Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
78 views
in Technique[技术] by (71.8m points)

android - Create and Share a File from Internal Storage

My goal is to create a XML file on internal storage and then send it through the share Intent.

I'm able to create a XML file using this code

FileOutputStream outputStream = context.openFileOutput(fileName, Context.MODE_WORLD_READABLE);
PrintStream printStream = new PrintStream(outputStream);
String xml = this.writeXml(); // get XML here
printStream.println(xml);
printStream.close();

I'm stuck trying to retrieve a Uri to the output file in order to share it. I first tried to access the file by converting the file to a Uri

File outFile = context.getFileStreamPath(fileName);
return Uri.fromFile(outFile);

This returns file:///data/data/com.my.package/files/myfile.xml but I cannot appear to attach this to an email, upload, etc.

If I manually check the file length, it's proper and shows there is a reasonable file size.

Next I created a content provider and tried to reference the file and it isn't a valid handle to the file. The ContentProvider doesn't ever seem to be called a any point.

Uri uri = Uri.parse("content://" + CachedFileProvider.AUTHORITY + "/" + fileName);
return uri;

This returns content://com.my.package.provider/myfile.xml but I check the file and it's zero length.

How do I access files properly? Do I need to create the file with the content provider? If so, how?

Update

Here is the code I'm using to share. If I select Gmail, it does show as an attachment but when I send it gives an error Couldn't show attachment and the email that arrives has no attachment.

public void onClick(View view) {
    Log.d(TAG, "onClick " + view.getId());

    switch (view.getId()) {
        case R.id.share_cancel:
            setResult(RESULT_CANCELED, getIntent());
            finish();
            break;

        case R.id.share_share:

            MyXml xml = new MyXml();
            Uri uri;
            try {
                uri = xml.writeXmlToFile(getApplicationContext(), "myfile.xml");
                //uri is  "file:///data/data/com.my.package/files/myfile.xml"
                Log.d(TAG, "Share URI: " + uri.toString() + " path: " + uri.getPath());

                File f = new File(uri.getPath());
                Log.d(TAG, "File length: " + f.length());
                // shows a valid file size

                Intent shareIntent = new Intent();
                shareIntent.setAction(Intent.ACTION_SEND);
                shareIntent.putExtra(Intent.EXTRA_STREAM, uri);
                shareIntent.setType("text/plain");
                startActivity(Intent.createChooser(shareIntent, "Share"));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }

            break;
    }
}

I noticed that there is an Exception thrown here from inside createChooser(...), but I can't figure out why it's thrown.

E/ActivityThread(572): Activity com.android.internal.app.ChooserActivity has leaked IntentReceiver com.android.internal.app.ResolverActivity$1@4148d658 that was originally registered here. Are you missing a call to unregisterReceiver()?

I've researched this error and can't find anything obvious. Both of these links suggest that I need to unregister a receiver.

I have a receiver setup, but it's for an AlarmManager that is set elsewhere and doesn't require the app to register / unregister.

Code for openFile(...)

In case it's needed, here is the content provider I've created.

public ParcelFileDescriptor openFile(Uri uri, String mode) throws FileNotFoundException {
    String fileLocation = getContext().getCacheDir() + "/" + uri.getLastPathSegment();

    ParcelFileDescriptor pfd = ParcelFileDescriptor.open(new File(fileLocation), ParcelFileDescriptor.MODE_READ_ONLY);
    return pfd;
}
question from:https://stackoverflow.com/questions/12170386/create-and-share-a-file-from-internal-storage

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

It is possible to expose a file stored in your apps private directory via a ContentProvider. Here is some example code I made showing how to create a content provider that can do this.

Manifest

<manifest xmlns:android="http://schemas.android.com/apk/res/android"
  package="com.example.providertest"
  android:versionCode="1"
  android:versionName="1.0">

  <uses-sdk android:minSdkVersion="11" android:targetSdkVersion="15" />

  <application android:label="@string/app_name"
    android:icon="@drawable/ic_launcher"
    android:theme="@style/AppTheme">

    <activity
        android:name=".MainActivity"
        android:label="@string/app_name">
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />
            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>

    <provider
        android:name="MyProvider"
        android:authorities="com.example.prov"
        android:exported="true"
        />        
  </application>
</manifest>

In your ContentProvider override openFile to return the ParcelFileDescriptor

@Override
public ParcelFileDescriptor openFile(Uri uri, String mode) throws FileNotFoundException {       
     File cacheDir = getContext().getCacheDir();
     File privateFile = new File(cacheDir, "file.xml");

     return ParcelFileDescriptor.open(privateFile, ParcelFileDescriptor.MODE_READ_ONLY);
}

Make sure you have copied your xml file to the cache directory

    private void copyFileToInternal() {
    try {
        InputStream is = getAssets().open("file.xml");

        File cacheDir = getCacheDir();
        File outFile = new File(cacheDir, "file.xml");

        OutputStream os = new FileOutputStream(outFile.getAbsolutePath());

        byte[] buff = new byte[1024];
        int len;
        while ((len = is.read(buff)) > 0) {
            os.write(buff, 0, len);
        }
        os.flush();
        os.close();
        is.close();

    } catch (IOException e) {
        e.printStackTrace(); // TODO: should close streams properly here
    }
}

Now any other apps should be able to get an InputStream for your private file by using the content uri (content://com.example.prov/myfile.xml)

For a simple test, call the content provider from a seperate app similar to the following

    private class MyTask extends AsyncTask<String, Integer, String> {

    @Override
    protected String doInBackground(String... params) {

        Uri uri = Uri.parse("content://com.example.prov/myfile.xml");
        InputStream is = null;          
        StringBuilder result = new StringBuilder();
        try {
            is = getApplicationContext().getContentResolver().openInputStream(uri);
            BufferedReader r = new BufferedReader(new InputStreamReader(is));
            String line;
            while ((line = r.readLine()) != null) {
                result.append(line);
            }               
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try { if (is != null) is.close(); } catch (IOException e) { }
        }

        return result.toString();
    }

    @Override
    protected void onPostExecute(String result) {
        Toast.makeText(CallerActivity.this, result, Toast.LENGTH_LONG).show();
        super.onPostExecute(result);
    }
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...