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I want to configure windows 10 Action center(which comes up with Winkey + A) to a touchpad gesture of my laptop.
My touchpad program allows me to configure its actions(launch different programs for different gestures) but I need to find out which .exe file in system32 directory is for the Action center. Does anybody have a clue? Thanks in advance.
Ex: To launch windows calculator for a touchpad gesture, I would have to link to "system32/calc.exe"
I just find that the process C:\Windows\SystemApps\ShellExperienceHost_cw5n1h2txyewy\ShellExperienceHost.exe called file C:\Windows\ShellExperiences\Windows.UI.ActionCenter.dll for Action Center, not sure if it's what you would like to find. Also the shell commands to open Actions center could run "ms-actioncenter:" – Kattee Lee – 2017-12-07T08:05:54.020
I don’t the location of Action Center, even don’t know the .exe file name of it. But, we can create a shortcut to the Action Center. What about linking the shortcut to your touchpad program, if your touchpad program can launch the shortcut, shortcut will launch Action Center. – Origami – 2017-12-07T08:16:48.263
@KateLi between yours and hanjo's answers i was able to link to action center to my touchpad. Thanks – Smiley – 2017-12-08T10:21:33.183
@Origami "But, we can create a
shortcut
to the Action Center". Are you referring to the same method as KateLi? – Smiley – 2017-12-08T10:23:25.437