Future.then必須回傳回傳的未來型別的值

我不斷收到此錯誤

Unhandled Exception: Invalid argument(s) (onError): The error handler of Future.then must return a value of the returned future's type

網路介面

class NetworkApi {
  NetworkApi(this.url);

  final String url;

  Future getData() async {
    http.Response response = await http.get(Uri.parse(url));

    if (response.statusCode == 200) {
      String data = response.body;

      return jsonDecode(data);
    } else {
      if (kDebugMode) {
        print(response.statusCode);
      }
    }
  }
}

介面選擇器

class ApiSelector {
  Future getApiData({String categoryName = ""}) async {
    late NetworkApi _api;
    if(categoryName != ""){
     _api = NetworkApi('api');
    }else{
      _api = NetworkApi('myapi');
    }
    var Data = await _api.getData();
    if (Data['status'] == 'error') {
      throw'Error while fetching data';
    }
    return MyModel.fromJson(Data);
  }
}

MyFunction 呼叫 API 選擇器

  Future getDataFromApi() async {
    await _apiSelector
        .getApiData(categoryName: "Some Category")
        .then((value) {
      myModel = value as MyModel;
      print(myModel);
      return value;
    } ,onError: (error) {
     print(error);
      return error;
    },);
  }

I have tried everything to solve this issue. May you please guide me what to do. I have even tried with exception handling but still getting this error.

uj5u.com熱心網友回復：

指定回傳型別的未來，如Future<dynamic>

網路介面

class NetworkApi {
  NetworkApi(this.url);

  final String url;

  Future<dynamic> getData() async {
    http.Response response = await http.get(Uri.parse(url));

    if (response.statusCode == 200) {
      String data = response.body;

      return jsonDecode(data);
    } else {
      if (kDebugMode) {
        print(response.statusCode);
      }
    }
  }
}

介面選擇器

class ApiSelector {
  Future<MyModel> getApiData({String categoryName = ""}) async {
    late NetworkApi _api;
    if(categoryName != ""){
     _api = NetworkApi('api');
    }else{
      _api = NetworkApi('myapi');
    }
    var Data = await _api.getData();
    if (Data['status'] == 'error') {
      throw'Error while fetching data';
    }
    return MyModel.fromJson(Data);
  }
}

MyFunction 呼叫 API 選擇器

  Future<MyModel?> getDataFromApi() async {
   try {
     var value = await _apiSelector.getApiData(categoryName: "Some Category");
     myModel = value as MyModel;
     return myModel;

   } catch(error) {
      print(error);
      return error;
   }
  }

uj5u.com熱心網友回復：

問題出在getDataFromApi：

Future getDataFromApi() async {
    await _apiSelector
        .getApiData(categoryName: "Some Category")
        .then((value) {
      myModel = value as MyModel;
      print(myModel);
      return value;
    } ,onError: (error) {
     print(error);
      return error;
    },);
  }

正如錯誤訊息所解釋的，您的處理程式無法回傳與成功路徑onError回傳的型別相同的型別。Future.then（有關解釋，請參閱https://stackoverflow.com/a/66397222/。）這通常是編譯時錯誤，但最終會成為運行時錯誤，因為getDataFromApi它被宣告為回傳 a Future(相當于Future<dynamic>而不是Future<MyModel>.

使用帶有錯誤回呼的原始FutureAPI 直接使用是相當混亂的。您已經在使用await，因此您沒有理由使用Future.then. 使用await，您可以使用try-catch代替，這使問題更加清晰：

Future<MyModel> getDataFromApi() async {
  try {  
    var myModel = await _apiSelector.getApiData(categoryName: "Some Category");
    print(myModel);
    return myModel;
  } catch (error) {
    print(error);
    return error; // WRONG: This is not a `MyModel`!
  }
}

要解決此問題，您的catch塊需要回傳 aMyModel或必須拋出例外。

uj5u.com熱心網友回復：

class ApiSelector {
  Future<MyModel> getApiData({String categoryName = ""}) async {
    late NetworkApi _api;
    if(categoryName != ""){
     _api = NetworkApi('api');
    }else{
      _api = NetworkApi('myapi');
    }
    var Data = await _api.getData();
    if (Data['status'] == 'error') {

    //I had to replace throw with Future.value Now All Errors are gone.
             return Future.value( 'Error while fetching data');
    }
    return MyModel.fromJson(Data);
  }
}

我希望這也能幫助將來某個時候面臨這種錯誤的其他人。

標籤：flutter api dart exception dart-null-safety

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