TypeScript dynamic property type on class

Question

There is a way to declare the type of dynamic class property if it's accessed by [] operator, like on the following example:

class Foo {
  [key: string]: number;
}

let a = new Foo();
let b = a['bar']; //Here, the compiler knows that b is a number

But is there a way to declare the same thing, without [] operator? A way to write this:

let a = new Foo();
let b = a.someProperty;

And letting TypeScript knows that someProperty is of type number, because we say to it: All unknown properties on Foo are of type number.

Amid · Accepted Answer · 2016-03-03 10:57:47Z

2

I do not think its is possible. When you define class you define 'static' information about its properties and methods. If you specify indexer - it means just that - that the objects of the class will have indexer, not any properties. Thats what classes are for after all - to define structure of your business entities.

The only way I am aware of doing something similar to what you want is by using object literals. For example this will work:

let x: { PropA: number, [x: string]: number };
x = { PropA: 1, PropX: 2, PropY: 3, PropZ: 4 };

Hope this helps.

answered Mar 3, 2016 at 10:57

Amid

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1 Comment

Blackus Over a year ago

I totally agree with your point, classes are here to define a structure and I'm not surprised it's not possible, because it doesn't make a lot of sense. By the way, your suggestion about object literals might help, because at some point during my application lifetime, I have information about these dynamic properties, so maybe I could do something hackish with that. Thanks!

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