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

cartant's answer above works well, and answers the question that was asked (How can I define this in Typescript, so that I could import rxjs/Observable and this operator, like I do with rxjs operators?)

I recently discovered the let operator which if you don't actually need to have the function implemented as an operator, will still let you DRY up your code.

I was starting on implementing an angular 2 service to interface with my rails backend and knew that most of my api calls would look very similar so I wanted to try and put as much of the common stuff in a function.

Almost all the calls will do the following:

1. retry on an error (my function below needs more work on that front)
2. map the http response into a typescript locally defined class (via json-typescript-mapper)
3. handle errors

Here is an example of my use the let operator to my http responses through a common function (handleResponse) via the rxjs let operator.

  handleResponse({klass, retries=0} :{klass:any,retries?:number }) : (source: Observable) => Observable {
    return (source: Observable)  : Observable => {
      return source.retry(retries)
            .map( (res) => this.processResponse(klass,res))
            .catch( (res) => this.handleError(res));
    } 
  } 

  processResponse(klass, response: Response) {
    return deserialize(klass, response.json());
  }

  handleError(res: Response) {
    const error = new RailsBackendError(res.status, res.statusText);
    return  Observable.throw(error);
  }

  getUserList({page=1,perPage=30,retry=0}: { page?:number, perPage?:number, retry?:number }={}) : Observable {
    const requestURL = `/api/v1/users/?${this.apiTokenQueryString}&page=${page}&per_page=${perPage}`;
    return this.http.get(requestURL).let(this.handleResponse({klass: UserList}));
  }