这对我来说很有效:

else if(requestCode == GALLERY_ACTIVITY_NEW && resultCode == Activity.RESULT_OK)
{
    Uri uri = data.getData();
    Log.i(TAG, "old uri =  " + uri);
    dumpImageMetaData(uri);

    try {
        ParcelFileDescriptor parcelFileDescriptor =
                getContentResolver().openFileDescriptor(uri, "r");
        FileDescriptor fileDescriptor = parcelFileDescriptor.getFileDescriptor();
        Log.i(TAG, "File descriptor " + fileDescriptor.toString());

        final BitmapFactory.Options options = new BitmapFactory.Options();
        options.inJustDecodeBounds = true;
        BitmapFactory.decodeFileDescriptor(fileDescriptor, null, options);

        options.inSampleSize =
           BitmapHelper.calculateInSampleSize(options,
                                              User.PICTURE_MAX_WIDTH_IN_PIXELS,
                                              User.PICTURE_MAX_HEIGHT_IN_PIXELS);
        options.inJustDecodeBounds = false;

        Bitmap bitmap = BitmapFactory.decodeFileDescriptor(fileDescriptor, null, options);
        imageViewPic.setImageBitmap(bitmap);

        ByteArrayOutputStream stream = new ByteArrayOutputStream();
        bitmap.compress(Bitmap.CompressFormat.JPEG, 100, stream);
        // get byte array here
        byte[] picData = stream.toByteArray();
        ParseFile picFile = new ParseFile(picData);
        user.setProfilePicture(picFile);
    }
    catch(FileNotFoundException exc)
    {
        Log.i(TAG, "File not found: " + exc.toString());
    }
}

这对我来说很有效:

else if(requestCode == GALLERY_ACTIVITY_NEW && resultCode == Activity.RESULT_OK)
{
    Uri uri = data.getData();
    Log.i(TAG, "old uri =  " + uri);
    dumpImageMetaData(uri);

    try {
        ParcelFileDescriptor parcelFileDescriptor =
                getContentResolver().openFileDescriptor(uri, "r");
        FileDescriptor fileDescriptor = parcelFileDescriptor.getFileDescriptor();
        Log.i(TAG, "File descriptor " + fileDescriptor.toString());

        final BitmapFactory.Options options = new BitmapFactory.Options();
        options.inJustDecodeBounds = true;
        BitmapFactory.decodeFileDescriptor(fileDescriptor, null, options);

        options.inSampleSize =
           BitmapHelper.calculateInSampleSize(options,
                                              User.PICTURE_MAX_WIDTH_IN_PIXELS,
                                              User.PICTURE_MAX_HEIGHT_IN_PIXELS);
        options.inJustDecodeBounds = false;

        Bitmap bitmap = BitmapFactory.decodeFileDescriptor(fileDescriptor, null, options);
        imageViewPic.setImageBitmap(bitmap);

        ByteArrayOutputStream stream = new ByteArrayOutputStream();
        bitmap.compress(Bitmap.CompressFormat.JPEG, 100, stream);
        // get byte array here
        byte[] picData = stream.toByteArray();
        ParseFile picFile = new ParseFile(picData);
        user.setProfilePicture(picFile);
    }
    catch(FileNotFoundException exc)
    {
        Log.i(TAG, "File not found: " + exc.toString());
    }
}

问题

如何从URI中获得实际的文件路径

回答

据我所知，我们不需要从URI获取文件路径，因为对于大多数情况，我们可以直接使用URI来完成我们的工作(如1。获取位图2。向服务器发送文件，等等)

1. 发送到服务器

我们可以直接使用URI将文件发送到服务器。

使用URI，我们可以获得InputStream，我们可以使用MultiPartEntity直接将其发送到服务器。

例子

/**
 * Used to form Multi Entity for a URI (URI pointing to some file, which we got from other application).
 *
 * @param uri     URI.
 * @param context Context.
 * @return Multi Part Entity.
 */
public MultipartEntity formMultiPartEntityForUri(final Uri uri, final Context context) {
    MultipartEntity multipartEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE, null, Charset.forName("UTF-8"));
    try {
        InputStream inputStream = mContext.getContentResolver().openInputStream(uri);
        if (inputStream != null) {
            ContentBody contentBody = new InputStreamBody(inputStream, getFileNameFromUri(uri, context));
            multipartEntity.addPart("[YOUR_KEY]", contentBody);
        }
    }
    catch (Exception exp) {
        Log.e("TAG", exp.getMessage());
    }
    return multipartEntity;
}

/**
 * Used to get a file name from a URI.
 *
 * @param uri     URI.
 * @param context Context.
 * @return File name from URI.
 */
public String getFileNameFromUri(final Uri uri, final Context context) {

    String fileName = null;
    if (uri != null) {
        // Get file name.
        // File Scheme.
        if (ContentResolver.SCHEME_FILE.equals(uri.getScheme())) {
            File file = new File(uri.getPath());
            fileName = file.getName();
        }
        // Content Scheme.
        else if (ContentResolver.SCHEME_CONTENT.equals(uri.getScheme())) {
            Cursor returnCursor =
                    context.getContentResolver().query(uri, null, null, null, null);
            if (returnCursor != null && returnCursor.moveToFirst()) {
                int nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
                fileName = returnCursor.getString(nameIndex);
                returnCursor.close();
            }
        }
    }
    return fileName;
}

2. 从URI中获取位图

如果URI指向图像，我们将得到位图，否则为null:

/**
 * Used to create bitmap for the given URI.
 * <p>
 * 1. Convert the given URI to bitmap.
 * 2. Calculate ratio (depending on bitmap size) on how much we need to subSample the original bitmap.
 * 3. Create bitmap bitmap depending on the ration from URI.
 * 4. Reference - http://stackoverflow.com/questions/3879992/how-to-get-bitmap-from-an-uri
 *
 * @param context       Context.
 * @param uri           URI to the file.
 * @param bitmapSize Bitmap size required in PX.
 * @return Bitmap bitmap created for the given URI.
 * @throws IOException
 */
public static Bitmap createBitmapFromUri(final Context context, Uri uri, final int bitmapSize) throws IOException {

    // 1. Convert the given URI to bitmap.
    InputStream input = context.getContentResolver().openInputStream(uri);
    BitmapFactory.Options onlyBoundsOptions = new BitmapFactory.Options();
    onlyBoundsOptions.inJustDecodeBounds = true;
    onlyBoundsOptions.inDither = true;//optional
    onlyBoundsOptions.inPreferredConfig = Bitmap.Config.ARGB_8888;//optional
    BitmapFactory.decodeStream(input, null, onlyBoundsOptions);
    input.close();
    if ((onlyBoundsOptions.outWidth == -1) || (onlyBoundsOptions.outHeight == -1)) {
        return null;
    }

    // 2. Calculate ratio.
    int originalSize = (onlyBoundsOptions.outHeight > onlyBoundsOptions.outWidth) ? onlyBoundsOptions.outHeight : onlyBoundsOptions.outWidth;
    double ratio = (originalSize > bitmapSize) ? (originalSize / bitmapSize) : 1.0;

    // 3. Create bitmap.
    BitmapFactory.Options bitmapOptions = new BitmapFactory.Options();
    bitmapOptions.inSampleSize = getPowerOfTwoForSampleRatio(ratio);
    bitmapOptions.inDither = true;//optional
    bitmapOptions.inPreferredConfig = Bitmap.Config.ARGB_8888;//optional
    input = context.getContentResolver().openInputStream(uri);
    Bitmap bitmap = BitmapFactory.decodeStream(input, null, bitmapOptions);
    input.close();

    return bitmap;
}

/**
 * For Bitmap option inSampleSize - We need to give value in power of two.
 *
 * @param ratio Ratio to be rounded of to power of two.
 * @return Ratio rounded of to nearest power of two.
 */
private static int getPowerOfTwoForSampleRatio(final double ratio) {
    int k = Integer.highestOneBit((int) Math.floor(ratio));
    if (k == 0) return 1;
    else return k;
}

评论

Android没有提供任何从URI获取文件路径的方法，在上面的大多数答案中，我们都硬编码了一些常量，这可能会在功能发布中中断(对不起，我可能错了)。 在直接从URI获取文件路径的解决方案之前，尝试是否可以用URI和Android默认方法解决您的用例。

参考

https://developer.android.com/guide/topics/providers/content-provider-basics.html https://developer.android.com/reference/android/content/ContentResolver.html https://hc.apache.org/httpcomponents-client-ga/httpmime/apidocs/org/apache/http/entity/mime/content/InputStreamBody.html

对于这种类型的uri 内容:/ / % 3 a19298 com.android.providers.media.documents /文件/文档 或者uri.getAuthority()是这些中的任何一个

"com.google.android.apps.docs.storage".equals(uri.getAuthority()) || "com.google.android.apps.docs.storage.legacy".equals(uri.getAuthority());

使用这个函数

private static String getDriveFilePath(Uri uri, Context context) {
        Uri returnUri = uri;
        Cursor returnCursor = context.getContentResolver().query(returnUri, null, null, null, null);

        int nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
        int sizeIndex = returnCursor.getColumnIndex(OpenableColumns.SIZE);
        returnCursor.moveToFirst();
        String name = (returnCursor.getString(nameIndex));
        String size = (Long.toString(returnCursor.getLong(sizeIndex)));
        File file = new File(context.getCacheDir(), name);
        try {
            InputStream inputStream = context.getContentResolver().openInputStream(uri);
            FileOutputStream outputStream = new FileOutputStream(file);
            int read = 0;
            int maxBufferSize = 1 * 1024 * 1024;
            int bytesAvailable = inputStream.available();

            //int bufferSize = 1024;
            int bufferSize = Math.min(bytesAvailable, maxBufferSize);

            final byte[] buffers = new byte[bufferSize];
            while ((read = inputStream.read(buffers)) != -1) {
                outputStream.write(buffers, 0, read);
            }
            Log.e("File Size", "Size " + file.length());
            inputStream.close();
            outputStream.close();
            Log.e("File Path", "Path " + file.getPath());
            Log.e("File Size", "Size " + file.length());
        } catch (Exception e) {
            Log.e("Exception", e.getMessage());
        }
        return file.getPath();
    }

This is what I do: Uri selectedImageURI = data.getData(); imageFile = new File(getRealPathFromURI(selectedImageURI)); private String getRealPathFromURI(Uri contentURI) { Cursor cursor = getContentResolver().query(contentURI, null, null, null, null); if (cursor == null) { // Source is Dropbox or other similar local file path return contentURI.getPath(); } else { cursor.moveToFirst(); int idx = cursor.getColumnIndex(MediaStore.Images.ImageColumns.DATA); return cursor.getString(idx); } } NOTE: managedQuery() method is deprecated, so I am not using it.

这个答案是来自m3n0R的问题安卓得到真正的路径Uri.getPath()和我声称没有信用。我只是想那些还没有解决这个问题的人可以使用这个。

我在这里尝试了几个答案，我认为我有一个解决方案，每次都可以工作，并且还可以管理权限。

这是基于LEO的巧妙解决方案。这篇文章应该包含所有的代码，你需要使这个工作，它应该工作在任何手机和Android版本;)

为了能够从SD卡中选择文件，你需要在你的清单中这样做:

<uses-permission android:name="android.permission.READ_EXTERNAL_STORAGE" />

常量:

private static final int PICK_IMAGE = 456; // Whatever number you like
public static final int MY_PERMISSIONS_REQUEST_READ_EXTERNAL = 28528; // Whatever number you like
public static final String FILE_TEMP_NAME = "temp_image"; // Whatever file name you like

检查权限和launchImagePick如果可能的话

if (ContextCompat.checkSelfPermission(getThis(),
        Manifest.permission.READ_EXTERNAL_STORAGE)
        != PackageManager.PERMISSION_GRANTED) {

    ActivityCompat.requestPermissions(getThis(),
            new String[]{Manifest.permission.READ_EXTERNAL_STORAGE},
            MY_PERMISSIONS_REQUEST_READ_EXTERNAL);
}
else {
    launchImagePick();
}

允许响应

@Override
public void onRequestPermissionsResult(int requestCode,
                                       @NonNull
                                         String permissions[],
                                       @NonNull
                                         int[] grantResults) {

    if (manageReadExternalPermissionResponse(this, requestCode, grantResults)) {
        launchImagePick();
    }
}

管理权限响应

public static boolean manageReadExternalPermissionResponse(final Activity activity, int requestCode, int[] grantResults) {

    if (requestCode == MY_PERMISSIONS_REQUEST_READ_EXTERNAL) {

        // If request is cancelled, the result arrays are empty.

        if (grantResults.length > 0
                && grantResults[0] == PackageManager.PERMISSION_GRANTED) {

            // Permission was granted, yay! Do the
            // contacts-related task you need to do.
            return true;

        } else if (grantResults.length > 0 && grantResults[0] == PackageManager.PERMISSION_DENIED) {

            boolean showRationale = ActivityCompat.shouldShowRequestPermissionRationale(activity,
                    Manifest.permission.READ_EXTERNAL_STORAGE);

            if (!showRationale) {
                // The user also CHECKED "never ask again".
                // You can either enable some fall back,
                // disable features of your app
                // or open another dialog explaining
                // again the permission and directing to
                // the app setting.

            } else {
                // The user did NOT check "never ask again".
                // This is a good place to explain the user
                // why you need the permission and ask if he/she wants
                // to accept it (the rationale).
            }
        } else {
            // Permission denied, boo! Disable the
            // functionality that depends on this permission.
        }
    }
    return false;
}

发射图像选择

private void launchImagePick() {

    Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
    intent.setType("image/*");
    intent.addCategory(Intent.CATEGORY_OPENABLE);
    startActivityForResult(intent, PICK_IMAGE);

    // see onActivityResult
}

管理图像选择响应

@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);
    if (requestCode == PICK_IMAGE) {

        if (resultCode == Activity.RESULT_OK) {
            if (data != null && data.getData() != null) {

                try {
                     InputStream inputStream = getContentResolver().openInputStream(data.getData())
                     if (inputStream != null) {

                        // No special persmission needed to store the file like that
                        FileOutputStream fos = openFileOutput(FILE_TEMP_NAME, Context.MODE_PRIVATE);

                        final int BUFFER_SIZE = 1 << 10 << 3; // 8 KiB buffer
                        byte[] buffer = new byte[BUFFER_SIZE];
                        int bytesRead = -1;
                        while ((bytesRead = inputStream.read(buffer)) > -1) {
                            fos.write(buffer, 0, bytesRead);
                        }
                        inputStream.close();
                        fos.close();

                        File tempImageFile = new File(getFilesDir()+"/"+FILE_TEMP_NAME);

                        // Do whatever you want with the File

                        // Delete when not needed anymore
                        deleteFile(FILE_TEMP_NAME);
                    }
                }
                catch (Exception e) {
                    e.printStackTrace();
                }
            } else {
                // Error display
            }
        } else {
            // The user did not select any image
        }
    }
}

这就是所有的人;这在我所有的电话上都适用。