在奇巧(或新画廊)之前,意图。ACTION_GET_CONTENT返回一个这样的URI

内容:/ /媒体/外部/图片/媒体/ 3951。

使用ContentResolver并查询 media . data返回文件URL。

然而,在奇巧,画廊返回一个URI(通过“Last”)像这样:

内容:/ / com.android.providers.media.documents /文档/图片:3951

我该怎么处理呢?


当前回答

这个Android库处理奇巧的情况变化(包括旧版本- 2.1+): https://github.com/iPaulPro/aFileChooser

使用String path = FileUtils。getPath(context, uri)将返回的uri转换为可用于所有OS版本的路径字符串。 更多信息请访问:https://stackoverflow.com/a/20559175/860488

其他回答

请尽量避免使用takePersistableUriPermission方法,因为它会为我引发运行时异常。 /** 从图库中选择。 * /

public void selectFromGallery() {
    if (Build.VERSION.SDK_INT < AppConstants.KITKAT_API_VERSION) {

        Intent intent = new Intent(); 
        intent.setType("image/*");
        intent.setAction(Intent.ACTION_GET_CONTENT);
        ((Activity)mCalledContext).startActivityForResult(intent,AppConstants.GALLERY_INTENT_CALLED);

    } else {

        Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
        intent.setType("image/*");
        ((Activity)mCalledContext).startActivityForResult(intent, AppConstants.GALLERY_AFTER_KITKAT_INTENT_CALLED);
    }
}

OnActivity的结果处理图像数据:

@Override onActivityResult(int requestCode, int resultCode, Intent data) {

    //gallery intent result handling before kit-kat version
    if(requestCode==AppConstants.GALLERY_INTENT_CALLED 
            && resultCode == RESULT_OK) {

        Uri selectedImage = data.getData();
        String[] filePathColumn = {MediaStore.Images.Media.DATA};
        Cursor cursor = getContentResolver().query(selectedImage,filePathColumn, null, null, null);
        cursor.moveToFirst();
        int columnIndex = cursor.getColumnIndex(filePathColumn[0]);
        String filePath = cursor.getString(columnIndex);
        cursor.close();
        photoFile = new File(filePath);
        mImgCropping.startCropImage(photoFile,AppConstants.REQUEST_IMAGE_CROP);

    }
    //gallery intent result handling after kit-kat version
    else if (requestCode == AppConstants.GALLERY_AFTER_KITKAT_INTENT_CALLED 
            && resultCode == RESULT_OK) {

        Uri selectedImage = data.getData();
        InputStream input = null;
        OutputStream output = null;

        try {
            //converting the input stream into file to crop the 
            //selected image from sd-card.
            input = getApplicationContext().getContentResolver().openInputStream(selectedImage);
            try {
                photoFile = mImgCropping.createImageFile();
            } catch (IOException e) {
                e.printStackTrace();
            }catch(Exception e) {
                e.printStackTrace();
            }
            output = new FileOutputStream(photoFile);

            int read = 0;
            byte[] bytes = new byte[1024];

            while ((read = input.read(bytes)) != -1) {
                try {
                    output.write(bytes, 0, read);
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } catch (Exception e) {
            e.printStackTrace();
        }


    }

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());
    }
}

这是一个完全的hack,但这是我所做的…

因此,在设置DocumentsProvider时,我注意到样例代码(在getDocIdForFile中,大约在第450行)根据文件相对于指定根的路径(即在第96行中设置mBaseDir的路径)为所选文档生成了唯一的id。

所以URI最终看起来像这样:

内容:/ / com.example.provider /文档/根:路径/ / /文件

正如文档所说,它假设只有一个根(在我的情况下是Environment.getExternalStorageDirectory(),但你可以在其他地方使用…然后它获取文件路径,从根目录开始,并使其成为唯一的ID,前置“root:”。所以我可以通过消除uri.getPath()中的“/document/root:”部分来确定路径,通过这样做来创建一个实际的文件路径:

public void onActivityResult(int requestCode, int resultCode, Intent data) {
// check resultcodes and such, then...
uri = data.getData();
if (uri.getAuthority().equals("com.example.provider"))  {
    String path = Environment.getExternalStorageDirectory(0.toString()
                 .concat("/")
                 .concat(uri.getPath().substring("/document/root:".length())));
    doSomethingWithThePath(path); }
else {
    // another provider (maybe a cloud-based service such as GDrive)
    // created this uri.  So handle it, or don't.  You can allow specific
    // local filesystem providers, filter non-filesystem path results, etc.
}

我知道。这很可耻,但它奏效了。同样,这依赖于您在应用程序中使用自己的文档提供程序来生成文档ID。

(此外,还有一种更好的方法来构建路径,不假设“/”是路径分隔符,等等。但你懂的。)

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();
    }

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