How to use shell variables in perl command call in a bash shell script?
I have a perl command in my shell script to evaluate date -1.
How can i use $myDate
in perl command call?
This is the section in my script:
myDate='10/10/2012'
Dt=$(perl -e 'use POSIX;print strftime '%m/%d/%y', localtime time-86400;")
I want use $myDate
in place of %m/%d/%y
.
Any help will be appreciated.
Thank you.
The same way you pass values to any other program: Pass it as an arg. (You might be tempted to generate Perl code, but that's a bad idea.)
Dt=$( perl -MPOSIX -e'print strftime $ARGV[0], localtime time-86400;' -- "$myDate" )
Note that code doesn't always return yesterday's date (since not all days have 86400 seconds). For that, you'd want
Dt=$( perl -MPOSIX -e'my @d = localtime time-86400; --$d[4]; print strftime $ARGV[0], @d;' -- "$myDate" )
or
Dt=$( perl -MDateTime -e'print DateTime->today(time_zone => "local")->subtract(days => 1)->strftime($ARGV[0]);' -- "$myDate" )
or simply
Dt=$( date --date='1 day ago' +"$myDate" )
Variables from the shell are available in Perl's %ENV
hash. With bash
(and some other shells) you need to take the extra step of "exporting" your shell variable so it is visible to subprocesses.
mydate=10/10/2012
export mydate
perl -e 'print "my date is $ENV{mydate}\n"'
Using "
instead of '
also passes shell variables to perl in version 5.24.
mydate=22/6/2016
perl -e "print $mydate"
The same way you pass values to any other program: Pass it as an arg. (You might be tempted to generate Perl code, but that's a bad idea.)
Dt=$( perl -MPOSIX -e'my($m,$d,$y) = split qr{/}, $ARGV[0]; --$d; print strftime "%m/%d/%y", 0,0,0, $d,$m-1,$y-1900;' -- "$myDate" )
Why not something like:
$ENV{'PATH'} = $ENV{'PATH'}.":"."/additional/path";
For me, I need to use $ENV{'VARIABLE'}
to pass VARIABLE
in shell to Perl. I do not know why simply $ENV{VARIABLE}
did not work.
For example,
In Bash:
#!/bin/bash -l
VARIABLE=1
export VARIABLE
perl example.pl
In Perl:
#!/usr/bin/perl -w
print "my number is " . $ENV{'VARIABLE'}