Suppose I have the following json (structured as <String key, Map value>):
{
'A1': {'name': 'a'},
'B2': {'name': 'b'}
}
and I want to parse it to this class (notice that I use the key as the id for that user), using the fromJson factory method, which accepts two arguments:
Class User {
final String id;
final String name;
factory User.fromJson(Map<String, dynamic> json, String key) {
return User(
id: key,
name: json['name'],
);
}
}
Can I achieve it using json_serializable ?
The json Map expected by this factory method is just the values of the top-level JSON object you're parsing.
All you need to do is parse the JSON, extract all keys, then pass the values to the factory method.
Something like this:
import 'dart:convert';
const json = '''
{
"A1": {"name": "a"},
"B2": {"name": "b"}
}
''';
class User {
final String id;
final String name;
User({required this.id, required this.name});
factory User.fromJson(Map<String, dynamic> json, String key) {
return User(
id: key,
name: json['name'],
);
}
#override
String toString() => 'User(id=$id, name=$name)';
}
main() {
final map = jsonDecode(json);
map.forEach((id, userJson) {
final user = User.fromJson(userJson, id);
print(user);
});
}
Prints:
User(id=A1, name=a)
User(id=B2, name=b)
Now, to use json_serializable, just annotate it and replace your implementation with the generated one...
#JsonSerializable()
class User {
...
factory User.fromJson(Map<String, dynamic> json, String key) =>
// pass in only the relevant json Map!
_$UserFromJson(json[key]);
}
Related
Is it possible to somehow create an abstract factory method? Maybe what I'm trying to do is possible to implement differently?
abstract class ApiModel {
// Error: A function body must be provided.
factory ApiModel.fromJson(Map<String, dynamic> json);
}
class User extends ApiModel {
final int id;
final String name;
User({required this.id, required this.name});
#override
factory User.fromJson(Map<String, dynamic> json) {
return User(
id: json['id'] as int,
name: json['name'] as String,
);
}
}
class ApiResponse<Model extends ApiModel> {
final List<Model> results;
ApiResponse({required this.results});
factory ApiResponse.fromJson(Map<String, dynamic> json) {
return ApiResponse(results: (json['results'] as List).map((item) => Model.fromJson(item)).toList());
}
}
I solved it like this:
factory ApiResponse.fromJson(Map<String, dynamic> json, Model Function(dynamic) mapper) {
return ApiResponse(
info: Info.fromJson(json['info']),
results: (json['results'] as List).map(mapper).toList(),
);
}
I have a lot of JSON models in my dart project that I would like to have a string representation of. I know that I can override the .toString() method per class, but that feels like a lot of work to write basically the same thing a bunch of times. Is there a way that I can create a mixin or extention to override toString? Or is it better to use code generation? (I found this package, but it seems like it isn't maintained any more)
The string representation I am looking for is just a list of all parameters, for example:
#JsonSerializable()
class User {
UserOver(
this.userId,
this.name,
);
int userId;
/// The full name of the user.
String name;
factory UserOverview.fromJson(Map<String, dynamic> json) => _$UserOverviewFromJson(json);
Map<String, dynamic> toJson() => _$UserOverviewToJson(this);
}
should have the following string representation:
User(userId: 1, name: "Name")
You answered your own question there buddy. Just create a base class and outsource it. It simple.
For example:
#JsonSerializable()
class User extends BaseModel {
User({
required this.userId,
required this.name,
});
int userId;
/// The full name of the user.
String name;
factory User.fromJson(Map<String, dynamic> json) =>
_$UserOverviewFromJson(json);
#override
Map<String, dynamic> toJson() => _$UserOverviewToJson(this);
}
Create a base model class
abstract class BaseModel {
Map<String, dynamic> toJson();
#override
String toString() {
return toJson().toString();
}
}
Don't forget these
User _$UserOverviewFromJson(Map<String, dynamic> json) => User(name: json['name'] as String, userId: json['userId'] as int);
Map<String, dynamic> _$UserOverviewToJson(User instance) => <String, dynamic>{
'name': instance.name,
'userId': instance.userId,
};
Now to use:
final cool = User(userId: 1, name: "Name");
print(cool.toString()); //{name: Name, userId: 1}
So I have a class like Question like bellow:
#JsonSerializable()
class Question {
String id;
String content;
Question({this.id, this.content});
factory Question.fromJson(Map<String, dynamic> json) =>
_$QuestionFromJson(json);
Map<String, dynamic> toJson() => _$QuestionToJson(this);
}
Please keep in mind that those _$QuestionFromJson and _$QuestionToJson comes from this library https://pub.dev/packages/json_serializable
Say I have many class like that which have a fromJson factory and a toJson method. I want to create a base class that contains those 2 method. A base model is easy for toJson as bellow:
abstract class BaseModel {
Map<String, dynamic> toJson();
}
But what about the factory method, I have no idea how to declare them then override it simply like:
#override
factory Question.fromJson(Map<String, dynamic> json) =>
_$QuestionFromJson(json);
EDIT:
My idea of using this is because I want to create a converter utility that I only need to pass in the class of the result like Converter.listFromJson<MyClass>(jsonString). For now, the helper is:
static List<T> listFromJson<T>(jsonString, Function mappingFunction) {
return myJsonMap.map(mappingFunction).cast<T>().toList();
}
so I have to map each item by passing the map function every time I use this helper method:
Converter.listFromJson<Question>(
jsonMap, (item) => Question.fromJson(item));
There'are a few more class that needs to be convert to the list like this. I want to reuse the method without the (item) => Question.fromJson(item) method part. That's why I want to create a base class that have the factory fromJson method so that I can use it in the converter
return myJsonMap.map((item) => BaseModel.fromJson(item)).cast<T>().toList();
then I just simply call
Converter.listFromJson<Question>(jsonMap);
Thank you for your time.
i don't know if i got you correctly, that's what i understood from your question
abstract class BaseModel{
BaseModel();
BaseModel.fromJson(Map<String,dynamic> json);
}
class Question extends BaseModel{
final String id;
final String name;
Question({this.id,this.name}): super();
#override
factory Question.fromJson(Map<String, dynamic> json) {
return Question(
id: json['id'],
name: json['name']
);
}
}
void main(){
Map<String,dynamic> json = {'id': "dsajdas",'name': 'test'};
Question question = Question.fromJson(json);
print('question: ${question.id}');
}
That was my approach but you can't do such a thing. There is a workaround by declaring .fromJson(json) in a variable. Look at my sample codes, hope you can get an idea.
class Categories {
final String id;
String name;
String image;
Categories({this.id, this.name, this.image});
Categories.fromJson(dynamic json)
: id = json['id'],
name = json['name'],
image = json['image'];
}
class CategoriesModel extends AppModel<Categories> {
List<Categories> list = [];
Function fromJson = (dynamic json) => Categories.fromJson(json);
}
class AppModel<T> {
List<T> list = [];
Function fromJson;
List<T> getList() {
if (this.list.isNotEmpty) return this.list;
List<dynamic> list = GetStorage().read('tableName');
list.forEach((data) {
this.list.add(fromJson(data));
});
return this.list;
}
}
How to convert this class into JSON or List?
class cliente {
int id;
String nome;
String apelido;
String sexo;
String status;
}
Edit
I'm changed my class and works fine to my case:
class client {
Map<String, dynamic> fields => {
"id": "",
"name": "",
"nickname": "",
"sex": "",
"status": "",
}
Then I use:
client.fields["id"] = 1;
client.fields["name"] = "matheus";
sqlite.rowInsert("insert into client(id, name)", client.fields.Keys.toList(), client.fields.Values.toList());
Just create a method inside your class and return a Map<String, dynamic>
class cliente {
int id;
String nome;
String apelido;
String sexo;
String status;
Map<String, dynamic> toJson() => {
'id': id,
'nome': nome,
'apelido': apelido,
'sexo': sexo,
'status': status,
};
}
And use it for example :
final dataObject = new client();
...fill your object
final jsonData = dataObject.toJson();
Also you can try using this package to avoid writing all of your fields : https://pub.dartlang.org/packages/json_serializable
I asked a question before about Dart encoding/decoding to JSON, however, the libraries that were suggested were not complete and I decided to manually handle that.
The objective is to convert these objects to a map.
class Parent extends Object {
int id;
String name;
List<Child> listChild = new List<Child>();
Map toMap() => {"id":id, "name":name, "listChild":listChild};
}
class Child extends Object {
int id;
String childName;
Map toMap() => {"id":id, "childName":childName};
}
When doing
print(JSON.encode(parent.toMap()));
I am seeing it go here, any suggestion how to make this work?
if (!stringifyJsonValue(object)) {
checkCycle(object);
try {
var customJson = _toEncodable(object);
if (!stringifyJsonValue(customJson)) {
throw new JsonUnsupportedObjectError(object);
}
_removeSeen(object);
} catch (e) {
throw new JsonUnsupportedObjectError(object, cause : e);
}
}
}
Map toMap() => {"id":id, "name":name: "listChild": listChild.map((c) => c.toJson().toList())};
is valid for JSON.
import 'dart:convert' show JSON;
...
String json = JSON.encode(toMap());
You can also use the toEncodeable callback - see How to convert DateTime object to json
If your class structure does not contain's any inner class then follow
class Data{
String name;
String type;
Map<String, dynamic> toJson() => {
'name': name,
'type': type
};
}
If your class uses inner class structure
class QuestionTag {
String name;
List<SubTags> listSubTags;
Map<String, dynamic> toJson() => {
'name': name,
'listSubTags': listSubTags.map((tag) => tag.toJson()).toList()
};
}
class SubTags {
String tagName;
String tagDesc;
SubTags(this.tagName, this.tagDesc);
Map<String, dynamic> toJson() => {
'tagName': tagName,
'tagDesc': tagDesc,
};
}
Just rename Map toMap() into Map toJson() and it will work fine. =)
void encode() {
Parent p = new Parent();
Child c1 = new Child();
c1 ..id = 1 ..childName = "Alex";
Child c2 = new Child();
c2 ..id = 2 ..childName = "John";
Child c3 = new Child();
c3 ..id = 3 ..childName = "Jane";
p ..id = 1 ..name = "Lisa" ..listChild = [c1,c2,c3];
String json = JSON.encode(p);
print(json);
}
class Parent extends Object {
int id;
String name;
List<Child> listChild = new List<Child>();
Map toJson() => {"id":id, "name":name, "listChild":listChild};
}
class Child extends Object {
int id;
String childName;
Map toJson() => {"id":id, "childName":childName};
}