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

Question

I have string like this,

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

And I need to convert that string to json, how I can do it in dart flutter? thank you so much for your help.

Solution 1

You have to use json.decode. It takes in a json object and let you handle the nested key value pairs. I'll write you an example

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};
}

Solution 2

Assume that we have a simple JSON structure like this:

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

We will create a Dart class named User with fields: 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} }';
  }
}

You can see factory User.fromJson() method in the code above. It parses a dynamic object into User object. So how to get dynamic object from a JSON string?

We use dart:convert library’s built-in jsonDecode() function.

import 'dart:convert';

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

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

  print(user);

The result will look like this.

{ bezkoder, 30 }

Ref : Dart/Flutter parse JSON string into Object

Solution 3

You can also convert JSON array to list of Objects as following:

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

And MyModel is something like this:

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;
  }
}

Solution 4

You must need to use this sometimes

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

 };
}

This code give you a like string {"numberPhone":"[phone]", "country":"CI"}. So it's easy to decode it's after like that

     json.decode({"numberPhone":"[phone]", "country":"CI"})

View more solutions

Comments

[Ashtav]

I have been try this, Map res = jsonDecode(sharedPreferences.getString('jsonString')); but I got this error: FormatException (FormatException: Unexpected character

[Ernesto Campohermoso]

a Map is distinct of dynamic

[VipiN Negi]

Yet this doesn't answer the OP question. You just explained the right way to accessing key-value pair from actual json.

[rizerphe]

While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanations and give an indication of what limitations and assumptions apply.

[Kamlesh]

how can we use this toJson function?

[Mohsen Emami]

@Kamlesh you can access it as MyModel model = new MyModel(...); final json = model.toJson();

[Kamlesh]

will it work because toJson() is factory typed function? any suggestion will be welcomed. Thanks.

[Kamlesh]

I have a map like userinfo = { 'name': , 'phonenumber': '9829098290', 'city': 'california' } If I pass it it to my model like User.fromJson(userinfo) it does not work. I know name field is null. Kindly suggest how to use it to make model type value like User.name, User.phonenumber, User.city. Thanks.

[Jason Pascoe]

The question object does not contain quotation marks, decode will fail