Combine RxJS operators into new operator using TypeScript

Question

I frequently find my self adding the same sequence of operators to observables, e.g.

observable$
  .do(x => console.log(\'some text\', x))
  .publishRepla         


        

Answer 1

To implement the operator you have described, create a cache.ts file with the following content:

import { Observable } from "rxjs/Observable";
import "rxjs/add/operator/do";
import "rxjs/add/operator/publishReplay";

// Compose the operator:

function cache(this: Observable, text: string): Observable {
  return this
    .do(x => console.log(text, x))
    .publishReplay()
    .refCount();
}

// Add the operator to the Observable prototype:

Observable.prototype.cache = cache;

// Extend the TypeScript interface for Observable to include the operator:

declare module "rxjs/Observable" {
  interface Observable {
    cache: typeof cache;
  }
}

And consume it like this:

import { Observable } from "rxjs/Observable";
import "rxjs/add/observable/of";
import "./cache";

let cached = Observable.of(1).cache("some text");
cached.subscribe(x => console.log(x));