On Sat, Jul 28, 2018 at 4:37 AM, ToddAndMargo <toddandma...@zoho.com
<mailto:toddandma...@zoho.com>> wrote:
Hi All,
How do I get the bash return code ("$?") from
the following?
$ReturnStr = qqx ( curl $TimeOutStr -L $Url -o $FileName ).lines;
Many thanks,
-T
On 07/28/2018 06:14 AM, Paul Procacci wrote:
I'm not sure about qqx because I too am a fledgling perl6 programmer,
but the run routine returns a Proc object that has an exitcode method.
++++++++++++++++++++++++
my $proc = run 'ls', 'dir!';
$proc.exitcode.say;
++++++++++++++++++++++++
Right in the documentation the following is stated as well:
"See alsorun <https://docs.perl6.org/routine/run>andProc::Async
<https://docs.perl6.org/type/Proc::Async>for better ways to execute
external commands. "
Hi Paul,
I adore the run command and use it very frequently.
I this instance, I actually want to write STDERR to the
shell and only capture STDIN and the exit code.
I am trying to get "curl" to show its progress meter,
which writes to STDERR.
So far I have
$ p6 'my $x="cat /etc/hosts; echo \$\?"; my $y = qqx ( $x ); say "$y";'
which sends STDIN and the exit code to $y, which I can deal with.
Thank you for the help!
-T
