I am able to start git-bash via the Windows command prompt in several ways:
"C:\Program Files\Git\bin\sh.exe"
"C:\Program Files\Git\git-bash.exe"
"C:\Program Files\Git\usr\bin\mintty.exe"
I would like to start git-bash in a particular directory using such a call.
For context, I am doing this from within the SAS Enhanced Editor. The actual call is
%sysexec(start "" "C:\Program Files\Git\bin\sh.exe" && exit);
For those unfamiliar with SAS, %sysexec opens an instance of the Windows Command Prompt and issues whatever command is given as an argument.
It works surprisingly well. However, I would like to be able to start the git-bash in a particular directory, not just home.
To this end, I can create an entry in .bashrc:
cd /c/new/starting/dir
The trouble with doing this, though, is whenever I open a git-bash, such as via the context menu in a particular folder, the default directory is that of .bashrc.
I see that there exist git-bash start options like --cd-to-home or --cd=<path> but I can't get them to work. For example,
"C:\Program Files\Git\bin\sh.exe" --cd-to-home
"C:\Program Files\Git\bin\sh.exe --cd-to-home"
"C:\Program Files\Git\git-bash.exe" --cd-to-home
"C:\Program Files\Git\git-bash.exe --cd-to-home"
etc.
Is it possible to start git-bash in a particular directory from Windows Command Prompt? If so, what's the proper syntax?
BONUS: Bonus points for doing it in 50 characters or less
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