Passing arguments to perl using bash variables

Question

Ordinarily I would invoke Perl and supply required arguments from within a bash script simply using:

perl script.pl arg1

However there are cases when I want to store both the perl script directory and the arguments in bash variables:

PERLDIR = "/example/directory/script.pl"
ARG1 = "40"

When trying to call the perl script using:

perl "$PERLDIR"

It works, however when trying to provide the argument i'm not sure of the syntax to utilise. If I use:

perl "$PERLDIR $ARG1" it'll attempt to open the directory:

/example/directory/script.pl 40

And throw an error.

Is there a way to do this and if so, how?

Dmitry Ermolov · Accepted Answer · 2016-05-04 17:22:23Z

4

You should use:

 perl "$PERLDIR" "$ARG1"

When you use many variables in one string enclosed with " it becomes one argument.

answered May 4, 2016 at 17:22

Dmitry Ermolov

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2 Comments

@Borodin Will do - and will also remove my comment.

@AnnaSchumann: I realise that you're getting to know the ropes. Thank you for understanding