Intent chooseFile;
Intent intent;
chooseFile = new Intent(Intent.ACTION_GET_CONTENT);
chooseFile.setType("*/*");
intent = Intent.createChooser(chooseFile,"Choose a file");
startActivityForResult(intent,PICKFILE_RESULT_CODE);

然后在我的onActivityResult()

switch(requestCode){
 case PICKFILE_RESULT_CODE:
   if(resultCode==-1){
      Uri uri = data.getData();
      String filePath = uri.getPath();
      Toast.makeText(getActivity(),filePath,Toast.LENGTH_LONG).show();
    }
 break;
}

这是打开一个文件选择器,但它不是我想要的.例如,我想选择一个文件(.txt),然后获取该文件,然后使用它.有了这个代码,我以为我会得到完整的路径,但不会发生;例如我得到：/ document / 5318 /.但是用这个路径我无法获取文件.我创建了一个名为PathToFile()的方法返回一个文件：

private File PathToFile(String path) {
    File tempFileToUpload;
    tempFileToUpload = new File(path);
    return tempFileToUpload;
}

我想要做的是让用户从任何地方选择一个文件,意思是DropBox,Drive,SDCard,Mega等…我没有找到正确的方法,我试图让路径然后得到一个文件由这个路径…但它不工作,所以我认为最好是获取文件本身,然后以该文件编程方式我复制此或删除.

编辑(当前代码)

我的意图

Intent chooseFile = new Intent(Intent.ACTION_GET_CONTENT);
 chooseFile.addCategory(Intent.CATEGORY_OPENABLE);
 chooseFile.setType("text/plain");
 startActivityForResult(
      Intent.createChooser(chooseFile,"Choose a file"),PICKFILE_RESULT_CODE
 );

在那里我有一个问题,因为我不知道什么是文本/简单的支持,但我会调查它,但这并不重要.

在我的onActivityResult()我使用与@Lukas Knuth answer相同,但我不知道是否可以复制该文件到我的SD卡的另一部分我正在等待他的答案.

@Override
public void onActivityResult(int requestCode,int resultCode,Intent data) {
    super.onActivityResult(requestCode,resultCode,data);
    if (requestCode == PICKFILE_RESULT_CODE && resultCode == Activity.RESULT_OK){
        Uri content_describer = data.getData();
        //get the path 
        Log.d("Path???",content_describer.getPath());
        BufferedReader reader = null;
        try {
            // open the user-picked file for reading:
            InputStream in = getActivity().getContentResolver().openInputStream(content_describer);
            // now read the content:
            reader = new BufferedReader(new InputStreamReader(in));
            String line;
            StringBuilder builder = new StringBuilder();

            while ((line = reader.readLine()) != null){
                builder.append(line);
            }
            // Do something with the content in
            text.setText(builder.toString());



        } catch (FileNotFoundException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            if (reader != null) {
                try {
                    reader.close();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }
    }
}

getPath()from @ Y.S.

我在做这个：

String[] projection = { MediaStore.Files.FileColumns.DATA };
            Cursor cursor = getActivity().getContentResolver().query(content_describer,projection,null,null);
int column_index = cursor.getColumnIndexOrThrow(projection[0]);
cursor.moveToFirst();
cursor.close();
Log.d( "PATH-->",cursor.getString(column_index));

得到一个NullPointerException：

java.lang.RuntimeException: Failure delivering result ResultInfo{who=null,request=131073,result=-1,data=Intent { dat=file:///path typ=text/plain flg=0x3 }} to activity {info.androidhive.tabsswipe/info.androidhive.tabsswipe.MainActivity2}: java.lang.NullPointerException

编辑工作感谢@Y.S.,@Lukas Knuth和@CommonsWare编辑.

这是我只接受文件/简单的意图.

Intent chooseFile = new Intent(Intent.ACTION_GET_CONTENT);
chooseFile.addCategory(Intent.CATEGORY_OPENABLE);
chooseFile.setType("text/plain");
startActivityForResult(
    Intent.createChooser(chooseFile,PICKFILE_RESULT_CODE
);

在我的onActivityResult()我创建一个URI,我获取Intent的数据,我创建一个文件,我保存绝对路径做content_describer.getPath();然后我保留路径的名称在TextView with content_describer.getLastPathSegment(); (这真的很棒@Y.S.不知道那个功能),我创建了一个第二个File,我称之为目标,我发送AbsolutePath可以创建这个文件.

@Override
public void onActivityResult(int requestCode,data);
    if (requestCode == PICKFILE_RESULT_CODE && resultCode == Activity.RESULT_OK){
        Uri content_describer = data.getData();
        String src = content_describer.getPath();
        source = new File(src);
        Log.d("src is ",source.toString());
        String filename = content_describer.getLastPathSegment();
        text.setText(filename);
        Log.d("FileName is ",filename);
        destination = new File(Environment.getExternalStorageDirectory().getAbsolutePath() + "/Test/TestTest/" + filename);
        Log.d("Destination is ",destination.toString());
        SetToFolder.setEnabled(true);
    }
}

此外,我创建了一个函数,您必须发送源文件和我们之前创建的目标文件将其复制到新文件夹.

private void copy(File source,File destination) throws IOException {

    FileChannel in = new FileInputStream(source).getChannel();
    FileChannel out = new FileOutputStream(destination).getChannel();

    try {
        in.transferTo(0,in.size(),out);
    } catch(Exception e){
        Log.d("Exception",e.toString());
    } finally {
        if (in != null)
            in.close();
        if (out != null)
            out.close();
    }
}

此外,对我说如果这个文件夹存在或不存在(我必须发送目标文件,如果它不存在我创建这个文件夹,如果我不做任何事情.

private void DirectoryExist (File destination) {

    if(!destination.isDirectory()) {
        if(destination.mkdirs()){
            Log.d("Carpeta creada","....");
        }else{
            Log.d("Carpeta no creada","....");
        }
    }

再次感谢您的帮助,希望您喜欢与大家一起制作的代码:)