我有以下代码演示问题:
@Component
public class App {
@Autowired S1 s1;
@Autowired S2 s2;
int jobs = 0;
@Scheduled(cron = "0 * * * * ?")
void foo() {
System.out.println("schedule cron job: " + jobs++);
Observable<String> observable = Observable.just("bar");
ConnectableObservable<String> publishedObservable = observable.publish();
publishedObservable.subscribe(s1);
publishedObservable.subscribe(s2);
publishedObservable.connect();
}
}
订阅者1:
@Component
public class S1 extends Subscriber<String> {
private AtomicInteger counter = new AtomicInteger(0);
@Override
public void onCompleted() {
}
@Override
public void onError(Throwable e) {
}
@Override
public void onNext(String s) {
System.out.println("S1:::: Times called: " + counter.getAndIncrement() + ", input: " + s);
}
}
订阅者2:
@Component
public class S2 extends Subscriber<String> {
private AtomicInteger counter = new AtomicInteger(0);
@Override
public void onCompleted() {
}
@Override
public void onError(Throwable e) {
}
@Override
public void onNext(String s) {
System.out.println("S2:::: Times called: " + counter.getAndIncrement() + ", input: " + s);
}
}
输出将是:
schedule cron job: 0
S1:::: Times called: 0, input: bar
S2:::: Times called: 0, input: bar
schedule cron job: 1
schedule cron job: 2
schedule cron job: 3
schedule cron job: 4
......
为什么每次调用 foo 方法时不调用 S1 和 S2 ?如何做到这一点?
这是因为 rx 一些订阅逻辑还是因为这些 bean 是单例的?