Flutter json数据解析

手动序列化

使用dart:convert手动序列化数据

json数据:

{
  "name": "John Smith",
  "email": "john@example.com"
}

使用jsonDecode()方法来解析json数据

Map<String, dynamic> user = jsonDecode(jsonString);

print(user['name']);

但是这样解析会存在一个问题,就是会失去静态类型的语言特性就是类型安全

模型类中序列化json数据

import 'dart:convert';

class User {
  final String name;
  final String email;

  User({this.name, this.email});
  User.fromJson(Map<String, dynamic> json) 
     : name= json['name'],
      email= json['email'];

  Map<String, dynamic> toJson() => {
    'name': name,
    'email': email
  };
}

void main(List<String> args) {
  String jsonString = "{\"name\": \"ctl\", \"email\": \"1111222\"}";
  Map userMap = jsonDecode(jsonString);
  var user = User.fromJson(userMap);

  print('Howdy, ${user.name}!');
  print('We sent the verification link to ${user.email}.');
}

与dio库结合

模型类:

class OneModel {
  final id;

  OneModel({this.id});

  OneModel.fromJson(Map<String, dynamic> json) : id = json['id'];
}

请求类:

因为dio请求过来的数据已经是map类型了,所以不需要jsonDecode进行编码,直接调用模型类即可解析

class RequestImage {
	final String url = "http://v3.wufazhuce.com:8000/api/channel/one/0/北京";

	Future getImage() async {
		try {
			Response response = await Dio().get(url);
			print(response.data.runtimeType);
//			Map dataMap = jsonDecode(response.data['data']);

			var user =  OneModel.fromJson(response.data['data']);
			print(user.id);
		} catch (e) {
			print(e);
		}
	}
}