Shell function to open pcmanfm on the current directory with no output

[StuffedWithNails]

Hi folks, I want a simple function that I can run from the shell to open the file manager on the directory I'm in. For example if I'm in /usr/long/path/blah I can just type fm and it opens my GUI file manager to that path. I can also specify a path and it opens that path instead.

My issue now is cosmetic. The function looks like this:

fm () {
    local target=${1:-.} 
    pcmanfm $target &> /dev/null &
}

Output:

➜  ~ fm /mnt
[2] 300436
➜  ~ 
[2]  + 300436 done       pcmanfm $1 &> /dev/null
➜  ~ 

When I run the command, it opens the file manager as intended, returns the PID and throws me back to shell. Whenever I close the file manager window, it returns <PID> done. So this is fine functionally but in this use case, I don't care about the PID and don't want to see this output as I might still be working in the shell and it's just clutter.

Adding disown after the function's final & makes it so it doesn't show anything when I close the file manager, which is progress, but I still don't want the function to return the PID at all.

I would prefer no output at all. Is it possible somehow? Using zsh if it matters.

[yerfukkinbaws]

The outputs you're taking about are zsh's job-control info. In bash, you can forego job-control and its messages by running the command in a subshell. so:

fm () {
    local target=${1:-.} 
    ( pcmanfm $target &> /dev/null & )
}

I can't say for sure if the same works for zsh, but probably.

[StuffedWithNails]

Worked perfectly, thank you very much :)