Does someone know how to deal with the following situation? Very often I do the following pattern:
(1) rlogin to a foreign host (2) Invoke a subshell (for example because I'm setting a Clearcase View) (3) Logout from the host Step (3) needs two steps: First I have to type 'exit' to leave the subshell, and then either 'exit' or 'logout' to leave the login shell. Is it possible to automate this in such a way that I have to type only one command, to leave all subshells (in this case, only 1, but in general, I might be several subshells deep) AND then logout? One idea would be to use ps, locate the process to the current subshell, crawling upwards via the PPIDs to find the PID of the login shell, and kill it. But this seems to be such an awkward solution, that I thought maybe there is an easier way to do it. (Note: This posting goes to the bash and to the zsh mailing list, because this problem might be solved differently in both shells, and I'm interested in both solutions - of course a solution working in bash AND zsh would be preferable). Ronald -- Ronald Fischer (phone +49-89-63676431) mailto:[EMAIL PROTECTED] _______________________________________________ Bug-bash mailing list Bug-bash@gnu.org http://lists.gnu.org/mailman/listinfo/bug-bash