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Print only next pattern in a line after a pattern match

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# 1  
Old 06-12-2013
Print only next pattern in a line after a pattern match
I have
Code:
2013-06-11 23:55:14 [17780] 1Umexd-0004cm-IG <= user@domain.com

I need sed/awk operation on this, so that it should print the very next pattern only after the the pattern mach <=

ie only print user@domain.com
# 2  
Old 06-12-2013
Code:
awk 'BEGIN{FS="<= ";}{print $2}'

This User Gave Thanks to rajamadhavan For This Post:
anil510
# 3  
Old 06-12-2013
Different way to enter field separator
Code:
awk -F"<= " '{print $2}'
awk '{print $2}' FS="<= "

another way
awk '{split($0,a,"<= ");print a[2]}'
This User Gave Thanks to Jotne For This Post:
anil510
# 4  
Old 06-12-2013
Thanks,

Its printing patterns after user@domain.com. I just need the next coloumn to <=

Code:
]# echo "2013-06-11 23:55:14 [17780] 1Umexd-0004cm-IG <= user@domain.com  XXX YY ZZZ" |awk 'BEGIN{FS="<= ";}{print $2}'
user@domain.com  XXX YY ZZZ

# 5  
Old 06-12-2013
Code:
echo "2013-06-11 23:55:14 [17780] 1Umexd-0004cm-IG <= user@domain.com  XXX YY ZZZ" | awk -F"<= " '{split($2,a," ");print a[1]}'
user@domain.com

This User Gave Thanks to Jotne For This Post:
anil510
# 6  
Old 06-12-2013
Code:
 echo '2013-06-11 23:55:14 [17780] 1Umexd-0004cm-IG <= user@domain.com' | perl -nle 'print $2 if /(.*)\<\=\s*(\S+)\s*/;'

# 7  
Old 06-12-2013
Thanks Jotne. Its working fine Smilie
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