React Native: Getting the position of an element

JavascriptIosReactjsReact Native

Javascript Problem Overview


I am styling an Image component with flexbox to be in the center of the screen which works pretty well. Now I want a second Image component to be displayed directly on the top of the first one. The second image is using absolute positioning. Currently I'm just guessing pixels so that it fits, but of course this is not accurate and way too much maintainability effort.

I am pretty much looking for the React Native equivalent of jQuery's .offset(). Is there such a thing and if there isn't what's the best way to achieve this?

Javascript Solutions


Solution 1 - Javascript

React Native provides a .measure(...) method which takes a callback and calls it with the offsets and width/height of a component:

myComponent.measure( (fx, fy, width, height, px, py) => {

    console.log('Component width is: ' + width)
    console.log('Component height is: ' + height)
    console.log('X offset to frame: ' + fx)
    console.log('Y offset to frame: ' + fy)
    console.log('X offset to page: ' + px)
    console.log('Y offset to page: ' + py)
})

##Example...

The following calculates the layout of a custom component after it is rendered:

class MyComponent extends React.Component {
    render() {
        return <View ref={view => { this.myComponent = view; }} />
    }
    componentDidMount() {
        // Print component dimensions to console
        this.myComponent.measure( (fx, fy, width, height, px, py) => {
            console.log('Component width is: ' + width)
            console.log('Component height is: ' + height)
            console.log('X offset to frame: ' + fx)
            console.log('Y offset to frame: ' + fy)
            console.log('X offset to page: ' + px)
            console.log('Y offset to page: ' + py)
        })        
    }
}

##Bug notes

  • Note that sometimes the component does not finish rendering before componentDidMount() is called. If you are getting zeros as a result from measure(...), then wrapping it in a setTimeout should solve the problem, i.e.:

      setTimeout( myComponent.measure(...), 0 )
    

Solution 2 - Javascript

You can use onLayout to get the width, height, and relative-to-parent position of a component at the earliest moment that they're available:

<View
  onLayout={event => {
    const layout = event.nativeEvent.layout;
    console.log('height:', layout.height);
    console.log('width:', layout.width);
    console.log('x:', layout.x);
    console.log('y:', layout.y);
  }}
>

Compared to using .measure() as shown in the accepted answer, this has the advantage that you'll never have to fiddle around deferring your .measure() calls with setTimeout to make sure that the measurements are available, but the disadvantage that it doesn't give you offsets relative to the entire page, only ones relative to the element's parent.

Solution 3 - Javascript

I had a similar problem and solved it by combining the answers above

class FeedPost extends React.Component {
  constructor(props) {
    ...
    this.handleLayoutChange = this.handleLayoutChange.bind(this);
  }


handleLayoutChange() {
    this.feedPost.measure( (fx, fy, width, height, px, py) => {
      console.log('Component width is: ' + width)
      console.log('Component height is: ' + height)
      console.log('X offset to page: ' + px)
      console.log('Y offset to page: ' + py)
    })
  }

  render {
    return(
      <View onLayout={(event) => {this.handleLayoutChange(event) }} 
      ref={view => { this.feedPost = view; }} >
...

Now I can see the position of my feedPost element in the logs:

08-24 11:15:36.838  3727 27838 I ReactNativeJS: Component width is: 156
08-24 11:15:36.838  3727 27838 I ReactNativeJS: Component height is: 206
08-24 11:15:36.838  3727 27838 I ReactNativeJS: X offset to page: 188
08-24 11:15:36.838  3727 27838 I ReactNativeJS: Y offset to page: 870

Solution 4 - Javascript

I needed to find the position of an element inside a ListView and used this snippet that works kind of like .offset:

const UIManager = require('NativeModules').UIManager;
const handle = React.findNodeHandle(this.refs.myElement);
UIManager.measureLayoutRelativeToParent(
  handle, 
  (e) => {console.error(e)}, 
  (x, y, w, h) => {
    console.log('offset', x, y, w, h);
  });

This assumes I had a ref='myElement' on my component.

Solution 5 - Javascript

If you use function components and don't want to use a forwardRef to measure your component's absolute layout, you can get a reference to it from the LayoutChangeEvent in the onLayout callback.

This way, you can get the absolute position of the element:

<MyFunctionComp
  onLayout={(event) => {
    event.target.measure(
      (x, y, width, height, pageX, pageY) => {
        doSomethingWithAbsolutePosition({
          x: x + pageX, 
          y: y + pageY,
        });
      },
    );
  }}
/>

Tested with React Native 0.63.3.

Solution 6 - Javascript

This seems to have changed in the latest version of React Native when using refs to calculate.

Declare refs this way.

  <View
    ref={(image) => {
    this._image = image
  }}>

And find the value this way.

  _measure = () => {
    this._image._component.measure((width, height, px, py, fx, fy) => {
      const location = {
        fx: fx,
        fy: fy,
        px: px,
        py: py,
        width: width,
        height: height
      }
      console.log(location)
    })
  }

Solution 7 - Javascript

There is a measureInWindow property on the ref argument object that can be used like so:

const [offset, setOffset] = React.useState();

<View ref={(view) =>
    if(!view) return;
	view.measureInWindow((x, y) => {
		setOffset({ x, y });
	})
}>
</View>

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJohannes SteinView Question on Stackoverflow
Solution 1 - JavascripttohsterView Answer on Stackoverflow
Solution 2 - JavascriptMark AmeryView Answer on Stackoverflow
Solution 3 - JavascriptMislavoo7View Answer on Stackoverflow
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Solution 7 - JavascriptSkyler KnightView Answer on Stackoverflow