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Editorial Team
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Asked: 2026-05-27T13:36:14+00:00

I created the following awk command in order to print only the line that

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I created the following awk command in order to print only the line that match the host and the ETH parameters

my problem is that I don’t know which eth1-8 is the real argument

How to print the line from the file by awk if ETH could be eth0 or eth1 or eth2 ….etc until eth8

 HOSTNAME=linux1
 LAN=eth0|eth1|eth2|eth3|eth4|eth5|eth6|eth7|eth8

 awk -v host=$HOSTNAME -v ETH=$LAN  '$2 == host && $3 == ETH'  file



 more file


192.17.200.10  linux1 eth0
192.17.200.10  linux1 eth1
192.17.200.11  linux2 eth2
192.17.200.12  linux3 eth3
192.17.200.13  linux4 eth4
192.17.200.14  linux5 eth5
192.17.200.15  linux6 eth6
192.17.200.16  linux7 eth7
192.17.200.17  linux8 eth8
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1 Answer

  1. Editorial Team

    Use a regular expression to specify the possible matches, e.g.:

    awk '$2 == "linux1" && $3 ~ /^eth[0-8]$/'

    In terms of the shell variable, you’ll want something like:

    LAN='^eth[0-8]$'
awk -v host=$HOSTNAME -v ETH=$LAN '$2 == host && $3 ~ ETH' file
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