在安卓中将视频从SD卡分享到Twitter

Question

我必须从我的安卓应用程序将视频上传到推特(发布带有视频的推文(。我使用twitter4j上传图像，它工作得很好。这是我上传图像的代码，我想将视频从SD卡上传到Twitter。如果有人知道解决方案，请帮助我。

String status ="Check out this Tweak "+shareUrl;
try {
ConfigurationBuilder builder = new ConfigurationBuilder();
builder.setOAuthConsumerKey(consumerKey);
builder.setOAuthConsumerSecret(consumerSecret);
// Access Token
String access_token = prefs.getString(PREF_KEY_OAUTH_TOKEN, "");
// Access Token Secret
String access_token_secret = prefs.getString(PREF_KEY_OAUTH_SECRET, "");

twitter4j.auth.AccessToken token = new twitter4j.auth.AccessToken(access_token,
access_token_secret);

twitter4j.auth.AccessToken accessToken = new twitter4j.auth.AccessToken(access_token,
access_token_secret);
Twitter twitter = new TwitterFactory(builder.build()).getInstance(accessToken);
// Update status
StatusUpdate statusUpdate = new StatusUpdate(status);
InputStream is = null;
try {
is = new URL(ShowDialogforMore.twittImage).openStream();
}
catch (MalformedURLException e )
{
e.printStackTrace();
}
catch(IOException e)
{
e.printStackTrace();
}
statusUpdate.setMedia("test.jpg",is);
twitter4j.Status response = twitter.updateStatus(statusUpdate);

Answer 1

我的问题得到了解决方案。这是我的代码。现在我可以将视频分享到推特。

try {
Intent tweetIntent = new Intent(Intent.ACTION_SEND);
String path = Environment.getExternalStorageDirectory()
+ File.separator + "Video" + File.separator + "Sample_nw.mp4";

File f = new File(path);
// String filename = "twitter_image.jpg";
//  File imageFile = new File(getExternalStorageDirectory(), filename);
String tweetUrl ="Tweak Video";
// tweetIntent.putExtra(Intent., "good afternoon all");
tweetIntent.putExtra(Intent.EXTRA_TEXT, tweetUrl);
tweetIntent.putExtra(Intent.EXTRA_STREAM, Uri.fromFile(f));
tweetIntent.setType("video/*");

PackageManager pm = activity.getApplicationContext().getPackageManager();
List<ResolveInfo> lract = pm.queryIntentActivities(tweetIntent, PackageManager.MATCH_DEFAULT_ONLY);
boolean resolved = false;
for (ResolveInfo ri : lract) {
if (ri.activityInfo.name.contains("twitter")) {

tweetIntent.setClassName(ri.activityInfo.packageName,
ri.activityInfo.name);
resolved = true;
break;
}
}
activity.startActivityForResult(resolved ?
tweetIntent :
Intent.createChooser(tweetIntent, "Choose one"), 110);
} catch (final ActivityNotFoundException e) {
}