如何修复 Flutter 需要一个“Map<String, dynamic>”类型的值，但得到一个“List<dynamic>”类型的值

Posted 2023-02-23

【中文标题】如何修复 Flutter 需要一个“Map<String, dynamic>”类型的值，但得到一个“List<dynamic>”类型的值

【英文标题】：How to fix Flutter Expected a value of type 'Map<String, dynamic>', but got one of type 'List<dynamic>'

【发布时间】：2021-06-12 07:27:35

【问题描述】：

我正在开发一个使用 Flutter Web 和 RESTful API 作为后端的 Web 应用程序。 所以，我正在尝试从 api 中获取数据，使用 Flutter Models 对其进行序列化，然后返回结果。

问题是，我得到了这个结果

Expected a value of type 'Map<String, dynamic>', but got one of type 'List<dynamic>'

如何解决这个问题？

这是我的颤振代码：

模型

// To parse this JSON data, do
//
//     final medicalRecordsModel = medicalRecordsModelFromJson(jsonString);

import 'dart:convert';

class MedicalRecordsModel 
  MedicalRecordsModel(
    this.id,
    this.category,
    this.fileName,
    this.dateTimestamp,
    this.description,
    this.upload,
    this.patientName,
    this.age,
    this.address,
    this.userId,
    this.patientId,
    this.isActive,
  );

  final String id;
  final String category;
  final String fileName;
  final String dateTimestamp;
  final String description;
  final String upload;
  final String patientName;
  final String age;
  final String address;
  final dynamic userId;
  final int patientId;
  final bool isActive;

  factory MedicalRecordsModel.fromJson(Map<String, dynamic> json) 
    return MedicalRecordsModel(
      id: json["id"],
      category: json["category"],
      fileName: json["fileName"],
      dateTimestamp: json["dateTimestamp"],
      description: json["description"],
      upload: json["upload"],
      patientName: json["patientName"],
      age: json["age"],
      address: json["address"],
      userId: json["userId"],
      patientId: json["patientId"],
      isActive: json["isActive"],
    );
  

API 连接

import 'dart:convert';
import 'dart:developer';
import 'dart:async';
import 'package:app/src/constants/medical_records.dart';
import 'package:app/src/models/medical_records/medical_records.dart';
import 'package:app/src/pages/Medical-Records/medical_record.dart';
import 'package:http/http.dart' as http;

class MedicalRecordsManager 
  var client = http.Client();
  var url = ConstantMedicalRecords.medical_records_api;

  Future<MedicalRecordsModel> getRecords() async 
    var url = ConstantMedicalRecords.medical_records_api;
    log('$url');
    try 
      final response = await client.get(url);
      if (response.statusCode == 200) 
        return MedicalRecordsModel.fromJson(jsonDecode(response.body));
        // print(recordsModel);
      
     catch (Exception) 
      print(Exception);
      print("Error occured");
    
  

这是我要获取的 JSON 数据

 
        "id": "103",
        "category": "DOCUMENT",
        "fileName": "Check Up",
        "dateTimestamp": "2021-02-1012:59:46",
        "description": "string",
        "upload": "String",
        "patientName": "1",
        "age": "25",
        "address": "Earth",
        "userId": null,
        "patientId": 12,
        "isActive": true
    

请帮我解决这个问题。

【问题讨论】：

你确定你只返回一个 MedicalRecordsManager 吗？...基于命名 getRecords().. 它应该是 Future>?.. 尝试打印出 json 响应并调试好

@Reign 我想返回的不仅仅是数据

您可以使这个过程变得更容易。请看这个答案。仅使用一个函数调用，您就可以得到最终结果。此方法已经过测试并在测试中涵盖。 ***.com/a/66632608/1737201

这些答案都不起作用吗？

【参考方案1】：

你可以这样做

MedicalRecordsModel.fromJson(jsonDecode(response.body) as Map<String, dynamic>);

【参考方案2】：

如下更改getRecord

Future<MedicalRecordsModel> getRecords() async 
    var url = ConstantMedicalRecords.medical_records_api;
    log('$url');
    try 
      final response = await client.get(url);
      if (response.statusCode == 200) 
        return MedicalRecordsModel.fromJson(jsonDecode(response.body)[0]);
        // print(recordsModel);
      
     catch (Exception) 
      print(Exception);
      print("Error occured");
    
  

我认为jsonDecode 提供了地图列表，因此您的 json 地图是该列表的第一个元素。

【参考方案3】：

此代码将按您的预期工作：

import 'package:json_helpers/json_helpers.dart';

void main() 
  // responseBody is the same response.body

  // When response is a list of objects
  final list = responseBody1.jsonList((e) => MedicalRecordsModel.fromJson(e));
  var obj = list[0];
  print(obj.category);
  print(obj.fileName);

  // When response is an object
  obj = responseBody2.json((e) => MedicalRecordsModel.fromJson(e));
  print(obj.category);
  print(obj.fileName);


final responseBody1 = '''
[
   
      "id":"103",
      "category":"DOCUMENT",
      "fileName":"Check Up",
      "dateTimestamp":"2021-02-1012:59:46",
      "description":"string",
      "upload":"String",
      "patientName":"1",
      "age":"25",
      "address":"Earth",
      "userId":null,
      "patientId":12,
      "isActive":true
   
]''';

final responseBody2 = '''

   "id":"103",
   "category":"DOCUMENT",
   "fileName":"Check Up",
   "dateTimestamp":"2021-02-1012:59:46",
   "description":"string",
   "upload":"String",
   "patientName":"1",
   "age":"25",
   "address":"Earth",
   "userId":null,
   "patientId":12,
   "isActive":true
''';

class MedicalRecordsModel 
  final String id;

  final String category;
  final String fileName;
  final String dateTimestamp;
  final String description;
  final String upload;
  final String patientName;
  final String age;
  final String address;
  final dynamic userId;
  final int patientId;
  final bool isActive;
  MedicalRecordsModel(
    this.id,
    this.category,
    this.fileName,
    this.dateTimestamp,
    this.description,
    this.upload,
    this.patientName,
    this.age,
    this.address,
    this.userId,
    this.patientId,
    this.isActive,
  );

  factory MedicalRecordsModel.fromJson(Map<String, dynamic> json) 
    return MedicalRecordsModel(
      id: json['id'] as String,
      category: json['category'] as String,
      fileName: json['fileName'] as String,
      dateTimestamp: json['dateTimestamp'] as String,
      description: json['description'] as String,
      upload: json['upload'] as String,
      patientName: json['patientName'] as String,
      age: json['age'] as String,
      address: json['address'] as String,
      userId: json['userId'] as String,
      patientId: json['patientId'] as int,
      isActive: json['isActive'] as bool,
    );
  

输出：

DOCUMENT
Check Up
DOCUMENT
Check Up

也就是说，当response是一个对象列表时：

final list = response.body.jsonList((e) => MedicalRecordsModel.fromJson(e));

当响应是一个对象时：

final object = response.body.json((e) => MedicalRecordsModel.fromJson(e));

如果你不知道结果是什么，那么你可以尝试这两种方法。

response.body.json((e) => Model.fromJson(e));
response.body.jsonList((e) => Model.fromJson(e));

如果你已经解码了一个 JSON 字符串并且想要转换结果（或其中的一部分），你可以使用以下方法：

如果解码的value的类型是Map：

final object = value.json((e) => Model.fromJson(e));

如果解码的value的类型是List：

final objects = value.json((e) => Model.fromJson(e));

【参考方案4】：

每个响应都以文本形式发送和接收，可以转换为 使用 dart 内置核心库 import 'dart:convert'; 的地图格式。

所以请求的响应可以这样处理。

final res = await http.post(Uri.parse(url),  body: json.encode(
          'userId': uid,
          'email': email,
        ),
        head body: json.encode(
          'userId': uid,
          'email': email,
        ),
        headers: 'Content-Type': 'application/json', 'token64': token);

这里json.encode()用于从Map转换为String。

现在 res 变量包含响应，它也是一个字符串，可以像这样使用json.decode() 转换为 Map。

final data = json.decode(res);

在处理数据时，有时我们会发生错误，例如 Map is not a type of Map<String, String> 等。

这可以通过类型转换 res 来解决，像这样。

Map<String, String> notification = Map<String, String>.from(data['notification']);

我在 Youtube 的无聊的 Flutter 开发节目中看到了这些类型转换方法。