misc/101015: jexec lacks a -U flag.

Jesper Wallin jesper at ifconfig.se
Sat Jul 29 09:40:18 UTC 2006

>Number:         101015
>Category:       misc
>Synopsis:       jexec lacks a -U flag.
>Confidential:   no
>Severity:       non-critical
>Priority:       low
>Responsible:    freebsd-bugs
>State:          open
>Class:          change-request
>Submitter-Id:   current-users
>Arrival-Date:   Sat Jul 29 09:40:12 GMT 2006
>Originator:     Jesper Wallin
>Release:        FreeBSD 6.1-RELEASE-p3
FreeBSD omega.ifconfig.se 6.1-RELEASE-p3 FreeBSD 6.1-RELEASE-p3 #0: Sat Jul 29 08:06:51 CEST 2006     root at omega.ifconfig.se:/usr/obj/usr/src/sys/omega  i386
If I start a jail, I can specify which user the process inside the jail should be runned by. If I later on want to start another process (or the same) by another user inside the jail, I will have to add the same user on the host system. For instance, if I want to run a process as the user www inside the jail, I won't be able to start another process on-the-fly using jexec, as the user www, as the user www on the host system has no home or shell.
Not repeatable as it's a 'wish to be added'-problem and not a bug.
My suggestion would be to add a -U (and -u) flag to the jexec command. On another note, how come there's no start-up configuration for jexec? As the system is right now, it's impossible to start one jail, running 3 different processes inside the jail as 3 different users by only using /etc/rc.conf.

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