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how to add type information in Json serialization in Angular 7

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0

how to add type information in Json serialization in Angular 7

Hello everybody, I need to add type information at Json serialization in Angular 7 in order to get this result (adding the line "$type": "Person",):

 "$type": "Person",
 "id": 8,
 "firstName": "Anke",
 "lastName": "Winkler",
 …

I need this in order to deserialize in C# with Json.NET and knowing which type to use in case of properties that are typed with an interface and thus may be of different types. Thanks a lot for any advice!

json angular serialization

share|improve this question

asked Mar 24 at 8:41

FabianusFabianus

638

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The same way you add any other property to your JSON?
  
  – JB Nizet
  Mar 24 at 8:54
  

  
  
  
  

add a comment | 

0

how to add type information in Json serialization in Angular 7

Hello everybody, I need to add type information at Json serialization in Angular 7 in order to get this result (adding the line "$type": "Person",):

 "$type": "Person",
 "id": 8,
 "firstName": "Anke",
 "lastName": "Winkler",
 …

I need this in order to deserialize in C# with Json.NET and knowing which type to use in case of properties that are typed with an interface and thus may be of different types. Thanks a lot for any advice!

json angular serialization

share|improve this question

asked Mar 24 at 8:41

FabianusFabianus

638

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The same way you add any other property to your JSON?
  
  – JB Nizet
  Mar 24 at 8:54
  

  
  
  
  

add a comment | 

how to add type information in Json serialization in Angular 7

Hello everybody, I need to add type information at Json serialization in Angular 7 in order to get this result (adding the line "$type": "Person",):

 "$type": "Person",
 "id": 8,
 "firstName": "Anke",
 "lastName": "Winkler",
 …

I need this in order to deserialize in C# with Json.NET and knowing which type to use in case of properties that are typed with an interface and thus may be of different types. Thanks a lot for any advice!

json angular serialization

share|improve this question

asked Mar 24 at 8:41

FabianusFabianus

638

 "$type": "Person",
 "id": 8,
 "firstName": "Anke",
 "lastName": "Winkler",
 …

json angular serialization

json angular serialization

share|improve this question

asked Mar 24 at 8:41

FabianusFabianus

638

share|improve this question

asked Mar 24 at 8:41

FabianusFabianus

638

asked Mar 24 at 8:41

FabianusFabianus

638

asked Mar 24 at 8:41

FabianusFabianus

638

1 Answer
1

active

oldest

votes

3

It is not much an Angular serialization issue, but more a TypeScript/JavaScript issue.

You can customize your serialization by overriding the toJSON() method in your TypeScript classes.

class User 
 public id: number;
 public firstName: string;
 public lastName: string;
 public age: number;

 public toJSON(): User 
 return Object.assign(, this, 
 $type: 'User'
 );
 

What the toJSON() method will do, is simply create a new object using the current one, and the add the $type property to it. It will be called when you call the JSON.stringify() method. Thus, you don't need to create a $type variable in your class.

Example:

const newUser: User = new User();

newUser.id = 8;
newUser.firstName = "John";
newUser.lastName = "Doe";
newUser.age = 42;

const newUserAsJson: string = JSON.stringify(newUser);

console.log(newUserAsJson);
// Displays:
// firstName: "John", lastName: "Doe", id: 8, age: 42, $type: "User"

Hope it helps.

share|improve this answer

answered Mar 24 at 9:08

EastrallEastrall

911617

add a comment | 

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3

It is not much an Angular serialization issue, but more a TypeScript/JavaScript issue.

You can customize your serialization by overriding the toJSON() method in your TypeScript classes.

class User 
 public id: number;
 public firstName: string;
 public lastName: string;
 public age: number;

 public toJSON(): User 
 return Object.assign(, this, 
 $type: 'User'
 );
 

What the toJSON() method will do, is simply create a new object using the current one, and the add the $type property to it. It will be called when you call the JSON.stringify() method. Thus, you don't need to create a $type variable in your class.

Example:

const newUser: User = new User();

newUser.id = 8;
newUser.firstName = "John";
newUser.lastName = "Doe";
newUser.age = 42;

const newUserAsJson: string = JSON.stringify(newUser);

console.log(newUserAsJson);
// Displays:
// firstName: "John", lastName: "Doe", id: 8, age: 42, $type: "User"

Hope it helps.

share|improve this answer

answered Mar 24 at 9:08

EastrallEastrall

911617

add a comment | 

3

It is not much an Angular serialization issue, but more a TypeScript/JavaScript issue.

You can customize your serialization by overriding the toJSON() method in your TypeScript classes.

class User 
 public id: number;
 public firstName: string;
 public lastName: string;
 public age: number;

 public toJSON(): User 
 return Object.assign(, this, 
 $type: 'User'
 );
 

What the toJSON() method will do, is simply create a new object using the current one, and the add the $type property to it. It will be called when you call the JSON.stringify() method. Thus, you don't need to create a $type variable in your class.

Example:

const newUser: User = new User();

newUser.id = 8;
newUser.firstName = "John";
newUser.lastName = "Doe";
newUser.age = 42;

const newUserAsJson: string = JSON.stringify(newUser);

console.log(newUserAsJson);
// Displays:
// firstName: "John", lastName: "Doe", id: 8, age: 42, $type: "User"

Hope it helps.

share|improve this answer

answered Mar 24 at 9:08

EastrallEastrall

911617

add a comment | 

It is not much an Angular serialization issue, but more a TypeScript/JavaScript issue.

You can customize your serialization by overriding the toJSON() method in your TypeScript classes.

class User 
 public id: number;
 public firstName: string;
 public lastName: string;
 public age: number;

 public toJSON(): User 
 return Object.assign(, this, 
 $type: 'User'
 );
 

What the toJSON() method will do, is simply create a new object using the current one, and the add the $type property to it. It will be called when you call the JSON.stringify() method. Thus, you don't need to create a $type variable in your class.

Example:

const newUser: User = new User();

newUser.id = 8;
newUser.firstName = "John";
newUser.lastName = "Doe";
newUser.age = 42;

const newUserAsJson: string = JSON.stringify(newUser);

console.log(newUserAsJson);
// Displays:
// firstName: "John", lastName: "Doe", id: 8, age: 42, $type: "User"

Hope it helps.

share|improve this answer

answered Mar 24 at 9:08

EastrallEastrall

911617

class User 
 public id: number;
 public firstName: string;
 public lastName: string;
 public age: number;

 public toJSON(): User 
 return Object.assign(, this, 
 $type: 'User'
 );
 

const newUser: User = new User();

newUser.id = 8;
newUser.firstName = "John";
newUser.lastName = "Doe";
newUser.age = 42;

const newUserAsJson: string = JSON.stringify(newUser);

console.log(newUserAsJson);
// Displays:
// firstName: "John", lastName: "Doe", id: 8, age: 42, $type: "User"

share|improve this answer

answered Mar 24 at 9:08

EastrallEastrall

911617

answered Mar 24 at 9:08

EastrallEastrall

911617

answered Mar 24 at 9:08

EastrallEastrall

911617