Android共享映像不起作用
Android共享映像不起作用
提问于 2013-10-10 21:19:21
我试图使用以下代码共享应用程序的屏幕快照：
View content = findViewById(R.id.layoutHome);
content.setDrawingCacheEnabled(true);
Bitmap bitmap = content.getDrawingCache();
File sdCardDirectory = Environment.getExternalStorageDirectory();
File image = new File(sdCardDirectory,"temp.png");
// Encode the file as a PNG image.
FileOutputStream outStream;
try {
outStream = new FileOutputStream(image);
bitmap.compress(Bitmap.CompressFormat.PNG, 100, outStream);
outStream.flush();
outStream.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
String url = "file://" + sdCardDirectory.toString() + "Images/temp.png";
Intent sharingIntent = new Intent(android.content.Intent.ACTION_SEND);
sharingIntent.setType("image/*");
String shareBody = "Here is the share content body";
sharingIntent.putExtra(android.content.Intent.EXTRA_SUBJECT,"Subject Here");
sharingIntent.putExtra(android.content.Intent.EXTRA_STREAM, url);
sharingIntent.putExtra(android.content.Intent.EXTRA_TEXT,shareBody);
startActivity(Intent.createChooser(sharingIntent, "Share via"));逻辑猫：
10-10 14:20:16.631: W/Bundle(16349): Key android.intent.extra.STREAM expected Parcelable but value was a java.lang.String. The default value <null> was returned.
10-10 14:20:16.658: W/Bundle(16349): Attempt to cast generated internal exception:
10-10 14:20:16.658: W/Bundle(16349): java.lang.ClassCastException: java.lang.String cannot be cast to android.os.Parcelable
10-10 14:20:16.658: W/Bundle(16349): at android.os.Bundle.getParcelable(Bundle.java:1171)
10-10 14:20:16.658: W/Bundle(16349): at android.content.Intent.getParcelableExtra(Intent.java:4140)
10-10 14:20:16.658: W/Bundle(16349): at android.content.Intent.migrateExtraStreamToClipData(Intent.java:6665)
10-10 14:20:16.658: W/Bundle(16349): at android.content.Intent.migrateExtraStreamToClipData(Intent.java:6650)
10-10 14:20:16.658: W/Bundle(16349): at android.app.Instrumentation.execStartActivity(Instrumentation.java:1410)
10-10 14:20:16.658: W/Bundle(16349): at android.app.Activity.startActivityForResult(Activity.java:3351)
10-10 14:20:16.658: W/Bundle(16349): at android.app.Activity.startActivityForResult(Activity.java:3312)
10-10 14:20:16.658: W/Bundle(16349): at android.app.Activity.startActivity(Activity.java:3522)
10-10 14:20:16.658: W/Bundle(16349): at android.app.Activity.startActivity(Activity.java:3490)
10-10 14:20:16.658: W/Bundle(16349): at com.example.simplegraph.EconActivity$DrawerItemClickListener.onItemClick(EconActivity.java:182)
10-10 14:20:16.658: W/Bundle(16349): at android.widget.AdapterView.performItemClick(AdapterView.java:298)
10-10 14:20:16.658: W/Bundle(16349): at android.widget.AbsListView.performItemClick(AbsListView.java:1086)
10-10 14:20:16.658: W/Bundle(16349): at android.widget.AbsListView$PerformClick.run(AbsListView.java:2855)
10-10 14:20:16.658: W/Bundle(16349): at android.widget.AbsListView$1.run(AbsListView.java:3529)
10-10 14:20:16.658: W/Bundle(16349): at android.os.Handler.handleCallback(Handler.java:615)
10-10 14:20:16.658: W/Bundle(16349): at android.os.Handler.dispatchMessage(Handler.java:92)
10-10 14:20:16.658: W/Bundle(16349): at android.os.Looper.loop(Looper.java:137)
10-10 14:20:16.658: W/Bundle(16349): at android.app.ActivityThread.main(ActivityThread.java:4745)
10-10 14:20:16.658: W/Bundle(16349): at java.lang.reflect.Method.invokeNative(Native Method)
10-10 14:20:16.658: W/Bundle(16349): at java.lang.reflect.Method.invoke(Method.java:511)
10-10 14:20:16.658: W/Bundle(16349): at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:786)
10-10 14:20:16.658: W/Bundle(16349): at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:553)
10-10 14:20:16.658: W/Bundle(16349): at dalvik.system.NativeStart.main(Native Method)问题：当我尝试与gmail共享时，gmail是强制关闭的。当我试图与Facebook分享时，Facebook默默地拒绝了这篇文章。消息会弹出信使，但是是空的。共享工作而不添加图像。
回答 3
Stack Overflow用户
回答已采纳
发布于 2013-10-10 21:27:07
首先，不要使用连接来构建文件路径，更不用说Uri值了。
第二，EXTRA_STREAM应该包含一个Uri，而不是String。
第三，由于您知道正确的MIME类型(image/png)，请使用它，而不是通配符。
第四，不要两次构建同一条道路。这里以正确的方式创建File image，然后忽略该值。
因此，转储String url行，用image/png替换image/*，并修改：
sharingIntent.putExtra(android.content.Intent.EXTRA_STREAM, url);将是：
sharingIntent.putExtra(android.content.Intent.EXTRA_STREAM, Uri.fromFile(file));Stack Overflow用户
发布于 2013-10-11 01:11:55
另外，考虑使用android.support.v4.content.FileProvider类来使用内容URI而不是文件URI共享文件。更安全。见FileProvider参考文档
Stack Overflow用户
发布于 2015-10-14 09:30:47
您需要始终传递内容URI (至少在Android 5.1+中是这样)。以下是如何从位图中获取内容路径：
Bitmap bitmap;//this should be your bitmap
String MediaFilePath = Images.Media.insertImage(MainActivity.getContentResolver(), bitmap, FileName, null);然后分享：
public static void ShareFile(String ContentPath, String Mime)
{
Intent sharingIntent = new Intent(android.content.Intent.ACTION_SEND);
sharingIntent.setType(Mime);
Uri FileUri = Uri.parse( ContentPath );
sharingIntent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
sharingIntent.addFlags(Intent.FLAG_GRANT_READ_URI_PERMISSION);
sharingIntent.putExtra(Intent.EXTRA_STREAM, FileUri);
MainActivity.startActivity(Intent.createChooser(sharingIntent, "Share to..."));
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/19306417
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