Discussion:
[linux-lvm] udev rules do not create partition entries when an lvm lv is partitioned
Joe Julian
2015-09-04 18:18:26 UTC
Permalink
Yes, I'm aware of all the arguments against even doing this, but I have
my reasons for doing it this way.

When you create a partition on an lv:

sgdisk /dev/vg1/lv1 -n 0:0:0

there is no udev rule to create the partition entry, '/dev/vg1/lv1p1'
(for instance). For my immediate use case, this causes ceph-disk to fail
as it cannot find the partition as it's expecting to find the device
name with a new "1" at the end. Any suggestions on how I could make this
happen?
Bryn M. Reeves
2015-09-04 19:12:09 UTC
Permalink
Yes, I'm aware of all the arguments against even doing this, but I have my
reasons for doing it this way.
sgdisk /dev/vg1/lv1 -n 0:0:0
there is no udev rule to create the partition entry, '/dev/vg1/lv1p1' (for
instance). For my immediate use case, this causes ceph-disk to fail as it
cannot find the partition as it's expecting to find the device name with a
new "1" at the end. Any suggestions on how I could make this happen?
Why not just name the LV lv1p1 if ceph-disk is hard wired to expect that
naming?

There's not a lot of benefits to using partitions on top of logical
volumes and if you're not expecting them they can be problematic (e.g.
when they are the partitions of a running VM that is also accessing the
disk).

It is possible to create partitions on pretty much any device-mapper
device using the normal tools and then create new dm devices to map the
partitions using kpartx.

Multipath-tools supplies udev rules to automate this so that the
partition maps are created automatically by udev (since it works with
whole disks): you could use that as an example if you really want to try
that approach but it's fairly easy to get the rules wrong in subtle ways
that may prevent booting.

If you just want a temporary mapping to set things up run:

# kpartx -a /dev/vg1/lv1

And you should find partition maps at /dev/mapper/vg1-lv1p1 etc.

kpartx -d to remove.

Regards,
Bryn.

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