how to exclude the header in shell script using awk


 
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# 1  
Old 07-15-2009
how to exclude the header in shell script using awk

Hello Everyone
In my shell script, I am retrieving the cluster ID and node number of an LPAR using the following command -

lsclcfg -l

This command's output looks as follows -

CLUSTER_NAME CLUSTER_ID NODE_NR
sch1h004 6104567 3


I want to store only the values of cluster_id and node_nr in the 2nd row into two different variables in my shell script. I don't need the first row i.e. the header.

I was playing around with awk command to see if I can accomplish this but to no avail.
Any thoughts/ideas as to how to accomplish this would be appreciated.
Thanks
gates1580
# 2  
Old 07-16-2009
redirect output to tail

Code:
lsclcfg -l | tail +2

this will show 2nd line onwards.
# 3  
Old 07-16-2009
assume something like below:
Code:
sed '1d' yourfile | while read line;do
        set $line
        id=$2
        nr=$3
        done
        echo $id"--"$nr

# 4  
Old 07-16-2009
This might help

cat yourfile | sed '1d' | awk '{ print $2,$3 }'
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