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Pint 2nd line after particular pattern

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# 1  
Old 05-27-2013
Pint 2nd line after particular pattern
Hi All
I want to make script to find out row value after particular pattern from file
Sample Input file.Here pattern is ABCD

Code:
38403040
010459
20130527
173659
39703015
ABCD
15933998
38403040
000042
20130527
173659
39703015
ABCD
15934006
38403041

Output file will be like

Code:
38403040
38403041

output will be having 2row below pattern ABCD.
Can any one help ?
# 2  
Old 05-27-2013
Code:
awk '/ABCD/ { getline ; getline ; print }' inputfile

# 3  
Old 05-27-2013
A few examples I've used:
Code:
# Print the line immediately before 'string', but not the line containing the 'string'.
sed -n '/string/{g;1!p;};h'

# Print the line immediately after 'string', but not the line containing the 'string'.
sed -n '/string/{n;p;}'

# Print one line before and after 'string', also print the line matching 'string' and its line number.
sed -n -e '/string/{=;x;1!p;g;$!N;p;D;}' -e h

# 4  
Old 05-27-2013
Another awk approach:
Code:
awk '/ABCD/{c=NR}c&&(NR-c)==2' file

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