在应用程序中,我需要在输入去抖动时发送一个事件。
我正在尝试这个:
@ViewChild('messageInput') messageInput: ElementRef;
private inputTimeOutObservable: any;
setTypingTimeOut(){
this.inputTimeOutObservable = Observable.fromEvent(this.messageInput.nativeElement, 'input')
.map((event: Event) => (<HTMLInputElement>event.target).value)
.debounceTime(1000)
.distinctUntilChanged()
.subscribe(data => {
this.sendEvent();
});
}
ngOnDestroy() {
if (this.inputTimeOutObservable) {
this.inputTimeOutObservable.unsubscribe();
this.inputTimeOutObservable = null;
}
}
输入做其他事情但这里是:
<input #messageInput id="message" type="text" (ngModelChange)="inputTextChanges($event)"
[(ngModel)]="messageValue" (keyup)='keyUp.next($event)'>
事件没有触发,我在这里不明白为什么。有什么想法吗?
最佳答案
根据您的代码,这是一个工作示例:
component.html
<input #messageInput id="message" type="text" [(ngModel)]="messageValue">
component.ts
import {Component, ElementRef, OnInit, ViewChild} from '@angular/core';
import {Observable} from "rxjs/Rx";
@Component({
selector: 'app-input-debounce',
templateUrl: './input-debounce.component.html',
styleUrls: ['./input-debounce.component.css']
})
export class InputDebounceComponent implements OnInit {
@ViewChild('messageInput') messageInput: ElementRef;
public messageValue: string = "";
private inputTimeOutObservable: any;
constructor() {
}
ngOnInit() {
}
ngAfterViewInit() {
this.setTypingTimeOut();
}
setTypingTimeOut() {
this.inputTimeOutObservable = Observable.fromEvent(this.messageInput.nativeElement, 'input')
.map((event: Event) => (<HTMLInputElement>event.target).value)
.debounceTime(1000)
.distinctUntilChanged()
.subscribe(data => {
console.log(this.messageValue);
console.timeEnd("Input changed after."); // No matter how frequent you type, this will always be > 1000ms
console.time("Input changed after."); // start track input change fire time
});
}
}
关于javascript - Angular 为 2 的输入去抖动,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48992384/