Consider these two software: MyGame.exe and SocketTest.exe.
MyGame.exe doesn't depend on any files to run but SocketTest.exe is an actual software. It can only run if one or more of four other files are in the same folder as itself. These files are: metouia.jar, SocketTest.bat, SocketTest.jar, SocketTest.sh
I want to create a software using Windows Form in Microsoft Visual Studio using C#. It will have two buttons: "Launch MyGame" and "Launch SocketTest."
I go to the Solution Explorer tab and right-click my .csproj file to click Add Existing Item... so I can add MyGame.exe, SocketTest.exe and the other four files into the Solution Explorer.
For the File Properties of the 6 files, under "Copy to Output Directory" I choose "Copy Always."
For Build Action, I'm not sure. I use "None". My buttons can still launch the two .exe files.
However, I'm not sure which settings to use if I want other computers to be able to launch those .exe files with my software, especially for the .exe file that depends on other files. I've read the Build Action descriptions (online too) but still can't decide on the proper one for my situation.