pass variable to another process when invoking from command line

vineet2004 has asked for the wisdom of the Perl Monks concerning the following question:

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Re: pass variable to another process when invoking from command line by gaal (Parson) on Dec 27, 2006 at 12:41 UTC
The two easiest ways to pass information to another process when invoking it is on the command line, and through standard input. Say you call the second program from the shell, not the first one. How does it expect the input? It's probably one of these two: `second_program filename # as a command line option echo "filename" \| second_program # piped into its standard input` [download] When calling `second_program`, you need to do one of these: `system "./second_program", $filename and die ... # pass it on the comm +and line # note, system needs +"and" for error checking open my $fh, "\| ./second_program" or die ... # via stdin print $fh $filename; # (note, no comma on +this line)` [download]	[reply] [d/l] [select]
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Re: pass variable to another process when invoking from command line by jettero (Monsignor) on Dec 27, 2006 at 12:40 UTC
`bashprompt$ ./program.pl \| ./program.pl` — nothing could be simpler... Course, I went and assumed you meant bash. If you meant open, then you want: `open my $out, "\|", "./program.pl" or die "lol: $!"`. -Paul	[reply] [d/l] [select]
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