从图库中获取图像后，只需要在下面的方法中传递URI，仅适用于Android 4.4 (KitKat):

public String getPath(Uri contentUri) {// Will return "image:x*"

    String wholeID = DocumentsContract.getDocumentId(contentUri);

    // Split at colon, use second item in the array
    String id = wholeID.split(":")[1];

    String[] column = { MediaStore.Images.Media.DATA };

    // Where id is equal to
    String sel = MediaStore.Images.Media._ID + "=?";

    Cursor cursor = getContentResolver().query(
            MediaStore.Images.Media.EXTERNAL_CONTENT_URI, column, sel,
            new String[] { id }, null);

    String filePath = "";

    int columnIndex = cursor.getColumnIndex(column[0]);

    if (cursor.moveToFirst()) {
        filePath = cursor.getString(columnIndex);
    }

    cursor.close();
    return filePath;
}

只是对第一个答案进行了简单的更新:mActivity.managedQuery()现在已弃用。我已经用新方法更新了代码。

private String getRealPathFromURI(Uri contentUri) {
    String[] proj = { MediaStore.Images.Media.DATA };
    CursorLoader loader = new CursorLoader(mContext, contentUri, proj, null, null, null);
    Cursor cursor = loader.loadInBackground();
    int column_index = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
    cursor.moveToFirst();
    String result = cursor.getString(column_index);
    cursor.close();
    return result;
}

Android开发源码

现有的很好的答案，其中一些是我自己想出来的:

我必须从URI中获取路径，并从路径中获取URI，谷歌很难区分，因此对于任何有相同问题的人(例如，从MediaStore中获取视频的缩略图，其物理位置您已经拥有)。前:

/**
 * Gets the corresponding path to a file from the given content:// URI
 * @param selectedVideoUri The content:// URI to find the file path from
 * @param contentResolver The content resolver to use to perform the query.
 * @return the file path as a string
 */
private String getFilePathFromContentUri(Uri selectedVideoUri,
        ContentResolver contentResolver) {
    String filePath;
    String[] filePathColumn = {MediaColumns.DATA};

    Cursor cursor = contentResolver.query(selectedVideoUri, filePathColumn, null, null, null);
    cursor.moveToFirst();

    int columnIndex = cursor.getColumnIndex(filePathColumn[0]);
    filePath = cursor.getString(columnIndex);
    cursor.close();
    return filePath;
}

后者(我用于视频，但也可以用于音频或文件或其他类型的存储内容，通过替换MediaStore。音频(等)MediaStore.Video):

/**
 * Gets the MediaStore video ID of a given file on external storage
 * @param filePath The path (on external storage) of the file to resolve the ID of
 * @param contentResolver The content resolver to use to perform the query.
 * @return the video ID as a long
 */
private long getVideoIdFromFilePath(String filePath,
        ContentResolver contentResolver) {


    long videoId;
    Log.d(TAG,"Loading file " + filePath);

            // This returns us content://media/external/videos/media (or something like that)
            // I pass in "external" because that's the MediaStore's name for the external
            // storage on my device (the other possibility is "internal")
    Uri videosUri = MediaStore.Video.Media.getContentUri("external");

    Log.d(TAG,"videosUri = " + videosUri.toString());

    String[] projection = {MediaStore.Video.VideoColumns._ID};

    // TODO This will break if we have no matching item in the MediaStore.
    Cursor cursor = contentResolver.query(videosUri, projection, MediaStore.Video.VideoColumns.DATA + " LIKE ?", new String[] { filePath }, null);
    cursor.moveToFirst();

    int columnIndex = cursor.getColumnIndex(projection[0]);
    videoId = cursor.getLong(columnIndex);

    Log.d(TAG,"Video ID is " + videoId);
    cursor.close();
    return videoId;
}

基本上，MediaStore的DATA列(或正在查询的任何子部分)存储文件路径，因此您可以使用已知的内容查找DATA字段，也可以使用该字段查找所需的任何其他内容。

然后，我进一步使用上述方案来确定如何处理我的数据:

 private boolean  getSelectedVideo(Intent imageReturnedIntent, boolean fromData) {

    Uri selectedVideoUri;

    //Selected image returned from another activity
            // A parameter I pass myself to know whether or not I'm being "shared via" or
            // whether I'm working internally to my app (fromData = working internally)
    if(fromData){
        selectedVideoUri = imageReturnedIntent.getData();
    } else {
        //Selected image returned from SEND intent 
                    // which I register to receive in my manifest
                    // (so people can "share via" my app)
        selectedVideoUri = (Uri)getIntent().getExtras().get(Intent.EXTRA_STREAM);
    }

    Log.d(TAG,"SelectedVideoUri = " + selectedVideoUri);

    String filePath;

    String scheme = selectedVideoUri.getScheme(); 
    ContentResolver contentResolver = getContentResolver();
    long videoId;

    // If we are sent file://something or content://org.openintents.filemanager/mimetype/something...
    if(scheme.equals("file") || (scheme.equals("content") && selectedVideoUri.getEncodedAuthority().equals("org.openintents.filemanager"))){

        // Get the path
        filePath = selectedVideoUri.getPath();

        // Trim the path if necessary
        // openintents filemanager returns content://org.openintents.filemanager/mimetype//mnt/sdcard/xxxx.mp4
        if(filePath.startsWith("/mimetype/")){
            String trimmedFilePath = filePath.substring("/mimetype/".length());
            filePath = trimmedFilePath.substring(trimmedFilePath.indexOf("/"));
        }

        // Get the video ID from the path
        videoId = getVideoIdFromFilePath(filePath, contentResolver);

    } else if(scheme.equals("content")){

        // If we are given another content:// URI, look it up in the media provider
        videoId = Long.valueOf(selectedVideoUri.getLastPathSegment());
        filePath = getFilePathFromContentUri(selectedVideoUri, contentResolver);

    } else {
        Log.d(TAG,"Failed to load URI " + selectedVideoUri.toString());
        return false;
    }

     return true;
 }

完美地为我修复了这篇文章的代码:

  public static String getRealPathImageFromUri(Uri uri) {
        String fileName =null;
        if (uri.getScheme().equals("content")) {
            try (Cursor cursor = MyApplication.getInstance().getContentResolver().query(uri, null, null, null, null)) {
                if (cursor.moveToFirst()) {
                    fileName = cursor.getString(cursor.getColumnIndexOrThrow(ediaStore.Images.Media.DATA));
                }
            } catch (IllegalArgumentException e) {
                Log.e(mTag, "Get path failed", e);
            }
        }
        return fileName;
    }

这些答案在所有情况下都不适用。我必须直接去谷歌的文档https://developer.android.com/guide/topics/providers/document-provider.html关于这个主题，并发现这个有用的方法:

private Bitmap getBitmapFromUri(Uri uri) throws IOException {
    ParcelFileDescriptor parcelFileDescriptor =
    getContentResolver().openFileDescriptor(uri, "r");
    FileDescriptor fileDescriptor = parcelFileDescriptor.getFileDescriptor();
    Bitmap image = BitmapFactory.decodeFileDescriptor(fileDescriptor);
    parcelFileDescriptor.close();
    return image;
}

您可以使用此位图在图像视图中显示它。

简单易行。您可以像下面这样从URI执行此操作!

public void getContents(Uri uri)
{
    Cursor vidCursor = getActivity.getContentResolver().query(uri, null, null,
                                                              null, null);
    if (vidCursor.moveToFirst())
    {
        int column_index =
        vidCursor .getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
        Uri filePathUri = Uri.parse(vidCursor .getString(column_index));
        String video_name =  filePathUri.getLastPathSegment().toString();
        String file_path=filePathUri.getPath();
        Log.i("TAG", video_name + "\b" file_path);
    }
}