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是否有可能从restservice java消费rss feed？

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我试图通过简单的休息服务调用消费rss feed . 但我得到连接超时 . 示例rss feed：https://www.yahoo.com/news/rss/

我的代码：

公共课测试{

// http://localhost:8080/RESTfulExample/json/product/get
public static void main(String a[]){

      try {

        URL url = new URL("https://www.yahoo.com/news/rss/");
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();

        conn.setRequestMethod("GET");
        conn.setRequestProperty("Accept", "application/json");
        //conn.setConnectTimeout(999999);
/*
        if (conn.getResponseCode() != 200) {
            throw new RuntimeException("Failed : HTTP error code : "
                    + conn.getResponseCode());
        }*/

        BufferedReader br = new BufferedReader(new InputStreamReader(
            (conn.getInputStream())));

        String output;
        System.out.println("Output from Server .... \n");
        while ((output = br.readLine()) != null) {
            System.out.println(output);
        }

        conn.disconnect();

      } catch (MalformedURLException e) {

        e.printStackTrace();

      } catch (IOException e) {

        e.printStackTrace();

      }

    }

}

例外：

java.net.ConnectException：连接超时：连接java.net.DualStackPlainSocketImpl.connect0（本机方法）java.net.DualStackPlainSocketImpl.socketConnect（DualStackPlainSocketImpl.java:79）at java.net.AbstractPlainSocketImpl.doConnect（AbstractPlainSocketImpl.java ：java.com.AbstractPlainSocketImpl.connectToAddress（AbstractPlainSocketImpl.java：200）java.net.AbstractPlainSocketImpl.connect（AbstractPlainSocketImpl.java:182）at java.net.PlainSocketImpl.connect（PlainSocketImpl.java:172）at java .net.SocksSocketImpl.connect（SocksSocketImpl.java:392）at java..Socket.connect（Socket.java:579）at sun.security.ssl.SSLSocketImpl.connect（SSLSocketImpl.java:625）at sun.security . 位于sun.net.www.http.HttpClient.openServer（HttpClient.java:432）sun.net.NetworkClient.doConnect（NetworkClient.java:180）的ssl.BaseSSLSocketImpl.connect（BaseSSLSocketImpl.java:160） .http.HttpClient.openServer（HttpClient.java:527）at sun.net.www.protocol.https.HttpsClient . （HttpsClient.java） ：275）at sun.net.www.protocol.https.HttpsClient.New（HttpsClient.java:371）at sun.net.www.protocol.https.AbstractDelegateHttpsURLConnection.getNewHttpClient（AbstractDelegateHttpsURLConnection.java:191）at sun.net . www.protocol.http.HttpURLConnection.plainConnect(HttpURLConnection.java:933）at sun.net.www.protocol.https.AbstractDelegateHttpsURLConnection.connect（AbstractDelegateHttpsURLConnection.java:177）at sun.net.www.protocol.http.HttpURLConnection . testInmain上的sun.net.www.protocol.https.HttpsURLConnectionImpl.getInputStream（HttpsURLConnectionImpl.java:254）中的getInputStream（HttpURLConnection.java:1301）

这是错误的方法/代码吗？如果你可以建议更好的方法/代码 .

java rest rss

1 回答

  • 0

    最后能够通过设置代理来解决问题 . 这是适合我的代码 .

    公共课测试{

    public static void main(String a[]) throws ClientProtocolException, IOException{
    HttpClient httpClient = new DefaultHttpClient();
    HttpHost proxy=new HttpHost("hostname", portnumber, "scheme"); // ex: hostname="your org proxy host" portnumber=8080 scheme="http"
    httpClient.getParams().setParameter(ConnRouteParams.DEFAULT_PROXY,proxy );

    HttpGet getRequest = new HttpGet(
        "https://www.yahoo.com/news/rss/");

    //getRequest.addHeader("accept", "application/xml");
    final RequestConfig config = RequestConfig.custom().setProxy(proxy).build();
    getRequest.setConfig(config);


    HttpResponse response = httpClient.execute(getRequest);

    /*if (response.getStatusLine().getStatusCode() != 200) {
        throw new RuntimeException("Failed : HTTP error code : "
           + response.getStatusLine().getStatusCode());
    }*/

    BufferedReader br = new BufferedReader(
                     new InputStreamReader((response.getEntity().getContent())));

    String output;
    System.out.println("Output from Server .... \n");
    while ((output = br.readLine()) != null) {
        System.out.println(output);
    }

    httpClient.getConnectionManager().shutdown();

    }

    }
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