pass shell variable to perl one liner

I am trying to pass shell variable to a perl one liner but it didn't work.

Am I missing something here?

export mac1
mac1=`ifconfig|grep HWaddr|gawk '{print $5}'`
eth0_input=/etc/sysconfig/network-scripts/ifcfg-eth0
eth0_input=eth0test
perl -s -p -i -e "s/^HWADDR=.+?$/HWADDR=$mac1/g" $eth0_input

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MazdajaiAsked:
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ozoCommented:
In what way did it not work?
What was the value of $mac1?
Which shell are you using?
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MazdajaiAuthor Commented:
Hi,
Mac1 has the variable but the replacement didn't happen.
I am trying to replace mac address basically.
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woolmilkporcCommented:
You should first fill the variable, then export it.

Or use "export mac1=`ifconfig ... ... `"

EDIT: Please take the above just as a general hint! I just saw that in your case it should work as is.
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ozoCommented:
$Mac1 is different from $mac1, but it looks like you are setting $mac1
Are there lines in eth0test that match /^HWADDR=.+?$/
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ozoCommented:
export should not be necessary in the above shell code
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MazdajaiAuthor Commented:
Was typing on my cell phone - it is mac1.
There is a line begins with the match:
HWADDR=
Will try again when I have a chance!
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TintinCommented:
No real need for perl.

mac=$(ifconfig eth0|awk '/HWaddr/ {print $NF}')
sed -i "s/^HWADDR=.*/HWADDR=$mac/" /etc/sysconfig/ifcfg-eth0

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ozoCommented:
Note that /^HWADDR=.*/ can match where /^HWADDR=.+?$/ may not, and /etc/sysconfig/ifcfg-eth0 may not be the same as eth0test
Otherwise, there should be no difference between the sed method and the perl method
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MazdajaiAuthor Commented:
Looks like it was the regex, I thought .+? (non-greedy) is the same as .* (greedy) -  Is there a tool to troubleshoot regex / matching in perl?
NO
HWADDR= 

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YES
HWADDR=aa:bb:cc:dd::ee 

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Tintin,
Yes sed works but I am more familiar with the regex engine in Perl. I have seen issue with the regex support in GNU tools.
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ozoCommented:
perl -mYAPE::Regex::Explain -e 'print YAPE::Regex::Explain->new($_)->explain for qw/^HWADDR=.*/,qw/^HWADDR=.+?$/'
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