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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