有条件地执行.bashrc命令

Question

I wanted to attach my Google Drive to my local machine (linux) so that I could access it via my terminal.

After some googling, I saw that I could install google-drive-ocamlfuse to do this.

When I boot my machine, I have to type:

google-drive-ocamlfuse ~/google-drive

to mount Google Drive.

To avoid doing this every time, I added the line to my .bashrc. Which works fine. But then every subsequent terminal I open tries to run the line, and I get the message:

fuse: mountpoint is not empty
fuse: if you are sure this is safe, use the 'nonempty' mount option

I presume I get this message because it is trying to mount something that is already mounted. Is there any way I can make this line in my .bashrc only execute if it is the first time the terminal is opened in a session. Or some other way I can stop the warning?

Not a big deal, but it would be nice to learn something.

Answer 1

Try putting this command in a start up file:

mountpoint ~/google-drive || google-drive-ocamlfuse ~/google-drive

mountpoint tests to see if its argument is a mountpoint. If it isn't, then mountpoint returns false and that triggers || to execute the second command.

mountpoint is part of the util-linux collection of utilities.

You may put this command in one of your shell start up files but it is probably better placed in one of the system start up files. Depending on your distribution, that file might be /etc/rc.local .

Alternate form for conditional execution

If you prefer, conditional execution can be performed using the if-then statements:

if ! mountpoint ~/google-drive
then
  google-drive-ocamlfuse ~/google-drive
fi

Here, ! negates the exit code of mountpoint so that google-drive-ocamlfuse is only executed if mountpoint returns false.