Dart类，接收Map类型再返回Map类型，也就是Json返回数据接收和返回

1：使用类中的属性作为模板，实例化时接收参数，然后返回一个dart 中的Map类ing数据
2：接收的参数是什么格式？
直接一一对应dart类型–使用普通构造函数
接收的是一个Map类型数据，那么就需要提取Map中的value，需要定义命名构造函数
3：类中定义方法 用来输出一个Map类型数据

class Shangpin {
  String? goodId;
  int? amount; 
  String? goodImage;
  double? goodPrice; 
  String? goodName;

  Shangpin(this.goodId, this.amount, this.goodImage, this.goodPrice, this.goodName);

  Shangpin.fromJson(Map<String, dynamic> json) {
    goodId = json['goodId'];
    amount = json['amount'];
    goodImage = json['goodImage'];
    goodPrice = json['goodPrice'];
    goodName = json['goodName'];
  }

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = Map<String, dynamic>();
    data['goodId'] = this.goodId;
    data['amount'] = this.amount;
    data['goodImage'] = this.goodImage;
    data['goodPrice'] = this.goodPrice;
    data['goodName'] = this.goodName;
    return data;
  }
}

void main() {

  //s1实例和s2实例
  Shangpin s1 = Shangpin('000001', 999, 'http://192.168.2.168/images/1.png', 80.00, '商品1名称');

  Shangpin s2 = Shangpin.fromJson({'goodId': '000002','amount': 666,'goodImage': 'http://192.168.2.168/images/2.png','goodPrice': 688.00,'goodName': '商品2名称'});


  print(s1); //实例
  print(s2); //实例

  
  var a = s1.toJson();//a=={goodId: 000001, amount: 999, goodImage: http://192.168.2.168/images/1.png, goodPrice: 80.0, goodName: 商品1名称}
  var b = s2.toJson();//b=={goodId: 000002, amount: 666, goodImage: http://192.168.2.168/images/2.png, goodPrice: 688.0, goodName: 商品2名称}

}