Reputation: 1600
On the Windows Form application I have a Lamp image (a black and white one, and a bright one. For OFF and ON respectively).
Using the Button how can I achieve the scenario such that same button will turn the property of the image (pictureBox in my case) to show the Lamp as ON and pressing the same button again will turn the Lamp off.
I am accessing the 'Visible' property of picture box.
Upvotes: 2
Views: 10446
Reputation: 11
A bit late to the party, but you can use a checkbox and set the appearance to button. I think that would do what is expected by the original post.
Upvotes: 1
Reputation: 766
I'm not sure about the way to put 2 images over each other, but if you want to reach the same effect:
this would be in the click event:
Properties.Settings.Default.IsOptimizedForTracer !=Properties.Settings.Default.IsOptimizedForTracer;
if (!Properties.Settings.Default.IsOptimizedForTracer)
{
btnOptimizeForTracer.Image = Properties.Resources.TracerOFF;
return;
}
btnOptimizeForTracer.Image = Properties.Resources.TracerON;
Upvotes: 0
Reputation: 69280
Put two images on top of each other and get the button to switch which one of them is enabled.
In the form designer you make one of them visible and the other non-visible. The code in the button handler can then be something like:
lightImage.Visible = !lightImage.Visible;
darkImage.Visible = != lightImage.Visible;
That will swap which one is visible and eliminate the need to keep state elsewhere.
Upvotes: 3