fromJson

Learn to create objects with fromJson.

The fromJson factory

fromJson is a factory method that builds an instance from a JSON Map. It provides type-safe access; safer than dynamic.

Example: factory User.fromJson(Map<String, dynamic> m) => User(m['name'] as String, m['age'] as int);. Each JSON field is cast.

Usage: var u = User.fromJson(jsonDecode(s) as Map<String, dynamic>);. Now u.name works with type safety.

Null-safe fromJson

JSON fields may be missing; therefore fromJson is written null-safe. m['name'] as String? makes it nullable, or a default is given: m['name'] ?? 'guest'.

For numeric fields, (m['age'] as num).toInt() is safe; the int/double distinction in JSON can be ambiguous.

For nested objects, the inner class's fromJson is called: Address.fromJson(m['address'] as Map<String, dynamic>).

Exercise Define User.fromJson

Define class User { String name; int age; User(this.name, this.age); factory User.fromJson(Map<String, dynamic> m) => User(m['name'] as String, m['age'] as int); }. Inside main add var u = User.fromJson({'name': 'Ayşe', 'age': 30}); and print(u.name);.

Show example solution

class User {
  String name;
  int age;
  User(this.name, this.age);
  factory User.fromJson(Map<String, dynamic> m) =>
      User(m['name'] as String, m['age'] as int);
}

void main() {
  var u = User.fromJson({'name': 'Ayşe', 'age': 30});
  print(u.name);
}

Test yourself

Match the fromJson parts

factory

Factory keyword
Map<String, dynamic>

Input type
as String

Type cast
User(...)

Instance creation
What does fromJson return?
- String
- Map
- Class instance
- Future
Show answer

Class instance — fromJson returns a class instance.

Type cast

m['name'] ____ String

Show answer

m['name'] as String

Next lesson: toJson

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