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如何从setTimeout
返回可观察值?从setTimeout(Rxjs)返回可观察值
send(action):Observable<any>{
if(this.readyState === 0){
setTimeout(() => this.send(action), this.timeout);
}
else{
// observable is an Rxjs observable....
return this.observable$.take(1);
}
}
复制和粘贴例如:
let observable$ = Rx.Observable.fromArray([1, 2, 3, 4, 5]);
timeout = 40;
// if you switch this to 1 it works..
readyState = 0;
setTimeout(() => readyState = 1, 120);
send().subscribe(c => console.log(c));
function send(action){
if(readyState === 0){
setTimeout(() => send(action), timeout);
}
else{
return observable$.take(1);
}
}
给人的感觉并不权使用具有可观察性的'setTimeout'。 'delay'运算符怎么样? – Maxime