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 StyleCopy data model.');
  }

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