提问者:小点点

_InternalLinkedHashMap<String,动态>不是List<动态>类型的子类型


大家好,我需要你的帮助,我想显示来自api的信息,我有一个错误消息,告诉我_InternalLinkedHashMap

 Future<List<FactureModel>> getUser() async {
    var apiUrl = '/factures';

    http.Response response = await http.get(Uri.parse(_baseUrl + apiUrl));

    try {
      if (response.statusCode == 200) {
        List<dynamic> list = await json.decode(response.body);
        print(list);
        return list.map((e) => FactureModel.fromJson(e)).toList();
      } else {
        print("Something wrong");
        return <FactureModel>[];
      }
    } catch (e) {
      print(e);
      print("Api request failled");
      return <FactureModel>[];
    }

    //var info = rootBundle.loadString("json/data.json");
    //List<dynamic> list = json.decode(await info);
    //print(list);
    //return Future.delayed(
    //const Duration(seconds: 5), () => list.map((e) => e).toList());
  }
}

共1个答案

匿名用户

将您解码后的响应直接存储到您的FractreModel而不是List中。之后,只需从旅游FractreModel返回列表。