Question

我为test forkJoin制作了这段代码，但它不起作用！你能检查一下是什么问题吗？

const observables = [];

observables.push(new Observable(subscriber => subscriber.next('Hello')));
observables.push(new Observable(subscriber => subscriber.next(' ')));
observables.push(new Observable(subscriber => subscriber.next('World') ));
observables.push(new Observable(subscriber => subscriber.next('!')));

forkJoin(observables).subscribe(word => console.log(word.join('')));

Answer 1

试试这个

const observables = [];

observables.push(Observable.of('Hello'));
observables.push(Observable.of(' '));
observables.push(Observable.of('World'));
observables.push(Observable.of('!'));

Observable.forkJoin(observables).subscribe(word => console.log(word.join('')));

Answer 2

重要的是完成forkJoin工作的observable，就像示例一样！

const observables = [];

observables.push(new Observable(subscriber => {
  subscriber.next('Hello');
  subscriber.complete();
}));
observables.push(new Observable(subscriber => {
  subscriber.next(' ');
  subscriber.complete();
}));
observables.push(new Observable(subscriber => {
  subscriber.next('World');
  subscriber.complete();
}));
observables.push(new Observable(subscriber => {
  subscriber.next('!');
  subscriber.complete();
}));
forkJoin(observables).subscribe(word => console.log(word.join('')));

ForkJoin角度问题

2 个答案: