Well, right now, I’m on my father’s computer. Of course his computer is completely different from mine (I have a bunch of display and optimization tweaks), but I wanna know what influences how I right click a program AND not switch to it for display.
I’m assuming it would have something to do with the registry values, because I searched Display settings and Taskbar properties (in addition to Folder Options) and I do not find a spot for it.
It’s just curiosity, because if I find that my computer behavior changes, I would like to know how to change it back !! Thanks a lot, guys !!
Now, let’s say you have 2 programs open and showing up on the taskbar. I’m active with one of those programs. How can I right click the other program without having to switch to it? (Or am I just imagining things and this can’t be done?)
Effectively you can’t as when you right click on the taskbar item the menu pops-up and if you choose any option, restore, for instance the program will have focus, e.g. the active program.
Why would you want to do this anyway ?
There is also the Show desktop icon which should be in the quick launch area, this is a toggle type switch, so if you have multiple applications open it will close them. If you have multiple applications minimised to the task bar then it would open them all.
The problem being when you click this to open the remaining minimised applications it may not retain the focus on the one you were in, but the times I have used this it does save focus to the application that was active when you clicked it to open minimised applications.
Not dreaming though if what you were seeing was the show desktop at work opening all minimised applications, but not by right clicking on it as that transfers focus.
The issue isn’t getting the context menu, it is if you open it from that menu then that application has focus, is active and not the one that was originally open/active.
That was the question how to open a program with the right click menu but not switch to it (e.g. the opened application being active.