TS Serializer

> **TS Serializer** provides `TypeScript` decorators to help developers with serializing/deserializing TypeScript classes into and from JSON objects. ## Installation ### Using npm: `npm install --save ts-serializer` ## Usage You can find a [codesandbox playground here](https://codesandbox.io/s/ts-serializer-playground-gdbid)! ### TypeScript #### Deserialization: ```TypeScript import {Serialize, SerializeProperty, Serializable} from 'ts-serializer'; @Serialize({}) class MyClass extends Serializable { @SerializeProperty({ map: 'username' }) public name:string; } let instance:MyClass = new MyClass(); console.log(instance.name); // Prints `undefined` instance.deserialize({ username: 'Some Value' }); console.log(instance.name); // Prints 'Some Value' ``` --- #### Serialization: ```TypeScript import {Serialize, SerializeProperty, Serializable} from 'ts-serializer'; @Serialize({}) class MyClass extends Serializable { @SerializeProperty({ map: 'username' }) public name:string; } let instance:MyClass = new MyClass(); instance.name = 'Some Value'; console.log(instance.serialize()); // Prints {username:'Some Value'} ``` ### JavaScript > **Note:** Although the library was designed to be used as a decorator in TypeScript, it doesn't matter that it can't be used in plain old JavaScript. The syntax can be a little messy but the result is the same. #### Deserialization: ```JavaScript var Serialize = TSerializer.Serialize; var SerializeProperty = TSerializer.SerializeProperty; function MyClass(){ this.name; } Serialize({})(MyClass); SerializeProperty({ map: 'username' })(MyClass.prototype, 'name'); var instance = new MyClass(); console.log(instance.name); // Prints `undefined` instance.deserialize({ username: 'Some Value' }); console.log(instance.name); // Prints 'Some Value' ``` --- #### Serialization: ```JavaScript var Serialize = TSerializer.Serialize; var SerializeProperty = TSerializer.SerializeProperty; function MyClass(){ this.name = 'Some Value'; } Serialize({})(MyClass); SerializeProperty({ map: 'username' })(MyClass.prototype, 'name'); var instance = new MyClass(); console.log(instance.serialize()); // Prints {username:'Some Value'} ``` ## Library Options The library allows you to pass different serialization/deserialization options both on class level and on property level. ### Class Options #### root When you want to deserialize just a child object from the JSON you can use the `root` option. ```TypeScript @Serialize({ root: 'childObject' }) class MyClass extends Serializable { @SerializeProperty({}) public name:string; } let instance:MyClass = new MyClass(); instance.deserialize({ childObject: { name: 'Some Value' } }); console.log(instance.name); // Prints 'Some Value' ``` ### Property Options #### root The `root` option can also be used on a property. > **Note:** If `root` is already specified at class level the value is inherited to all class properties. If you want to override this, you can use hte `.` value. In this case, the property will be mapped up one level. ```TypeScript @Serialize({}) class MyClass extends Serializable { @SerializeProperty({ root: 'childObject' }) public name:string; } let instance:MyClass = new MyClass(); instance.deserialize({ childObject: { name: 'Some Value' } }); console.log(instance.name); // Prints 'Some Value' ``` --- #### map When the property name in the JSON doesn't match with your class properties, the `map` option can be used. This option maps a property from the JSON with a different property from your class. ```TypeScript @Serialize({}) class MyClass extends Serializable { @SerializeProperty({ map: 'username' }) public name:string; } let instance:MyClass = new MyClass(); instance.deserialize({ username: 'Some Value' }); console.log(instance.name); // Prints 'Some Value' ``` --- #### list The `list` option can be used when the JSON property value is a list of items. ```TypeScript @Serialize({}) class MyClass extends Serializable { @SerializeProperty({ list: true }) public items:string[]; } let instance:MyClass = new MyClass(); instance.deserialize({ items: ['a', 'b', 'c'] }); console.log(instance.items); // Prints ['a', 'b', 'c'] ``` --- #### type When you want to use non-primitive types for deserialization use the `type` option. > **Note:** The `type` object should also be a `Serializable` object. ```TypeScript @Serialize({}) class User extends Serializable { @SerializeProperty({}) public firstName:string; @SerializeProperty({}) public lastName:string; } @Serialize({}) class Profile extends Serializable { @SerializeProperty({ type: User }) public user:User; } let instance:Profile = new Profile(); instance.deserialize({ user: { firstName: 'John', lastName: 'Doe' } }); console.log(instance.user.firstName); // Prints 'John' console.log(instance.user.lastName); // Prints 'Doe' ``` #### optional The `optional` option can be used when the property or the value may not exist. ```TypeScript @Serialize({}) class User extends Serializable { @SerializeProperty({}) public name:string; @SerializeProperty({ optional: true }) public age:number; } @Serialize({}) class Profile extends Serializable { @SerializeProperty({ type: User }) public user:User; } let instance:Profile = new Profile(); instance.deserialize({ user: { firstName: 'John', } }); console.log(instance.user.firstName); // Prints 'John' console.log(instance.user.age); // Prints 'null' ``` # Contribute You can help improving this project sending PRs and helping with issues.

	Stars	Issues	Version	Updated	Created
ngrx-json-api			2.3.2	5 years ago	8 years ago
ts-jsonapi			2.1.3	4 years ago	8 years ago
ts-serializable			3.0.56	11 days ago	7 years ago
ts-serializer	20	19	2.2.6	4 years ago	8 years ago

ngrx-json-api vs ts-jsonapi vs ts-serializable vs ts-serializer

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