我正在尝试使用以下代码分享应用程序的屏幕截图:
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"));
logcat的:
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被强制关闭。当我尝试与Facebook分享时,Facebook默默拒绝该帖子。消息传递信使,但是空洞。共享无需添加图像即可工作。
答案 0 :(得分:17)
首先,永远不要使用连接来构建文件路径,更不用说Uri
值。
其次,EXTRA_STREAM
应该包含Uri
,而不是String
。
第三,既然您知道正确的MIME类型(image/png
),请使用它,而不是使用通配符。
第四,永远不要两次建立相同的路径。在这里,您以正确的方式创建File image
,然后忽略该值。
因此,转储String url
行,将image/*
替换为image/png
,然后修改:
sharingIntent.putExtra(android.content.Intent.EXTRA_STREAM, url);
是:
sharingIntent.putExtra(android.content.Intent.EXTRA_STREAM, Uri.fromFile(file));
答案 1 :(得分:1)
另外,请考虑使用android.support.v4.content.FileProvider类来使用内容URI而不是文件URI来共享文件。它更安全。请参阅reference documentation for FileProvider
答案 2 :(得分:1)
您需要始终传递内容URI(至少在Android 5.1+中)。这里是如何从位图获取内容路径:
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..."));
}