sendFeedback method

Implementation

@override
Future<void> sendFeedback(FeedbackModel feedback) async {
  final response = await http.post(
    Uri.parse(apiUrl),
    headers: {'Content-Type': 'application/json'},
    body: jsonEncode(feedback.toJson()),
  );

  if (response.statusCode != 200) {
    throw Exception('Failed to send feedback: ${response.body}');
  }
}