如何根据角度2中的按钮单击移动div滚动位置

在 app.component.html 文件中，我有一个带有水平滚动的 div 元素和两个按钮，即“下一个”和“上一个”。根据这些按钮单击我想移动滚动。

应用程序组件.html

<div id="th-infinite-scroll-tracker" style="overflow-y:scroll; height: 200px;" scrollTracker (scroll)="onScroll($event)" (mouseup)="searchLog($event)">
        <div *ngFor="let log of arr; let i = index" innerHTML="{{log}}" [id]="i"></div>
</div>
<button (click)="onPreviousSearchPosition()" [disabled]="disablePreviousBtn">Previous</button>
<button (click)="onNextSearchPosition()" [disabled]="disableNextBtn">Next</button>

应用程序组件.ts

@HostListener('scroll', ['$event'])
onScroll(event){
    this.scrollObject = event;
}
onPreviousSearchPosition(){
    this.disableNextBtn = false;

    this.scrollObject.target.scrollTop = 20 * --this.idCount;
  }
onPreviousNextPosition(){
    this.disableNextBtn = false;

    this.scrollObject.target.scrollTop = 20 * ++this.idCount;
  }

使用上面的代码我们可以移动滚动，但我们需要手动滚动后将获得的滚动事件对象。 请建议我如何在组件类中创建滚动事件对象

这是滚动 div 元素的方式 -https://plnkr.co/edit/2v0iVgkOZfISqlFAkrNX?p=preview

example:

import { Component, ViewChild, ElementRef } from '@angular/core';
@Component({
  selector: 'my-app',
  template: `
    <div #panel style="overflow-y:scroll; height: 20px;" >
      <div *ngFor="let log of arr; let i = index" innerHTML="{{log}}" [id]="i"></div>
    </div>
    <button (click)="onPreviousSearchPosition()">Previous</button>
    <button (click)="onNextSearchPosition()">Next</button>
  `
})
export class AppComponent { 
  public arr = ['foo', 'bar', 'baz'];
  @ViewChild('panel', { read: ElementRef }) public panel: ElementRef<any>;

  public onPreviousSearchPosition(): void {
    this.panel.nativeElement.scrollTop -= 20;
  }

  public onNextSearchPosition(): void {
    this.panel.nativeElement.scrollTop += 20;
  }
}