list - API工作正常，但返回类名的实例

Question

我有这个Api
打印项目oneJalsah是正确的，但是返回List时出现问题，进入控制台
ALL = ['JalsahTime'的实例，'JalsahTime'的实例，'JalsahTime'的实例，'JalsahTime'的实例，'JalsahTime'的实例，'JalsahTime'的实例，'JalsahTime'的实例，'JalsahTime'的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，实例“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例，一个“JalsahTime”的实例JalsahTime”，“JalsahTime”的实例，“JalsahTime”的实例，“JalsahTime”的实例]

这是我的鳕鱼

class Day1API{
  Future<List<JalsahTime>> getDay1()async{

    String url = "http://www.something.......";
    JalsahTime oneJalsah;
    List<JalsahTime> allJalsay =  List<JalsahTime>();
        print(url);
    var response = await http.get(url);
    if (response.statusCode == 200){
      var data = jsonDecode(response.body);
      print(data);
      for (var item in data){
        print("item = $item");
        oneJalsah = JalsahTime(item['id'], item['day'], item['start'], item['end'], item['date'], item['stateDay'], item['stateTime']);
        allJalsay.add(oneJalsah);
      }
      print("ALL = $allJalsay");
      print('jalsah done');
      return allJalsay;
    }
    else print(" day failed");
  }
}

这是我的课

class JalsahTime{
  String id,day,start,end,date,stateDay,stateTime;
  JalsahTime(this.id, this.day, this.start, this.end, this.date, this.stateDay,
      this.stateTime);
}

Answer 1

将toString方法添加到您的类(class)

class JalsahTime{
  String id,day,start,end,date,stateDay,stateTime;
  JalsahTime(this.id, this.day, this.start, this.end, this.date, this.stateDay,
      this.stateTime);

  @override
  String toString(){
      return "Some representation of that instance: ${this.id}"   
  }
}