Re: zsh ignores the arguments on its first command.

[Martin Buchholz <martinrb@google.com>]

Thanks, Peter.

You are right that this one is not a bug in zsh (or at least not just zsh).
Sigh... I guess I fell into a shell-quoting trap,
even though I consider myself an expert on that...

Here's one right way to do it:

$ ssh localhost "zsh -fc 'echo 1 2 3; echo 4 5 6'"
1 2 3
4 5 6

The openssh documentation doesn't seem to specify
exactly how a remote command is executed.

Martin

On Tue, Jun 1, 2010 at 14:45, Peter Stephenson
<p.w.stephenson@ntlworld.com> wrote:
> Martin Buchholz wrote:
>> Hi zsh maintainers,
>>
>> This is a bug report.
>>
>> It seems that when the shell is invoked using
>> zsh -i -c 'COMMAND1; COMMAND2'
>> then COMMAND1 is executed without its arguments!
>
> I think your terminal is being screwed up somehow.  That explains both
> why you can't see the echo output and can't see the ${+terminfo}.
>
> It usually helps if you can boil bugs down to what happens with the "-f"
> option, so we're not relying on unseen side effects of initialisation
> scripts.
>
> Also, I'm afraid we don't have the spare time to go poking around in
> distribution's zshrc files, but if you think there's some effect we
> (rather than the distribution) should know about, please do report it.
>
> --
> Peter Stephenson <p.w.stephenson@ntlworld.com>
> Web page now at http://homepage.ntlworld.com/p.w.stephenson/
>