Dear monks, I have a shell script with a perl oneliner in it, and want to pass a shell variable to this oneliner.
The idea is this: 
#!/bin/bash
file=$1
var=$2
cat $1 | perl -e 'perlcodeinhere' >output

[download]
The question is how to get the $var variable into the perl code since I apparently can not get in is as regular argument while using the -e parameter.

Rgds, Ole C.

Update:
Thanks for the input, several nice ways of doing it.
I will also be sure to utilize the perl pie construct in the future, that is most certainly usefull for quite a few of my daily tasks.

Another Update:
As Bart points out it is possible to have the variable as a regular argument. What kept me from realising that was that I was using a while(<>) loop inside my code, my entire oneliner was like this: 
cat $i | perl -e 'while(<>){$_=~ s/$ARGV[0]//g;print;}' $var >$i.new

[download]
and that did not work. I guess that is because the while(<>) expects to find files to work on in the @ARGV list. When I changed it to 
cat $i | perl -e 'while(<STDIN>){$_=~ s/$ARGV[0]//g;print;}' $var >$i.
+new

[download]
it worked the way I expected.

In reply to arguments to perl -e by olecs

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