From: DervishD <zsh@dervishd.net>
To: Zsh Users <zsh-users@sunsite.dk>
Subject: Silent shell but not silent script
Date: Fri, 2 Sep 2005 13:09:12 +0200 [thread overview]
Message-ID: <20050902110911.GA158@DervishD> (raw)
Hi all :)
This one is tricky ;) I have to invoke a command which writes its
diagnostic messages to stderr and its normal output to stdout. This
is important, because sometimes I will want to separate the output to
two files using redirections.
The problem is this:
$ ./command
zsh: no such file or directory: ./command
Sometimes the command I have to run won't exist, so zsh will
issue that error. How the heck can I make zsh not to spill the error
message but at the same time let the command to use stderr and
stdout? I mean, I can silent zsh doing this:
$ ./command 2> /dev/null
but that will silent the stderr output from the command if the
command exists.
I cannot use MULTIOS and I cannot use any special option of zsh
because of portability, so... can this be done? Can I use any
subshell trick to do the job?
The only solution I've found so far is to pipe stderr thru a sed
script and get rid of any shell message, but it is too much... Other
solution is doing something like this:
$ { ./command 2>&1 } 2> /dev/null
but then I completely loose the ability of separating the stdout
output from the stderr one :(
Thanks a lot in advance :) and sorry for the weird question O:)
Raúl Núñez de Arenas Coronado
--
Linux Registered User 88736 | http://www.dervishd.net
http://www.pleyades.net & http://www.gotesdelluna.net
It's my PC and I'll cry if I want to...
next reply other threads:[~2005-09-02 11:05 UTC|newest]
Thread overview: 5+ messages / expand[flat|nested] mbox.gz Atom feed top
2005-09-02 11:09 DervishD [this message]
2005-09-02 11:30 ` Mike Hernandez
2005-09-02 14:22 ` DervishD
2005-09-02 15:34 ` Bart Schaefer
2005-09-02 16:03 ` DervishD
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