How do I use the generic argument as an option in typescript

Issue

I have this code:

private descargar<T>(url: string, type: "arraybuffer" | "document" | "blob" | "json" | "text"): Promise<T | null>{
            return new Promise(function (resolve) {
                // Get file name from url.
                var xhr = new XMLHttpRequest();
                xhr.responseType = "blob"
                xhr.onload = function () {                              
                    resolve(xhr.response);
                    return;
                };
                xhr.onerror = ()=>resolve(null);
                xhr.open("GET", url);
                xhr.send();
            })
        }

And it gets called like this:

const response = await this.descargar<Blob>(url, "blob")

It's redundant because you need to specify Blob twice. What's the alternative?

Solution

Say you define a type that maps the string that might be supplied to the type parameter to its associated type:

type DescargarMap = {
  arraybuffer: ArrayBuffer;
  document: Document;
  blob: Blob;
  json: Object;
  text: string;
}

(I've taken a best-guess at how these strings map to types... may need fixing)

now, using this map type, you can define your function as follows:

descargar<T extends keyof DescargarMap>(
  url: string, 
  type: T): Promise<DescargarMap[T] | null>{...}

and call it

// `buf` is `ArrayBuffer | null`
const buf = await descargar("", "arraybuffer");

Now, appropriate options for the type parameter are shown in code-completion hints of the IDE and the return-type is correctly mapped without having to supply the generic parameter.

Playground Link

Answered By - spender