angular - @ViewChild in *ngIf

Question

Question

What is the most elegant way to get @ViewChild after corresponding element in template was shown?

Below is an example. Also Plunker available.

Component.template.html:

<div id="layout" *ngIf="display">
  <div #contentPlaceholder></div>
</div>

Component.component.ts:

export class AppComponent {

    display = false;
    @ViewChild('contentPlaceholder', { read: ViewContainerRef }) viewContainerRef;

    show() {
        this.display = true;
        console.log(this.viewContainerRef); // undefined
        setTimeout(() => {
            console.log(this.viewContainerRef); // OK
        }, 1);
    }
}

I have a component with its contents hidden by default. When someone calls show() method it becomes visible. However, before Angular 2 change detection completes, I can not reference to viewContainerRef. I usually wrap all required actions into setTimeout(()=>{},1) as shown above. Is there a more correct way?

I know there is an option with ngAfterViewChecked, but it causes too much useless calls.

ANSWER (Plunker)

Answer 1

Use a setter for the ViewChild:

 private contentPlaceholder: ElementRef;

 @ViewChild('contentPlaceholder') set content(content: ElementRef) {
    if(content) { // initially setter gets called with undefined
        this.contentPlaceholder = content;
    }
 }

The setter is called with an element reference once *ngIf becomes true.

Note, for Angular 8 you have to make sure to set { static: false }, which is a default setting in other Angular versions:

 @ViewChild('contentPlaceholder', { static: false })

Note: if contentPlaceholder is a component you can change ElementRef to your component Class:

  private contentPlaceholder: MyCustomComponent;

  @ViewChild('contentPlaceholder') set content(content: MyCustomComponent) {
     if(content) { // initially setter gets called with undefined
          this.contentPlaceholder = content;
     }
  }