返回空的 Observable 时不调用订阅回调

Question

我希望在返回空的 Observable 时调用subscribe回调，类似于Promise.resolve([]) ：

import { EMPTY } from 'rxjs';

function funcToTest(): Observable<any[]> {
  return EMPTY;
};

test('returns empty array', (done) => {
  funcToTest().subscribe(() => {
    done();
  });
});

相反，它返回此错误：

Error: Timeout - Async function did not complete within 5000ms (set by jasmine.DEFAULT_TIMEOUT_INTERVAL)

Answer 1

Observables 的行为与 promises 不同。 RxJS EMPTY Observable 不会调用“成功”回调，只会调用“完成”回调。 done函数应该在“完成”而不是“成功”中调用：

funcToTest().subscribe({
  success()  { /* Called when 'x' is returned, e.g. after the subscriber calls 'next' */ },
  error(err) { /* Called on an error. */ },
  complete() {
    /* Called after the subscriber calls 'complete'. No more values can be returned */
    done();
  }
});

请参阅文档中的示例： https : //rxjs-dev.firebaseapp.com/guide/observable