如何在 dart flutter 中将 json 字符串转换为 json 对象？

Question

我有这样的字符串，

{id:1, name: lorem ipsum, address: dolor set amet}

而且我需要将该字符串转换为 json，如何在 dart flutter 中完成？非常感谢您的帮助。

Answer 1

您必须导入 dart:encode 库。然后使用 jsonDecode 函数，这将产生类似于 Map

的动态

https://api.dartlang.org/stable/2.2.0/dart-convert/dart-convert-library.html

Answer 2

您必须使用 json.decode。它接收一个 json 对象并让您处理嵌套的键值对。我给你写个例子

import 'dart:convert';

// actual data sent is {success: true, data:{token:'token'}}
final response = await client.post(url, body: reqBody);

// Notice how you have to call body from the response if you are using http to retrieve json
final body = json.decode(response.body);

// This is how you get success value out of the actual json
if (body['success']) {
  //Token is nested inside data field so it goes one deeper.
  final String token = body['data']['token'];

  return {"success": true, "token": token};
}

Answer 3

您还可以将 JSON 数组转换为对象列表，如下所示：

String jsonStr = yourMethodThatReturnsJsonText();
Map<String,dynamic> d  = json.decode(jsonStr.trim());
List<MyModel> list = List<MyModel>.from(d['jsonArrayName'].map((x) => MyModel.fromJson(x)));

而MyModel是这样的：

class MyModel{

  String name;
  int age;

  MyModel({this.name,this.age});

  MyModel.fromJson(Map<String, dynamic> json) {
    name= json['name'];
    age= json['age'];
  }

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

    return data;
  }
}

Answer 4

你有时肯定需要用到这个

Map<String, dynamic> toJson() {
  return {
    jsonEncode("phone"): jsonEncode(numberPhone),
    jsonEncode("country"): jsonEncode(country),

 };
}

此代码给你一个喜欢的字符串 {"numberPhone":"+225657869", "country":"CI"}。这样就很容易解码了

     json.decode({"numberPhone":"+22565786589", "country":"CI"})

Answer 5

假设我们有一个像这样的简单 JSON 结构：

{
  "name": "bezkoder",
  "age": 30
}

我们将创建一个名为 User 的 Dart class，其字段为：name & age.

class User {
  String name;
  int age;

  User(this.name, this.age);

  factory User.fromJson(dynamic json) {
    return User(json['name'] as String, json['age'] as int);
  }

  @override
  String toString() {
    return '{ ${this.name}, ${this.age} }';
  }
}

你可以在上面的代码中看到factory User.fromJson()方法。它将动态对象解析为 User 对象。那么如何从 JSON 字符串中获取 dynamic 对象呢？

我们使用 dart:convert 库的 built-in jsonDecode() 函数。

import 'dart:convert';

main() {
  String objText = '{"name": "bezkoder", "age": 30}';

  User user = User.fromJson(jsonDecode(objText));

  print(user);

结果将如下所示。

{ bezkoder, 30 }

参考：Dart/Flutter parse JSON string into Object

Answer 6

 String name =  "{click_action: FLUTTER_NOTIFICATION_CLICK, sendByImage: https://ujjwalchef.staging-server.in/uploads/users/1636620532.png, status: done, sendByName: mohittttt, id: HM11}";
  List<String> str = name.replaceAll("{","").replaceAll("}","").split(",");
  Map<String,dynamic> result = {};
  for(int i=0;i<str.length;i++){
    List<String> s = str[i].split(":");
    result.putIfAbsent(s[0].trim(), () => s[1].trim());
  }
  print(result);
}

Answer 7

为了将字符串转换为JSON，我们必须使用自定义逻辑对其进行修改，在这里我们首先删除数组和对象的所有符号，然后我们用特殊字符拆分文本并附加键和值（对于地图）。请在 dartpad.dev

中尝试此代码片段

import 'dart:developer';

void main() {
  String stringJson = '[{product_id: 1, quantity: 1, price: 16.5}]';

  stringJson = removeJsonAndArray(stringJson);

  var dataSp = stringJson.split(',');
  Map<String, String> mapData = {};
  for (var element in dataSp) {
    mapData[element.split(':')[0].trim()] = element.split(':')[1].trim();
  }

  print("jsonInModel: ${DemoModel.fromJson(mapData).toJson()}");
}

String removeJsonAndArray(String text) {
  if (text.startsWith('[') || text.startsWith('{')) {
    text = text.substring(1, text.length - 1);
    if (text.startsWith('[') || text.startsWith('{')) {
      text = removeJsonAndArray(text);
    }
  }
  return text;
}

class DemoModel {
  String? productId;
  String? quantity;
  String? price;

  DemoModel({this.productId, this.quantity, this.price});

  DemoModel.fromJson(Map<String, dynamic> json) {
    log('json: ${json['product_id']}');
    productId = json['product_id'];
    quantity = json['quantity'];
    price = json['price'];
  }

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = <String, dynamic>{};
    data['product_id'] = productId;
    data['quantity'] = quantity;
    data['price'] = price;
    return data;
  }
}

如何在 dart flutter 中将 json 字符串转换为 json 对象？

how to convert json string to json object in dart flutter?

string

json

dart

flutter

jsonconvert