0

If a component has multiple elements, the triggered hover state of child A seems to remain active until the mouse is removed from the scope of the entire component (its boundary), not just from that child.

enter image description here

6
  • I'm not an XD user, but I know web building.... I really don't understand this question at all... and I've read everything multiple times. Are you wanting the child element (presumably the white rectangle on the right) to also activate the hover state? -- What you've essentially written, as I read it, is... "When I move my mouse over things the hover state is on, when I move the mouse away, hover is off" which is how hover is supposed to work.
    – Scott
    Sep 20, 2020 at 20:21
  • @Scott i think they want the hover to be stateful, omce you hover keep state 1 but odnt kiill state on exit but clear it onb exit of other element
    – joojaa
    Sep 21, 2020 at 9:58
  • The component is composed of three elements: thr brown rectangle, and top1 and bottom white. The goal was to trigger the top one to change to black in hover, but go back to normal when the mouse leaves that top square (because the trigger is only set to the top one). However, what is happening is that if I remove th4 mouse from the top square but no from the area of the brown square, the hover state remains active, until my mouse leaves the entire component area. I want the state to stop the momnet the mouse leaves the top mini rectangle. Sep 21, 2020 at 21:05
  • Continued: I want the state to stop the momnet the mouse leaves the top mini rectangle. I think xd considers the state of the entire component rather than any ofv its children for hovers, even though only part of the component is changed on that state. Sep 21, 2020 at 21:15
  • Continued: I think the only solution to my particular situation is to essentially make the top white and bottom white rectangles be components themselves. Thus the parent component will have two children, and that way the mouseOut behavior will work. Sep 27, 2020 at 2:02

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.