Question

//This is my code for returning json response from twitter api which works fine when i am not behind any proxy server 

import oauth.signpost.OAuthConsumer;
import oauth.signpost.commonshttp.CommonsHttpOAuthConsumer;
import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.util.EntityUtils;
import org.json.JSONArray;

public class TwitterFeeds {

    static String AccessToken = "xxxx";
    static String AccessSecret = "xxxx";
    static String ConsumerKey = "xxxx";
    static String ConsumerSecret = "xxxx";

    public static void main(String[] args) throws Exception { 
        OAuthConsumer consumer = new CommonsHttpOAuthConsumer(ConsumerKey, ConsumerSecret);
        consumer.setTokenWithSecret(AccessToken, AccessSecret);

        HttpGet request = new     HttpGet("https://api.twitter.com/1.1/statuses/user_timeline.json?screen_name=xxxx&count=2"); 

        consumer.sign(request);
        HttpClient client = new DefaultHttpClient();       
        HttpResponse response = client.execute(request);
        String json_string = EntityUtils.toString(response.getEntity());
        System.out.println(json_string);
        JSONArray  obj = new JSONArray (json_string);
        System.out.println(obj.toString());
    }
}

//我在阻止我的请求的代理服务器后面，当我不在任何代理服务器后面时，代码工作正常。并且收到以下错误

Exception in thread "main" org.apache.http.conn.HttpHostConnectException: Connection to https://api.twitter.com refused
    at org.apache.http.impl.conn.DefaultClientConnectionOperator.openConnection(DefaultClientConnectionOperator.java:158)
    at org.apache.http.impl.conn.AbstractPoolEntry.open(AbstractPoolEntry.java:149)
    at org.apache.http.impl.conn.AbstractPooledConnAdapter.open(AbstractPooledConnAdapter.java:121)
    at org.apache.http.impl.client.DefaultRequestDirector.tryConnect(DefaultRequestDirector.java:562)
    at org.apache.http.impl.client.DefaultRequestDirector.execute(DefaultRequestDirector.java:415)
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:820)
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:754)
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:732)
    at TwitterFeeds.main(TwitterFeeds.java:47)
Caused by: java.net.ConnectException: Connection timed out: connect
    at java.net.DualStackPlainSocketImpl.connect0(Native Method)
    at java.net.DualStackPlainSocketImpl.socketConnect(Unknown Source)
    at java.net.AbstractPlainSocketImpl.doConnect(Unknown Source)
    at java.net.AbstractPlainSocketImpl.connectToAddress(Unknown Source)
    at java.net.AbstractPlainSocketImpl.connect(Unknown Source)
    at java.net.PlainSocketImpl.connect(Unknown Source)
    at java.net.SocksSocketImpl.connect(Unknown Source)
    at java.net.Socket.connect(Unknown Source)
    at org.apache.http.conn.ssl.SSLSocketFactory.connectSocket(SSLSocketFactory.java:374)
    at org.apache.http.impl.conn.DefaultClientConnectionOperator.openConnection(DefaultClientConnectionOperator.java:148)
    ... 8 more

这是我显示连接被拒绝的错误。以及用户获取推文（json格式）的代码，当我不在任何代理服务器后面时工作，当我在代理服务器后面时不起作用

Answer 1

我认为您的HttpGet请求缺少代理设置。

您应该执行以下操作：

// Here you set your proxy host and port
HttpHost proxy = new HttpHost(<your_proxy_addess>, <port>); 
DefaultProxyRoutePlanner routePlanner = new DefaultProxyRoutePlanner(proxy);
HttpClient client = HttpClients.custom().setRoutePlanner(routePlanner).build();
HttpResponse response = client.execute(request);

如果您认为提供任何类型的代理凭证是必要的，那么应采取以下方法：

创建HttpGet请求

HttpHost proxy = new HttpHost(<your_proxy_addess>, <port>);
RequestConfig config = RequestConfig.custom().setProxy(proxy).build();
HttpGet request = new HttpGet("https://api.twitter.com/1.1/statuses/user_timeline.json?screen_name=xxxx&count=2");
request.setConfig(config);

将您的凭据提供给HttpClient：

CredentialsProvider credsProvider = new BasicCredentialsProvider();
credsProvider.setCredentials(
        new AuthScope(<your_proxy_addess>, <port>),
        new UsernamePasswordCredentials("user", "passwd"));

HttpClient client = HttpClients.custom()
        .setDefaultCredentialsProvider(credsProvider).build();
HttpResponse response = client.execute(request);

Answer 2

我认为你因代理错误而走上正轨。

在java中，至少有2-3种方法来绕过＆＃34;绕过＆＃34;代理人。他们每个人都有优点和不便：

将参数传递给JVM：

$ java -Dhttp.proxyHost=webcache.example.com -Dhttp.proxyPort=8080 -Dhttp.proxyUser=someUserName -Dhttp.proxyPassword=somePassword

通过这样做，所有HTTP连接都受此设置的影响。按照下一点，你可以采用一种稍微动态的方法。

使用System.setProperty（String，String）：

与上面的方法一样，您可以为JVM设置和取消设置属性，但现在可以直接在代码中设置和取消设置。

//Set the http proxy to webcache.example.com:8080
System.setProperty("http.proxyHost", "webcache.example.com");
System.setProperty("http.proxyPort", "8080");

// Now, let's 'unset' the proxy.
System.setProperty("http.proxyHost", null);

如果您希望保持一切简单并且您的应用程序不是多线程，那么这是一个很好的解决方案。就像我之前说的那样，为所有JVM设置了属性。

使用Proxy类：

这个解决方案比另外两个解决方案更灵活。

SocketAddress addr = new InetSocketAddress("webcache.example.com", 8080);
Proxy proxy = new Proxy(Proxy.Type.HTTP, addr);      
URL url = new URL("http://java.example.org/");
URConnection conn = url.openConnection(proxy);

注意：对于HTTPS，只需使用Proxy.Type.HTTPS即可。对于其他方法，请使用https.proxyPort代替http.proxyPort（例如）

运行我的代码时拒绝连接，这可能是与proxy.how相关的问题。

2 个答案: