deserialize method

  1. @override
void deserialize(
  1. Map<String, dynamic>? json
)
override

Implementation

@override
void deserialize(Map<String, dynamic>? json) {
  if (json == null) {
    throw ApiException(
        400, 'Failed to deserialize ClassificationResult data model.');
  }

  if (json.containsKey('ClassName')) {
    className = json['ClassName'] as String;
  } else {
    className = null;
  }

  if (json.containsKey('ClassProbability')) {
    classProbability = json['ClassProbability'] as double;
  } else {
    classProbability = null;
  }
}