I am trying to work out why .bashrc is not executing when I invoke ssh with
the -t option and _does_ execute when I invoke ssh without the -t option.
ssh -qt remote-host which rsync # indicates ~/.bashrc has not executed on
remote host
ssh -q remote-host which rsync # indicates ~/.bashrc has executed on
remote host
ssh -qt remote-host tty # reports /dev/pts/1
ssh -q remote-host tty # reports not a tty
ssh -qt remote-host echo '$-' # reports hBc
ssh -q remote-host echo '$-' # reports hBc
ssh -q remote-host ps -o "pid,ppid,args" -u xjsrs
PID PPID COMMAND
8704 8702 sshd: [EMAIL PROTECTED]
8705 8704 ps -o pid,ppid,args -u xjsrs
ssh -qt remote-host ps -o "pid,ppid,args" -u xjsrs
PID PPID COMMAND
8733 8731 sshd: [EMAIL PROTECTED]/1
8734 8733 ps -o pid,ppid,args -u xjsrs
According to echo '$-' neither shell is interactive.Yet, the one that is
started without a pseudo terminal does have .bashrc executed.
I added an debug statements to .bash_profile and it not getting executed in
either case.
There is no /etc/sshrc file and I don't have a ~/.ssh/rc file.
The bash version is:
GNU bash, version 3.00.15(1)-release (x86_64-redhat-linux-gnu)
Copyright (C) 2004 Free Software Foundation, Inc.
Upon reading the manual, the rule that bash seems to be using to decide that
.bashrc should be executed if -t is not specified, is this one:
Bash attempts to determine when it is being run by the remote shell
daemon, usually rshd. If bash determines it is being run by rshd, it
reads and executes
commands from ~/.bashrc, if that file exists and is readable. It
will not do this if invoked as sh. The --norc option may be used to inhibit
this behavior, and
the --rcfile option may be used to force another file to be read, but
rshd does not generally invoke the shell with those options or allow them to
be specified.
However, when I use the ssh -t option, it would seem that allocation of a
pseudo-tty is causing bash to assume that it is not being invoked by a
remote shell daemon.
Is there any way I can have an ssh pseudo-tty and get bash to execute
~/.bashrc?
jon seymour.
