Question

import 'dart:async';
import 'question.dart';
import 'package:http/http.dart' as http;
import 'dart:convert';

String opentdb = 'https://opentdb.com/api.php?amount=15&type=boolean';

class QuestionServices {
  Future<List<Question>> getData() async {
    List<Question> questions;
    String link = opentdb;
    var res = await http
        .get(Uri.encodeFull(link), headers: {"Accept": "application/json"});
    print(res.body);
    if (res.statusCode == 200) {
      var data = json.decode(res.body);
      var rest = data['results'] as List;
      print(rest);
      questions =
          rest.map<Question>((json) => Question.fromJson(json)).toList();
    }
    print("List Size: ${questions.length}");
    // _questions = questions;
    return questions;
  }

  List<Question> newQuestions = getData();
}

class Question {
  final String question;
  final bool answer;

  Question({this.question, this.answer});

  factory Question.fromJson(Map<String, dynamic> json) {
    return Question(
      question: json['question'] as String,
      answer: json['correct_answer'] as bool,
    );
  }
}

我正在尝试从JSON数据库创建问题列表，但是每当尝试获取返回的列表时，都会出现错误：

"A value of type 'Future<List<Question>>' can't be assigned to a variable of type 'List<Question>'."

我不确定为什么我要返回的列表给出了该错误。也许有其他方法可以将json放入列表中？

Answer 1

List<Question> getData() async { // remove Future
    List<Question> questions;
    String link = opentdb;
 ...


//how to access now you will get instance of List<Question>

getData().then((List<Question> newQuestions){

})

Answer 2

getData返回一个Future，因此您需要执行以下操作：

Future<List<Question>> newQuestions = getData();

Answer 3

Future<List<Question>>Listfetchdata() async{
   final responsein = await http.get('https://opentdb.com/api.php?amount=15&type=boolean');
   final getindata = json.decode(responsein.body);
   final getindata1 = json.encode(getindata['results']);
   return compute(listdata,getindata1);
 }

 List<Question> listdata(String creation){
   final parsed = jsonDecode(creation).cast<Map<String,dynamic>>();
   return parsed.map<Question>((json)=>Question.fromJson(json)).toList();
 }

List listitem;
 var maketolistarry=[''];
 void lastgettingdata() async{
   final king=await Listfetchdata();
listitem=king.map((e) => e.category).toList();
maketolistarry=listitem;
 }

在maketolistarry的帮助下，您将获得一个列表

类型'Future <List <Question >>'的值不能分配给'List <Question>'类型的变量

3 个答案: