On 07/28/2018 01:37 AM, ToddAndMargo 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 01:50 AM, Benji wrote:
I don't think you can with `qqx`. It returns the string output directly.
You can with `run` though.
my $p = run 'ls';
say $p.exitcode;
Hi,
I am writing exitcode down in my run notes! Thank you.
I had been using
$RtnCode = $proc.status;
What I am trying to do is run curl and read STDIN and the
return code, but leave the STDERR progress bar on the
shell. I can't use "run" for this as I want the program
I called to write STDERR to the shell.
so far I have
$ p6 'my $x="cat /etc/hosts; echo \$\?"; my $y = qqx ( $x ); say "$y";'
which sends the exit code to the last line, which is pretty easy to
separate from the STDIN.
-T
