How to make always on top ?
-
Is there a way in QML to specify that an image when visible will always be on top ?
I have images that represent when a data item is stale (stopped being updated for a specific period of time). These images will be made visible when the stale state is true, is there a way that along with setting
visible
totrue
I can also make sure that the image appears on top? -
@SPlatten yes and no
there's
z
property which influences the stacking order of siblings. By default it's 0 so everything is drawn from top (qml file) to bottom.You can set one z to 1 and it will be drawn on top of every sibling. but there is not automatic to top of stacking order.
If you end up with multiple siblings with z = 1 you end up with top to bottom order again.
-
Is there a way in QML to specify that an image when visible will always be on top ?
I have images that represent when a data item is stale (stopped being updated for a specific period of time). These images will be made visible when the stale state is true, is there a way that along with setting
visible
totrue
I can also make sure that the image appears on top? -
@J-Hilk , found this:
https://stackoverflow.com/questions/68707402/how-can-i-raise-a-qml-widget-to-the-top
Is there any other way of performing the same as
raise
inQML
? -
@SPlatten yes and no
there's
z
property which influences the stacking order of siblings. By default it's 0 so everything is drawn from top (qml file) to bottom.You can set one z to 1 and it will be drawn on top of every sibling. but there is not automatic to top of stacking order.
If you end up with multiple siblings with z = 1 you end up with top to bottom order again.
-
@SPlatten yes and no
there's
z
property which influences the stacking order of siblings. By default it's 0 so everything is drawn from top (qml file) to bottom.You can set one z to 1 and it will be drawn on top of every sibling. but there is not automatic to top of stacking order.
If you end up with multiple siblings with z = 1 you end up with top to bottom order again.