Question

我有一个Web应用程序，我使用Rx.js来处理事件流。该应用使用后端提供的rest api。

我主要订阅api电话，当请求完成后，我会渲染结果并重置其他控制状态（隐藏进度元素等）。

当身份验证令牌过期时，某些api次呼叫可能会失败，我必须让用户再次登录（显示登录弹出窗口等）。

我很好奇有没有办法恢复＆＃34;成功登录后的api呼叫流？用户无需提供其他操作来获取服务器响应的内容。

当前工作流程的原始示例：

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

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

Answer 1

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.

Rx.js和应用程序工作流程

1 个答案: