Question

我使用aFileChooser提供的代码来获取应用程序内的共享图像。来自图库的图片工作正常，但如果我在谷歌浏览器中使用图片说并尝试共享它，它会给我一个NPE，因为我的imagePath为空。

String imagePath = getPath(getActivity(), imageUri);

我的uri在此代码中被识别为MediaStore（和）general：

else if ("content".equalsIgnoreCase(uri.getScheme())) {

        if (isGooglePhotosUri(uri))
            return uri.getLastPathSegment();

        return getDataColumn(context, uri, null, null);
    }

但是在getDataColumn()内我的光标转储如下：

08-24 12:00:58.196  13186    13256    ReceivePhotos  D  Cursor is: >>>>> Dumping cursor android.content.ContentResolver$CursorWrapperInner@e110803
08-24 12:00:58.196  13186    13256    ReceivePhotos  D  0 {
08-24 12:00:58.196  13186    13256    ReceivePhotos  D  _data=null
08-24 12:00:58.196  13186    13256    ReceivePhotos  D  }
08-24 12:00:58.196  13186    13256    ReceivePhotos  D  <<<<<
08-24 12:00:58.196  13186    13256    ReceivePhotos  D  Cursor column index is: 0

getDataColumn（）方法：

public static String getDataColumn(Context context, Uri uri, String selection, String[] selectionArgs) {

    Cursor cursor = null;
    final String column = "_data";
    final String[] projection = {
        column
    };

    try {
        cursor = context.getContentResolver().query(uri, projection, selection, selectionArgs,null);
        Log.d("ReceivePhotos", " Cursor is: " + DatabaseUtils.dumpCursorToString(cursor));

        if (cursor != null && cursor.moveToFirst()) {
            final int column_index = cursor.getColumnIndexOrThrow(column);
            Log.d("ReceivePhotos", " Cursor column index is: " + column_index);
            return cursor.getString(column_index);
        }
    } finally {
        if (cursor != null)
            cursor.close();
    }

    return null;
}

ImageUri日志

08-24 12:07:32.780  13629    13696    ReceivePhotos  D  Image uri: content://com.android.chrome.FileProvider/images/screenshot/1472011649310784004280.jpg

电话＆amp;操作系统详细信息

Android 6.0.1上的Sony E5823

Answer 1

您不能也不应该尝试获取与URI对应的基础路径 - 在绝大多数情况下，您的应用程序永远不会访问路径本身，而只能通过URI。

值得庆幸的是，您可以直接从URI获取图像的二进制数据：

InputStream in;
Bitmap bitmap = null;
try {
  in = getContentResolver().openInputStream(imageUri);
  // You could do anything with the InputStream.
  // Here we'll just get the Bitmap at full size
  bitmap = BitmapFactory.decodeStream(in);
} catch (IOException e) {
  // Something went wrong
} finally {
  if (in != null) {
    try {
      in.close();
    } catch (IOException ignored) {}
  }
}
// Now you have a Bitmap.

无法接收共享图片 - Android 6.0

1 个答案: