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过去，我使用返回Future<Response>. 我注意到我可以在.then()方法内返回其他类型，最终类型结果将根据我在 then 中返回的类型而改变。

但是当我更改为使用返回的包时，在其中返回Future<dynamic>不同的类型.then()不会再改变最终结果类型。它不断导致Future<dynamic>. 但我可以保证动态类型是一种Response类型。这是我尝试过的：

第一个：

Future<Response> get(String url) =>
    fetcher.get(url).then((response) => response as Response); // error: Unhandled Exception: type 'Future<dynamic>' is not a subtype of type 'Future<Response>'

第二：

Future<Response> get(String url) =>
    fetcher.get(url) as Future<Response>; // error: Unhandled Exception: type 'Future<dynamic>' is not a subtype of type 'Future<Response>'

第三：

Future<Response> get(String url) =>
    fetcher.get(url).then((response) { 
        final result = response as Response;
        return result;
    }); // error: Unhandled Exception: type 'Future<dynamic>' is not a subtype of type 'Future<Response>'

在过去：

Future<List<Model>> get(String url) =>
    client.get(url).then((response) => Model.fromJson(response.body)); // working.

我用于此的包：https ://pub.dev/packages/twitter_api#-example-tab-

编辑：澄清提取器：

import 'package:twitter_api/twitter_api.dart';
import "package:http/http.dart";


twitterApi fetcher = twitterApi(
    consumerKey: consumerApiKey,
    consumerSecret: consumerApiSecretKey,
    token: locker.session?.token,
    tokenSecret: locker.session?.secret
);

Future<Response> get(String path) {
    final split = url.split("?");
    final onlyPath = split[0];
    final onlyParam = split.length > 1 ? split[1] : "";
    final options = Map.fromIterable(onlyParam.split("&").where((it) => it.contains("=")), key: (e) => e.split("=")[0] as String, value: (e) => e.split("=")[1] as String);
    return fetcher.getTwitterRequest("GET", onlyPath.replaceFirst("/", "", options: options).then((response) {
        debugPrint ("GET ${response.request.url} Response: ${response.body}");
        return response as Response;
    }); 
}

输出：

[ERROR:flutter/lib/ui/ui_dart_state.cc(157)] Unhandled Exception: type 'Future<dynamic>' is not a subtype of type 'Future<Response>'
#0      TwitterApiProvider.fetchUser (package:feazy/src/resources/twitter_api_provider.dart:11:70)
#1      Repository.fetchUser (package:feazy/src/resources/repository.dart:19:27)
#2      _HomeState.twitterLogin (package:feazy/src/ui/home.dart:50:21)
...
GET https://api.twitter.com/1.1/users/show.json?screen_name=somescreenname Response: {"id":<int>,"id_str":<string>,"name":"Some name","screen_name":"somescreenname",...

编辑2：进一步检查，我注意到在IDE上，IDE告诉我返回类型fetcher.getTwitterRequest()不是Future<dynamic>，而是dynamic。所以我像这样更改我的代码，但它仍然无法正常工作。

Future<Response> get (String path) {
    ...
    final future = authClient.getTwitterRequest(...).then((response) {
        debugPrint ("POST ${response.request.url} Response: ${response.body}");
        return response as Response;
    });
    final futureResponse = future as Future<Response>;
    return futureResponse;
}

在这种情况下，错误在 上final futureResponse = future as Future<Response>;，说的是同样的事情：Unhandled Exception: type 'Future<dynamic>' is not a subtype of type 'Future<Response>' in type cast。

4

1 回答 1

0

恕我直言，您的所有示例都可以，但错误实际上告诉我们您正试图将其Future<Response>转换为Future<dynamic>

更新

你可以使用async await，像这样：

Future<Response> get(String path) async {
    final split = url.split("?");
    final onlyPath = split[0];
    final onlyParam = split.length > 1 ? split[1] : "";
    final options = Map.fromIterable(onlyParam.split("&").where((it) => it.contains("=")), key: (e) => e.split("=")[0] as String, value: (e) => e.split("=")[1] as String);
    final response = await fetcher.getTwitterRequest("GET", onlyPath.replaceFirst("/", "", options: options); 
    return response as Response;
}
于 2020-01-29T07:26:45.567 回答