API 19及以上，图像文件路径从Uri工作完美。我还检查了最新的PIE API 28。

public String getImageFilePath(Uri uri) {
    String path = null, image_id = null;

    Cursor cursor = getContentResolver().query(uri, null, null, null, null);
    if (cursor != null) {
        cursor.moveToFirst();
        image_id = cursor.getString(0);
        image_id = image_id.substring(image_id.lastIndexOf(":") + 1);
        cursor.close();
    }

    cursor = getContentResolver().query(android.provider.MediaStore.Images.Media.EXTERNAL_CONTENT_URI, null, MediaStore.Images.Media._ID + " = ? ", new String[]{image_id}, null);
    if (cursor!=null) {
        cursor.moveToFirst();
        path = cursor.getString(cursor.getColumnIndex(MediaStore.Images.Media.DATA));
        cursor.close();
    }
    return path;
}

由于managedQuery已弃用，您可以尝试:

CursorLoader cursorLoader = new CursorLoader(context, uri, proj, null, null, null);
Cursor cursor = cursorLoader.loadInBackground();

@PercyPercy的轻微修改版本-它不会抛出，如果有任何错误，它只返回null:

public String getPathFromMediaUri(Context context, Uri uri) {
    String result = null;

    String[] projection = { MediaStore.Images.Media.DATA };
    Cursor cursor = context.getContentResolver().query(uri, projection, null, null, null);
    int col = cursor.getColumnIndex(MediaStore.Images.Media.DATA);
    if (col >= 0 && cursor.moveToFirst())
        result = cursor.getString(col);
    cursor.close();

    return result;
}

在这里，我将向您展示如何创建一个BROWSE按钮，当您单击它时，它将打开SD卡，您将选择一个文件，结果您将获得所选文件的文件名和文件路径:

一个你要按的按钮

browse.setOnClickListener(new OnClickListener()
{
    public void onClick(View v)
    {
        Intent intent = new Intent();
        intent.setAction(Intent.ACTION_PICK);
        Uri startDir = Uri.fromFile(new File("/sdcard"));
        startActivityForResult(intent, PICK_REQUEST_CODE);
    }
});

获取结果文件名和文件路径的函数

protected void onActivityResult(int requestCode, int resultCode, Intent intent)
{
    if (requestCode == PICK_REQUEST_CODE)
    {
        if (resultCode == RESULT_OK)
        {
            Uri uri = intent.getData();

            if (uri.getScheme().toString().compareTo("content")==0)
            {
                Cursor cursor =getContentResolver().query(uri, null, null, null, null);
                if (cursor.moveToFirst())
                {
                    int column_index = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);//Instead of "MediaStore.Images.Media.DATA" can be used "_data"
                    Uri filePathUri = Uri.parse(cursor.getString(column_index));
                    String file_name = filePathUri.getLastPathSegment().toString();
                    String file_path=filePathUri.getPath();
                    Toast.makeText(this,"File Name & PATH are:"+file_name+"\n"+file_path, Toast.LENGTH_LONG).show();
                }
            }
        }
    }
}

既然上面的答案对我不起作用，下面是对我有效的解决方案:

对于>19和<=19 API级别。

这个方法涵盖了从uri中获取filePath的所有情况

/**
 * Get a file path from a Uri. This will get the the path for Storage Access
 * Framework Documents, as well as the _data field for the MediaStore and
 * other file-based ContentProviders.
 *
 * @param context The activity.
 * @param uri The Uri to query.
 * @author paulburke
 */
public static String getPath(final Context context, final Uri uri) {

    // DocumentProvider
    if ( Build.VERSION.SDK_INT >= Build.VERSION_CODES.KITKAT && DocumentsContract.isDocumentUri(context, uri)) {
        // ExternalStorageProvider
        if (isExternalStorageDocument(uri)) {
            final String docId = DocumentsContract.getDocumentId(uri);
            final String[] split = docId.split(":");
            final String type = split[0];

            if ("primary".equalsIgnoreCase(type)) {
                return Environment.getExternalStorageDirectory() + "/" + split[1];
            }else{
                Toast.makeText(context, "Could not get file path. Please try again", Toast.LENGTH_SHORT).show();
            }
        }
        // DownloadsProvider
        else if (isDownloadsDocument(uri)) {

            final String id = DocumentsContract.getDocumentId(uri);
            final Uri contentUri = ContentUris.withAppendedId(
                    Uri.parse("content://downloads/public_downloads"), Long.valueOf(id));

            return getDataColumn(context, contentUri, null, null);
        }
        // MediaProvider
        else if (isMediaDocument(uri)) {
            final String docId = DocumentsContract.getDocumentId(uri);
            final String[] split = docId.split(":");
            final String type = split[0];

            Uri contentUri = null;
            if ("image".equals(type)) {
                contentUri = MediaStore.Images.Media.EXTERNAL_CONTENT_URI;
            } else if ("video".equals(type)) {
                contentUri = MediaStore.Video.Media.EXTERNAL_CONTENT_URI;
            } else if ("audio".equals(type)) {
                contentUri = MediaStore.Audio.Media.EXTERNAL_CONTENT_URI;
            } else {
                contentUri = MediaStore.Files.getContentUri("external");
            }

            final String selection = "_id=?";
            final String[] selectionArgs = new String[] {
                    split[1]
            };

            return getDataColumn(context, contentUri, selection, selectionArgs);
        }
    }
    // MediaStore (and general)
    else if ("content".equalsIgnoreCase(uri.getScheme())) {
        return getDataColumn(context, uri, null, null);
    }
    // File
    else if ("file".equalsIgnoreCase(uri.getScheme())) {
        return uri.getPath();
    }

    return null;
}

这个解决方案适用于所有情况:

在某些情况下，从URL中获取路径太难了。那你为什么需要路径?把文件复制到其他地方?你不需要路径。

public void SavePhotoUri (Uri imageuri, String Filename){

    File FilePath = context.getDir(Environment.DIRECTORY_PICTURES,Context.MODE_PRIVATE);
    try {
        Bitmap selectedImage = MediaStore.Images.Media.getBitmap(context.getContentResolver(), imageuri);
        String destinationImagePath = FilePath + "/" + Filename;
        FileOutputStream destination = new FileOutputStream(destinationImagePath);
        selectedImage.compress(Bitmap.CompressFormat.JPEG, 100, destination);
        destination.close();
    }
    catch (Exception e) {
        Log.e("error", e.toString());
    }
}