Anyone have code for replicating a node's position?

Adobe XD UXP Plugin API
1

I need to create a scenegraph node of a different type that completely mimics another node’s position (rotation, translation, etc.).

Does anyone have any code for this, or any hints?

2

You can read every property of an existing node, create a new node with a different type, and apply those properties to the new node. I don’t have an example code for this. Other on the forum might…?

3

Yes, that’s what I’m trying, but that involves setting both rotation and placingInParent–you can’t just set the read-only translation matrix–, which together aren’t obvious how to use.

4

@stevekwak helped me greatly by fixing my broken code.

For the record, here’s the very useful resulting snippet.

// Make the new node correspond to the existing
// node's position and rotation, etc.
// Assumes both share their parent.
// We've already taken care of scale by using
// the original node's size as a basis for
// newNode creation, so the only components left
// are translation and rotation.
const makeNodeCongruentTo = (newNode, node) => {
  newNode.placeInParentCoordinates(
    {
      x: newNode.localBounds.x,
      y: newNode.localBounds.y,
    },
    { x: node.topLeftInParent.x, y: node.topLeftInParent.y }
  )
  if (newNode.rotation != node.rotation)
    newNode.rotateAround(node.rotation - newNode.rotation, {
      x: newNode.localBounds.x,
      y: newNode.localBounds.y,
    })
}