bugReport method

Future<void> bugReport({ required String description, required String name, required String contact, List<File>? files, })

Send bug report with the given

Throws an exception if the request fails.

Implementation

Future<void> bugReport({
  required String description,
  required String name,
  required String contact,
  List<File>? files,
}) async {
  await _send(
    {
      'bug_report': jsonEncode({
        'description': description,
        'user': {
          'name': name,
          'contact': contact,
        },
      }),
      'files': files,
    },
    formData: true,
    throwOnError: true,
  );
}