Typescript: Create interface based on a class

Question

I got a class called MyClass. It has a bunch of properties. I want to create an interface that contains the options of that class. Many options have the same name and typing than MyClass, but not all.

Edit: Most options properties are optional.

The goal is to duplicate the least possible.

Right now, I'm using a dummy object to avoid code duplication. Is there a cleaner solution than this?

class MyClass {
    callback:(p0:number,p1:string,p2:string[]) => number[];
    myAttr:number;
    composed:string;

    constructor(options:MyClassOptions){
        if(options.callback !== undefined)
            this.callback = options.callback;
        if(options.myAttr !== undefined)
            this.myAttr = options.myAttr;
        if(options.composedP1 !== undefined && options.composedP2 !== undefined)
            this.composed = options.composedP1 + options.composedP2;
    }
}

var dummy = <MyClass>null; 
interface MyClassOptions {
    callback?:typeof dummy.callback;
    myAttr?:typeof dummy.myAttr;
    composedP1:string;
    composedP2:string;
}

Ryan Cavanaugh · Accepted Answer · 2016-08-24 03:56:47Z

1

Many options have the same name and typing than MyClass, but not all.

If this is the case, there's not a good option.

If you want to have exactly the same members, and your class doesn't have any private or protected properties/methods, you can write

interface MyClassOptions extends MyClass {
  extraProp: string;
}

One thing you could do is make a base class with just the options you want to share, make the interface from there with the "real" class you use being a derivative of the base class.

answered Aug 24, 2016 at 3:56

Ryan Cavanaugh

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1 Comment

RainingChain Over a year ago

One thing I forgot to mention explicitly is that all the properties in MyClassOptions are optional. This solution makes all the properties non-optional.

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