第一次发布海报，但我真的被卡住了。 我正在做一个小项目，我正在尝试使用netbeans项目发送推文。我正在使用twitter4j，似乎最近事情发生了变化，你必须使用他们的OAuth功能。我在twitter上创建了一个应用程序并尝试了一些代码，但我一直收到这个错误：线程“main”中的异常连接超时相关讨论可以在Internet上进行：
http://www.google.co.jp/search?q=b2b52c28 or
http://www.google.co.jp/search?q=1b442895
TwitterException{exceptionCode=[b2b52c28-1b442895 b2b52c28-1b44286b], statusCode=-1, retryAfter=-1, rateLimitStatus=null, featureSpecificRateLimitStatus=null, version=2.2.5}
at twitter4j.internal.http.HttpClientImpl.request(HttpClientImpl.java:200)
at twitter4j.internal.http.HttpClientWrapper.request(HttpClientWrapper.java:65)
at twitter4j.internal.http.HttpClientWrapper.post(HttpClientWrapper.java:102)
at twitter4j.TwitterImpl.post(TwitterImpl.java:1929)
at twitter4j.TwitterImpl.updateStatus(TwitterImpl.java:433)
at login.Login.start(Login.java:36)
at login.Login.main(Login.java:63)
Caused by: java.net.SocketTimeoutException: connect timed out
at java.net.DualStackPlainSocketImpl.waitForConnect(Native Method)
at java.net.DualStackPlainSocketImpl.socketConnect(DualStackPlainSocketImpl.java:75)
at java.net.AbstractPlainSocketImpl.doConnect(AbstractPlainSocketImpl.java:339)
at java.net.AbstractPlainSocketImpl.connectToAddress(AbstractPlainSocketImpl.java:200)
at java.net.AbstractPlainSocketImpl.connect(AbstractPlainSocketImpl.java:182)
at java.net.PlainSocketImpl.connect(PlainSocketImpl.java:157)
at java.net.SocksSocketImpl.connect(SocksSocketImpl.java:391)
at java.net.Socket.connect(Socket.java:579)
at sun.net.NetworkClient.doConnect(NetworkClient.java:175)
at sun.net.www.http.HttpClient.openServer(HttpClient.java:388)
at sun.net.www.http.HttpClient.openServer(HttpClient.java:483)
at sun.net.www.http.HttpClient.<init>(HttpClient.java:213)
at sun.net.www.http.HttpClient.New(HttpClient.java:300)
at sun.net.www.http.HttpClient.New(HttpClient.java:316)
at sun.net.www.protocol.http.HttpURLConnection.getNewHttpClient(HttpURLConnection.java:992)
at sun.net.www.protocol.http.HttpURLConnection.plainConnect(HttpURLConnection.java:928)
at sun.net.www.protocol.http.HttpURLConnection.connect(HttpURLConnection.java:846)
at sun.net.www.protocol.http.HttpURLConnection.getOutputStream(HttpURLConnection.java:1087)
at twitter4j.internal.http.HttpClientImpl.request(HttpClientImpl.java:158)
... 6 more
Java Result: 1
我不完全确定我做错了什么。以下是我尝试过的代码。
package login;
import java.io.IOException;
import twitter4j.ResponseList;
import twitter4j.Status;
import twitter4j.Twitter;
import twitter4j.TwitterException;
import twitter4j.TwitterFactory;
import twitter4j.auth.AccessToken;
public class Login {
private final static String CONSUMER_KEY = "******";
private final static String CONSUMER_KEY_SECRET =
"******";
public void start() throws TwitterException, IOException {
Twitter twitter = new TwitterFactory().getInstance();
twitter.setOAuthConsumer(CONSUMER_KEY, CONSUMER_KEY_SECRET);
// here's the difference
String accessToken = getSavedAccessToken();
String accessTokenSecret = getSavedAccessTokenSecret();
AccessToken oathAccessToken = new AccessToken(accessToken,
accessTokenSecret);
twitter.setOAuthAccessToken(oathAccessToken);
// end of difference
twitter.updateStatus("Hi, im updating status again from Namex Tweet for Demo");
System.out.println("\nMy Timeline:");
// I'm reading your timeline
ResponseList list = twitter.getHomeTimeline();
/* for (Status each : list) {
System.out.println("Sent by: @" + each.getUser().getScreenName()
+ " - " + each.getUser().getName() + "\n" + each.getText()
+ "\n");
}*/
}
private String getSavedAccessTokenSecret() {
// consider this is method to get your previously saved Access Token
// Secret
return "oC8tImRFL6i8TuRkTEaIcWsF8oY4SL5iTGNkG9O0Q";
}
private String getSavedAccessToken() {
// consider this is method to get your previously saved Access Token
return "102333999-M4W1Jtp8y8QY8RH7OxGWbM5Len5xOeeTUuG7QfcY";
}
public static void main(String[] args) throws Exception {
new Login().start();
}
}
答案 0 :(得分：4)
我可以建议一条替代路线..
我最近一直在搞乱twitter4j而且我的处理方式略有不同 - 我发现了一种使用ConfigurationBuilder对象验证客户端的简单方法，并将其传递给工厂，获取所需的Twitter对象实例
package main;
import twitter4j.Twitter;
import twitter4j.TwitterFactory;
import twitter4j.TwitterStream;
import twitter4j.TwitterStreamFactory;
import twitter4j.conf.ConfigurationBuilder;
public class Base {
protected Twitter twitter;
//protected TwitterStream twitterStream;
private ConfigurationBuilder configBuilder;
public Base(){
configBuilder = new ConfigurationBuilder();
configBuilder.setDebugEnabled(true);
configBuilder.setOAuthConsumerKey("[consumer key here]");
configBuilder.setOAuthConsumerSecret("[consumer secret key here]");
configBuilder.setOAuthAccessToken("[OAuthAccessToken here]");
configBuilder.setOAuthAccessTokenSecret("[secret OAuthAccessToken here]");
//use the ConfigBuilder.build() method and pass the result to the TwitterFactory
TwitterFactory tf = new TwitterFactory(configBuilder.build());
//you can now get authenticated instance of Twitter object.
twitter = tf.getInstance();
}
}
然后，您可以使用实现所需功能的子类扩展此类，或者在代码中的其他位置创建ConfigurationBuilder / TwitterFactory / Twitter对象。
下面我实现了一个创建status的类，并且可以返回包含其他信息的Status对象，例如createdAt（）和ID等。
package main;
import twitter4j.Status;
import twitter4j.TwitterException;
public class StatusUpdater extends Base{
public StatusUpdater(){}
public Status updateStatus(String statusToUpdate) throws TwitterException{
Status status = twitter.updateStatus(statusToUpdate);
System.out.println("statusToUpdate: " + status + ".");
return status;
}
}
然后，您可以使用以下语句来创建状态。这可以从mbean / ejb / servlet等完成。
try {
StatusUpdater statusUpdater = new StatusUpdater();
String statusTextToSet = "test status";
Status updatedStatus = statusUpdater.updateStatus(statusTextToSet);
System.out.println("Created at: " + updatedStatus.getCreatedAt());
} catch (TwitterException tex) {
System.out.println(tex.getErrorMessage());
}
有关配置过程的更多信息here
答案 1 :(得分：0)
您的代码看起来不错。您是否确认使用了正确的消费者密钥和应用秘密？请尝试twitter.verifyCredentials()查看是否收到相同的错误或更具体的错误。
如果您通过代理更正网络，则需要包含代理设置以允许连接成功，详情请参见http://twitter4j.org/en/configuration.html#HTTP%20proxy%20server
顺便说一句，您可能希望从帖子中删除您的访问令牌和密码，如果它们仍然有效，则有人可以登录您的帐户。