system command question

[Himal]

I there a reason why a command will not work if i do system "$cmd &>
xyz.log" even though it works if I do system "$cmd"

[J. Gleixner]

Himal wrote:

> I there a reason why a command will not work if i do system "$cmd &>
> xyz.log" even though it works if I do system "$cmd"

This is an OS/Shell issue.

From your command line try:

whateverCommandYouAreTryingToRun & > xyz.log

and you should see the error.

[Chris Mattern]

Himal wrote:

> I there a reason why a command will not work if i do system "$cmd &>
> xyz.log" even though it works if I do system "$cmd"

Most likely, the shell perl is using to parse your command is
not csh. Exactly why this is so would depend on your configuration.
Try this, which will work for sh/ksh/bash: system "$cmd >xyz.log 2>&1"

Chris Mattern

[Andreas Kahari]

In article <3F61EB8C.8000604@gwu.edu>, Chris Mattern wrote:

> Himal wrote:

>> I there a reason why a command will not work if i do system "$cmd &>
>> xyz.log" even though it works if I do system "$cmd"

>
> Most likely, the shell perl is using to parse your command is
> not csh. Exactly why this is so would depend on your configuration.
> Try this, which will work for sh/ksh/bash: system "$cmd >xyz.log 2>&1"

There is no reason for it to be csh. The manual for the system
command says:

If there is only one scalar argument, the argument is
checked for shell metacharacters, and if there are any, the
entire argument is passed to the system's command shell for
parsing (this is "/bin/sh -c" on Unix platforms, but varies
on other platforms). If there are no shell metacharacters
in the argument, it is split into words and passed directly
to "execvp", which is more efficient.

So it's either /bin/sh (on Unix systems) or no shell at all.,
if you haven't done some pretty freakish changes in the default
configuration.

--
Andreas Kähäri

[Chris Mattern]

Andreas Kahari wrote:

> In article <3F61EB8C.8000604@gwu.edu>, Chris Mattern wrote:
>

>>Himal wrote:
>>

>>>I there a reason why a command will not work if i do system "$cmd &>
>>>xyz.log" even though it works if I do system "$cmd"

>>
>>Most likely, the shell perl is using to parse your command is
>>not csh. Exactly why this is so would depend on your configuration.
>>Try this, which will work for sh/ksh/bash: system "$cmd >xyz.log 2>&1"

>
>
> There is no reason for it to be csh. The manual for the system
> command says:
>
> If there is only one scalar argument, the argument is
> checked for shell metacharacters, and if there are any, the
> entire argument is passed to the system's command shell for
> parsing (this is "/bin/sh -c" on Unix platforms, but varies
> on other platforms). If there are no shell metacharacters
> in the argument, it is split into words and passed directly
> to "execvp", which is more efficient.
>
> So it's either /bin/sh (on Unix systems) or no shell at all.,
> if you haven't done some pretty freakish changes in the default
> configuration.
>

Ah, OK, I thought it used the user's shell. OK, then,
system "$cmd >xyz.log 2>&1" should be the answer, assuming
the OP is on a Unix system (which seems logical, since he's
trying to use csh syntax).

Chris Mattern

[Anno Siegel]

Chris Mattern <syscjm@gwu.edu> wrote in comp.lang.perl.misc:

> Andreas Kahari wrote:

> > In article <3F61EB8C.8000604@gwu.edu>, Chris Mattern wrote:
> >

> >>Himal wrote:
> >>
> >>>I there a reason why a command will not work if i do system "$cmd &>
> >>>xyz.log" even though it works if I do system "$cmd"

[...]

> Ah, OK, I thought it used the user's shell. OK, then,
> system "$cmd >xyz.log 2>&1" should be the answer, assuming
> the OP is on a Unix system (which seems logical, since he's
> trying to use csh syntax).

Except that he isn't. "$cmd >& xyz.log" would be csh syntax.
"$cmd &> xyz.log" is neither here nor there.

Anno