Angular 2: Get position of HTML element

Question

I'm trying to implement a custom directive in Angular 2 for moving an arbitrary HTML element around. So far everything is working except that I don't now how to get the initial position of the HTML element when I click on it and want to start moving. I'm binding to the top and left styles of my HTML element with those two host bindings:

/** current Y position of the native element. */
@HostBinding('style.top.px') public positionTop: number;

/** current X position of the native element. */
@HostBinding('style.left.px') protected positionLeft: number;

The problem is that both of them are undefined at the beginning. I can only update the values which will also update the HTML element but I cannot read it? Is that suppose to be that way? And if yes what alternative do I have to retrieve the current position of the HTML element.

Vassal

Answer 1

<div (click)="move()">xxx</div>

// get the host element
constructor(elRef:ElementRef) {}

move(ref: ElementRef) {
  console.log(this.elRef.nativeElement.offsetLeft);
}

See also https://mcmap.net/q/40543/-retrieve-the-position-x-y-of-an-html-element

Stichomythia

But offset doesn't give me the relative position of the element? – Vassal

Your question doesn't say anything about "relative". offsetLeft is just an example. The rest is the same as in plain JavaScript, this is why I added the link to my answer. – Woollen

Answer 2

In typeScript you can get the position as follows:

@ViewChild('ElementRefName') element: ElementRef;

const {x, y} = this.element.nativeElement.getBoundingClientRect();

Comfit

This solution won't work when the browser is not at 100% size – Decile

What will work when the browser is not at 100% size? – Ji

@ViewChild('ElementRefName') element: ElementRef; is an error now in Angular 13: error TS2564: Property 'element' has no initializer and is not definitely assigned in the constructor. – Copaiba

@b264: that's just formality. You can solve it like that: @ViewChild('ElementRefName') element!: ElementRef; or @ViewChild('ElementRefName') element: ElementRef|undefined; – Rata

Answer 3

in html:

<div (click)="getPosition($event)">xxx</div>

in typescript:

getPosition(event){
    let offsetLeft = 0;
    let offsetTop = 0;

    let el = event.srcElement;

    while(el){
        offsetLeft += el.offsetLeft;
        offsetTop += el.offsetTop;
        el = el.parentElement;
    }
    return { offsetTop:offsetTop , offsetLeft:offsetLeft }
}

Stop

Testing this in an Angular app now and it's giving odd values (very high numbers). I don't believe a child's offset is the simple aggregation of all of it's parents offsets. I believe the correct property is offsetParent. – Crocodilian

@Crocodilian Yes, I'm using offsetParent instead of parentElement and it appears to be working. – Flasket