 
    {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    finally
    {

    }
}

然而，在运行之后，我没有得到URL的xml：

HTTP/1.1 404 Unkown host
Server: Varnish
Content-Type: text/html; charset=utf-8
Content-Length: 395
Accept-Ranges: bytes
Date: Wed, 10 Sep 2014 09:16:22 GMT
X-Varnish: 1432508106
Age: 0
Via: 1.1 varnish
Connection: close


<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
 #http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html>
  <head>
    <title>404 Unkown host</title>
  </head>
  <body>
    <h1>Error 404 Unkown host</h1>
    <p>Unkown host</p>
    <h3>Guru Meditation:</h3>
    <p>XID: 1432508106</p>
    <hr>
    <p>Varnish cache server</p>
  </body>
</html>

所以我猜测我的GET请求在某种程度上是错误的，但是我无法找到问题所在。关于如何构建对此XML文档的请求的任何线索？

Answer 1

如果你得到＆＃34;未知的主人＆＃34;回来，可能会有多个主机在同一个IP后面。在这种情况下，添加一个Host-Header，即

pw.print("GET " + "/weatherapi/locationforecast/1.9/?lat=60.10;lon=9.58;msl=70" + " HTTP/1.0\r\n");
pw.print("Host: api.yr.no\r\n");
pw.print("\r\n");

Answer 2

请尝试以下代码。

try
{
    URL url = new URL("http://api.yr.no/weatherapi/locationforecast/1.9/?lat=60.10;lon=9.58;msl=70");
    URLConnection yc = url.openConnection();
    BufferedReader in = new BufferedReader(new InputStreamReader(
                                yc.getInputStream()));
    String inputLine;
    while ((inputLine = in.readLine()) != null) {
        System.out.println(inputLine);
    }
    in.close();
} catch(Exception e) {

}

从URL获取xml

2 个答案: