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我使用Twitter4j来制作一个Twitter应用程序,用于学习如何制作应用程序。我通过复制sample code from the twitter4j site为ICS制作了一个非常基本的应用程序。 twitter4j api发送OAuth进程的http请求,然后打开授权过程发生的网页。问题是应用程序卡在代码的http请求部分。我已在清单文件中添加了许可<uses-permission android:name="android.permission.INTERNET"/>
。代码片段如下。使用Twitter4j的Http请求不起作用并卡住
Log.v("storeCredentials", "Started");
System.out.println("storeCredentials"+" Started");
Twitter t1 = TwitterFactory.getSingleton();
t1.setOAuthConsumer(twitter.getConsumerKey(), twitter.getConsumerSecret());
Log.v("storeCredentials", "consumer set");
System.out.println("storeCredentials"+" consumer set");
//ConnectionDetector detective = new ConnectionDetector(this);
System.out.println("storeCredentials"+" debug 5");
Log.v("storeCredentials", "connection present");
//Progress Dialog
System.out.println("storeCredentials"+" debug 6");
final ProgressDialog d1 = new ProgressDialog(this);
System.out.println("storeCredentials"+" debug 7");
d1.setProgressStyle(ProgressDialog.STYLE_SPINNER);
d1.setCancelable(false);
d1.show();
new Thread(new Runnable(){
public void run()
{
Twitter temp = TwitterFactory.getSingleton();
temp.setOAuthConsumer(twitter.getConsumerKey(), twitter.getConsumerSecret());
try
{
RequestToken tempToken = temp.getOAuthRequestToken();
AccessToken accessToken = null;
System.out.println("storeCredentials"+" debug 10");
Log.v("storeCredentials", "tokens init");
//BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//while (null == accessToken) return true;
String url = tempToken.getAuthorizationURL();
//String url = "http://www.google.co.in";
if (!url.startsWith("https://") && !url.startsWith("http://"))
{
url = "http://" + url;
}
Uri webpage = Uri.parse(url);
Intent webIntent = new Intent(Intent.ACTION_VIEW, webpage);
Log.v("storeCredentials", "Intent created");
System.out.println("storeCredentials"+" Intent created");
// Verify it resolves
PackageManager packageManager = getPackageManager();
List<ResolveInfo> activities = packageManager.queryIntentActivities(webIntent, 0);
boolean isIntentSafe = activities.size() > 0;
Log.v("Number of browsers", String.valueOf(activities.size()));
System.out.println("storeCredentials"+" Number of browsers" + " String.valueOf(activities.size())");
// Start an activity if it's safe
if (isIntentSafe)
{
startActivity(webIntent);
}
我敢肯定的应用卡在RequestToken tempToken = temp.getOAuthRequestToken();
线监守我没有收到,要求我选择浏览器的提示。 twitter4j网站的示例代码正常工作。我通过在java应用程序中使用它进行验证。
为什么会有这种情况发生?我还必须包含其他许可以及返回请求吗?
如果您在IDE中执行此操作,请尝试设置断点并点击它。 – karmanaut 2013-03-29 13:23:37