Java 类型不兼容,在twitter4j中找不到符号?
这就是代码,我在标记行(粗体)中发现了问题,我认为这是因为jar版本,但我对此不确定。如果这是因为jar版本,请务必让我知道正确的版本Java 类型不兼容,在twitter4j中找不到符号?,java,twitter4j,Java,Twitter4j,这就是代码,我在标记行(粗体)中发现了问题,我认为这是因为jar版本,但我对此不确定。如果这是因为jar版本,请务必让我知道正确的版本 import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import twitter4j.Twitter; import twitter4j.TwitterException; import
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import twitter4j.Twitter;
import twitter4j.TwitterException;
import twitter4j.TwitterFactory;
import twitter4j.auth.AccessToken;
import twitter4j.auth.RequestToken;
public class NamexTweet {
private final static String CONSUMER_KEY = "xxxxxxxxxxxxx";
private final static String CONSUMER_KEY_SECRET = "yyyyyyyyyyyyyyy";
public void start() throws TwitterException, IOException {
Twitter twitter = new TwitterFactory().getInstance();
twitter.setOAuthConsumer(CONSUMER_KEY, CONSUMER_KEY_SECRET);
**RequestToken requestToken = twitter.getOAuthRequestToken();**
System.out.println("Authorization URL: \n"
+ requestToken.getAuthorizationURL());
AccessToken accessToken = null;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
while (null == accessToken) {
try {
System.out.print("Input PIN here: ");
String pin = br.readLine();
**accessToken = twitter.getOAuthAccessToken(requestToken, pin);**
} catch (TwitterException te) {
System.out.println("Failed to get access token, caused by: "
+ te.getMessage());
System.out.println("Retry input PIN");
}
}
System.out.println("Access Token: " + accessToken.getToken());
System.out.println("Access Token Secret: "
+ accessToken.getTokenSecret());
twitter.updateStatus("hi.. im updating this using Namex Tweet for Demo");
}
public static void main(String[] args) throws Exception {
new NamexTweet().start();// run the Twitter client
}
}
twitter4j 2.2.4是可靠的版本,可以使用。您必须附加以下库:
- twitter4j-core-4.0.4.jar
- twitter4j-stream-4.0.4.jar