错误：需要一个“int”类型的值，但得到一个“String”类型的值；扑

Question

我正在尝试使用 http mathod“GET”从实时数据库中获取测验数据。 正在检索数据，但它没有显示在 listview 中，当我打印列表的长度时，它为 0。这就是我的终端中显示的错误：错误：预期类型为“int”的值，但得到了“字符串”类型之一

我无法弄清楚问题是什么。请帮我解决这个问题，因为我尝试了大约 5 天但无法解决。

谢谢。

这些是我在列表中获取数据的代码。

Future<void> getList() async {
    list.clear();
    final url = Uri.parse(
        'https://testisnotesttheisthelearningapp-default-rtdb.firebaseio.com/Schools/${widget.uid}/ResultsFolder/${widget.id}.json');
    final response = await get(url);
    print(response.body);
    var map = json.decode(response.body) as Map<String, dynamic>;
    // print(map);
    if (map != null) {
      map.forEach((key, value) {
        print(value['Name']);
        var temp = Result(
            value['Percent'],
            (value['choice'] as List<dynamic>)
                .map((e) => Model(e['index'], e['question'], []))
                .toList(),
            value['Score'],
            (value['question'] as List<dynamic>).map((e) => e).toList(),
            value['Name']);
        list.add(temp);
      });
    }
  }

这是我的实时数据库中的 JSON 格式数据：

{
  "Name" : "Miets Digital",
  "Percent" : 100,
  "Score" : 2,
  "choice" : [ "Correct", "Correct" ],
  "question" : [ {
    "question" : "WHo is Elon Musk?"
  }, {
    "question" : "How did Elon musk got rich?"
  } ]
}

Answer 1

试试这个：-

Future<void> getList() async {
    list.clear();
    final url = Uri.parse(
        'https://testisnotesttheisthelearningapp-default-rtdb.firebaseio.com/Schools/${widget.uid}/ResultsFolder/${widget.id}.json');
    final response = await get(url);
    print(response.body);
    var map = json.decode(response.body) as Map<String, dynamic>;
    // print(map);
    if (map != null) {
     print(map['Name']);
        var temp = Result(
            map['Percent'],
            (map['choice'] as List<dynamic>)
                .map((e) => Model(e['index'], e['question'], []))
                .toList(),
            map['Score'],
            (map['question'] as List<dynamic>).map((e) => e).toList(),
            map['Name']);
        list.add(temp);
    }
  }

forEach() 循环遍历映射中的每个键值对。变量 key 和 value 为您提供循环当前正在处理的键和值。如果要访问地图的值，则必须使用语法map['key_name']。

这也解释了错误，因为您的 Result() 构造函数需要一个整数但得到一个字符串值。

Answer 2

解析 JSON 映射不是一个好习惯。你只需调用模型类

尝试按照以下代码进行操作

Future<QuizModelResponse> getList() async {
 // here write your code 
 var map = json.decode(response.body);
 return QuizModelResponse.fromJsonMap(map);
}

列表响应类

class QuizModelResponse {
  List<QuizItemModel> content;

  QuizModelResponse.fromJsonMap(dynamic data) :
        content = List<QuizItemModel>.from(
            data.map((it) => QuizItemModel.fromJson(it)));
}

模型类

class QuizItemModel {
  String name;
  int percent;
  int score;
  List<String> choice;
  List<Question> question;

  QuizItemModel(
      {this.name, this.percent, this.score, this.choice, this.question});

  QuizItemModel.fromJson(Map<String, dynamic> json) {
    name = json['Name'];
    percent = json['Percent'];
    score = json['Score'];
    choice = json['choice'].cast<String>();
    if (json['question'] != null) {
      question = new List<Question>();
      json['question'].forEach((v) {
        question.add(new Question.fromJson(v));
      });
    }
  }

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = new Map<String, dynamic>();
    data['Name'] = this.name;
    data['Percent'] = this.percent;
    data['Score'] = this.score;
    data['choice'] = this.choice;
    if (this.question != null) {
      data['question'] = this.question.map((v) => v.toJson()).toList();
    }
    return data;
  }
}

class Question {
  String question;

  Question({this.question});

  Question.fromJson(Map<String, dynamic> json) {
    question = json['question'];
  }

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = new Map<String, dynamic>();
    data['question'] = this.question;
    return data;
  }
}