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Date:      Sun, 13 Dec 2020 17:04:26 +0100
From:      mj-mailinglist@gmx.de
To:        freebsd-questions@freebsd.org, freebsd-jail@freebsd.org
Subject:   Questions about the output of jls
Message-ID:  <trinity-1eddb95f-d7d8-4b02-be03-0296774a8434-1607875466295@3c-app-gmx-bap80>

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Hi,

I habe a current system, where i have current and 12.2-STABLE jails. Checking with jls, i get this output:

root@fbsd13:~ # jls -h jid name ip4.addr host.hostname vnet osrelease path | column -t
jid  name  ip4.addr      host.hostname  vnet  osrelease     path
8    j0    192.168.0.10  j0.local       2     13.0-CURRENT  /jails/j0
10   j1    -             j1.local       1     13.0-CURRENT  /jails/j1
12   j2    -             j2.local       1     13.0-CURRENT  /jails/j2

the jails are running this versions:

root@fbsd13:~ # jexec -l j0 freebsd-version -u
12.2-STABLE
root@fbsd13:~ # jexec -l j1 freebsd-version -u
13.0-CURRENT
root@fbsd13:~ # jexec -l j2 freebsd-version -u
12.2-STABLE


What is "osrelease"? Looking at the name, i would have guessed, it is the
version of the freebsd userland, running in the jail. But it does't seem so.
j1 and j2 are VNET jails, so it seems the 1 in the vnet column signifies this,
j0 is a "standard" jail using the hosts network stack, so the 2 stands for standard?

Is ist possible for jls to get the ip address and the userland version of/in the jail?
Or is the only way to get this information to jexec ifconfig and freebsd-version?

--
Martin




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