Date: Tue, 24 May 2016 18:52:33 +0000 From: bugzilla-noreply@freebsd.org To: freebsd-ports-bugs@FreeBSD.org Subject: [Bug 209737] sysutils/qjail [Maintainer update] fix bug Message-ID: <bug-209737-13@https.bugs.freebsd.org/bugzilla/>
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https://bugs.freebsd.org/bugzilla/show_bug.cgi?id=3D209737 Bug ID: 209737 Summary: sysutils/qjail [Maintainer update] fix bug Product: Ports & Packages Version: Latest Hardware: Any OS: Any Status: New Severity: Affects Many People Priority: --- Component: Individual Port(s) Assignee: freebsd-ports-bugs@FreeBSD.org Reporter: qjail1@a1poweruser.com Created attachment 170616 --> https://bugs.freebsd.org/bugzilla/attachment.cgi?id=3D170616&action= =3Dedit port_diff Summary of 4.7 update 1. When running "qjail create -c" option for creating a jail with ssh and a user id / password of the jail name. On first start the user id=20 gets created in the jail. This first start status was not being=20 turned off. Added code to start logic to fix this. 2. Did not like the way console function was handling console commands being passed to the jail for execution. Changed the logic so it works= =20 like this. If -c option used alone or if -c & -u used together, them default jexec command format used. If no -c option then=20 "jexec jailname login -f <root or -u value>" format used.=20 This is real login causing the Welcome message to display.=20=20 3. Corrected the qjail manual console function to document usage of=20 -c and -u flags. --=20 You are receiving this mail because: You are the assignee for the bug.=
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