I have a server on which I use Vagrant to install and manage VirtualBox
VMs for various operating systems. Over time I have installed 3
different OpenBSD VMs. The box for each of these was originally
obtained from the listings at vagrantup.com. Each box holds a different
OpenBSD version.
I believe the following command accurately lists the Vagrant box versions:
###
$ ls -l .vagrant.d/boxes/generic-VAGRANTSLASH-openbsd6/
total 2
drwxr-xr-x 3 vmuser vmuser 3 Nov 4 2018 1.8.40
drwxr-xr-x 3 vmuser vmuser 3 Oct 7 2019 1.9.34
drwxr-xr-x 3 vmuser vmuser 3 Feb 1 2020 2.0.6
###
This data correlates with the output of this command:
###
$ vagrant box list | grep -i openbsd
generic/openbsd6 (virtualbox, 1.8.40)
generic/openbsd6 (virtualbox, 1.9.34)
generic/openbsd6 (virtualbox, 2.0.6)
###
Problem: I want to make sure that I am using the correct syntax to
remove *one* of the boxes but not the others. Suppose that I want to
remove 1.8.40. Is the following the correct way to remove that one and
leave the other two untouched?
###
vagrant box remove generic-openbsd6 --provider virtualbox --box-version
1.8.40
###
Thank you very much.
Jim Keenan
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