Regular expression to extract ipv6 address

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# 1  
Old 06-05-2011
Regular expression to extract ipv6 address

Hi all ,

I have a string in my weblog xheader v6-day-2011:xx:yy:zz:qq:qq:ww:ee:rr

My requirement is to lookup the sting v6-day-2011 in this header and if found would like to extract the V6 ip part .

v6-day-2011 is always constant for a ipv6 entry so i would like to extract every thing that is after the v6-day-2011: sting.

Regular expression i tried some thing like to extract the second part

^[v6-day-2011] : ([^[:space:]]+)

INPUT : v6-Day-2011:420:4:e7ed:45b0:ccc6:776b:62c9
Expected Output : 420:4:e7ed:45b0:ccc6:776b:62c9

Some how my regx is not working and erring out ...

Any suggestion what i have missed ?

# 2  
Old 06-05-2011
echo "v6-Day-2011:420:4:e7ed:45b0:ccc6:776b:62c9" |sed '/^v6-Day-2011/ s/v6-Day-2011://'

# 3  
Old 06-06-2011
echo  $INPUT | awk -F"2011:" ' /^v6-Day-2011/ { print $2 }'

# 4  
Old 06-06-2011
echo 'v6-Day-2011:420:4:e7ed:45b0:ccc6:776b:62c9' |  perl -nle 'print $2 if /v6-Day-(.+?):(.*)$/'

# 5  
Old 06-06-2011
Similar sed..
 echo "v6-Day-2011:420:4:e7ed:45b0:ccc6:776b:62c9" |sed '/^v6-Day-2011:/s///'

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