Does TypeScript support constructor with object initializer？

Question

For example, in c#, we have this snippet:

 class OrderLine
    {
        public OrderLine(){
          OrderId ="1",
          ItemTitle ="2"
        Console.WriteLine("constructing");
        }
        public string Platform { get; set; }
        public string OrderId { get; set; }
        public string ItemTitle { get; set; }
    }

    public static void Main()
    {
        var ol = new OrderLine 
        {
                Platform ="p"
        };
        Console.WriteLine(ol.Platform); 
    }

In Typescript, if I use {} to initialize an object, I won't be able to call the constrcuor which can give me some default value and do something else. if I use new key word to do some default construction, my compiler doesn't allow me to use object initializer(I am using angular 6), and I have to call ol.Platform to assign property value. When I have several properties to set value, writing multiple "ol." is not as fast as to use the object initializer syntax. is there a better way?

CertainPerformance · Accepted Answer · 2020-04-16 21:58:53Z

3

You can have the constructor accept an object for which each of its properties get assigned to the instance with Object.assign:

type Init = {
    age: number;
    gender: string;
};
class Foo {
    public name: string;
    public age: number | undefined;
    public gender: string | undefined;
    constructor(name: string, init?: Partial<Init>) {
        this.name = name;
        if (init) {
          Object.assign(this, init);
        }
    }
}
const f = new Foo('bob', { age: 10 });

answered Apr 16, 2020 at 21:58

CertainPerformance

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2 Comments

yww325 Over a year ago

it's tricky, but for my large model with dozens of property, this works. But I prefer not to define a Partial Init type, can I use any here?

CertainPerformance Over a year ago

You can always use any to get around having to do type-checking, but it's usually not a good idea, because then you might as well not be using Typescript at all for that point in the code. If you use any, then the keys and values of the new instance could be literally anything, and you lose all of the type safety Typescript was made to help you with.

qujck · Accepted Answer · 2022-04-27 12:09:38Z

0

Adding this to the original answer ...

You can define init as the same type, i.e. Partial<Foo>

@RoutingKey('abc')
class Foo {
    public name: string | undefined;
    public age: number | undefined;
    public gender: string | undefined;
    constructor(init?: Partial<Foo>) {
        if (init) {
            Object.assign(this, init);
        }
    }
}

const foo = new Foo({ name: 'something' });

This was useful to me as I had to call new to get the constructor decorator @RoutingKey() to fire

edited Apr 27, 2022 at 12:09

answered Apr 27, 2022 at 12:02

qujck

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