简体   繁体   中英

Rx.js and application workflow

I've got a web application where I'm using the Rx.js for handling event streams. The app uses a rest api provided by a backend.

Mostly I make a subscription for an api call and when a request is done I render results and reset other controls states (hide progress elements and so on).

Some api calls can be failed when an auth token is expired and I have to make a user to login again (show a login popup or so).

I'm curious is there a way to "restore" an api call stream after a successful login? Where a user has not to provide additional actions to get a server response.

Primitive example of the current workflow:

var apiCallStream = $.ajaxAsObservable(params):
apiCallStream.subscribe(
  result => renderResult(result),
  err => handleError(err));

function handleError(err) {
  if (err.xhr.error === 401) {
    LoginPopup();
  } else {
    ErrorPopup(err);
  }
}

Here some some (very rough pseudo-code), but you could do this with retryWhen :

// use Rx.DOM.get for an Observable Ajax GET
var source = Rx.DOM.get('/some/end/point')
  .retryWhen(function(errors) {
    // retryWhen: errors is a stream of errors
    // whatever observable you return, when it emits, the
    // observable you're operating on will be retried. (the entire thing)
    return errors.filter(function(e) { return e.status === 401; })
       // if status is 401, then tell the user to login 
      .flatMapLatest(function() { return doUserLogin; });
  });

// a contrived observable that shows a form and
// return a stream of successful logins
var doUserLogin = Observable.create(function(observer) {
  // show a form
  var form = $('.my-form').show();
  // when submit is fired on said form...
  return Rx.Observable.fromEvent(form, 'submit')
     // send a login over ajax
    .flatMap(e => Rx.DOM.post('some/url/login', form.serialize()))
});

Hopefully that gives you something to start with.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM