Re: Is it possible to capture stdout and stderr to separate variables in Zsh?

[Han Pingtian <hanpt@linux.vnet.ibm.com>]

On Tue, Mar 06, 2012 at 11:01:11PM -0800, Bart Schaefer wrote:
> On Mar 6,  9:16am, Philippe Troin wrote:
> }
> } On Tue, 2012-03-06 at 09:09 +0100, Nikolai Weibull wrote:
> } > Is it possible to capture stdout and stderr to separate variables in Zsh?
> } 
> } All I can think of is:
> } 
> }         coproc cat &
> }         pid=$!
> }         stdout="$( ( print "printed on stdout"; print -u 2 "printer on stderr" ) 2>&p )"
> }         sleep 1 
> }         kill "$pid"
> }         stderr="$(cat <&p)"
> } 
> } You'll notice the very ugly sleep+kill hack I had to use as I could not
> } find how you can close a coprocess's standard input cleanly.  Removing
> } the sleep+kill makes the cat <&p hang forever.
> 
> You need this:  http://www.zsh.org/mla/users/2011/msg00095.html  :-)
> 
Hi,

I'm using zsh 5.0, but looks like still cannot close the prior coproc
with "coproc exit":

% coproc while read line;do print -r -- "$line:u";done
[1] 31518
% coproc exit
[2] 31519
%
[2]  + done       exit
% jobs
[1]  + running    while read line; do; print -r -- "$line:u"; done
%

or I'm missing something here?

Thanks.