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Shell Programming and Scripting

Print the first n line in each section

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# 1  
Old 02-01-2013
Print the first n line in each section
Hi,
i have a file like this:

Code:
...
11111111
22222222
33333333
#
4444444
5555555
6666666
7777777
#
...

i want just print the 2 first line between each section (each section is separated with "#"). so desired output would be like this:

Code:
...
11111111
22222222
#
4444444
5555555
#
...

i wrote this code, but it doesn't work!
Code:
awk 'BEGIN{RS="#";FS="\n"}{print $1,$2}' in > out

# 2  
Old 02-01-2013
Code:
awk '/^#/{print;c=0}c<2&&!/^#/{print;++c;}' filename

This User Gave Thanks to Yoda For This Post:
saeed.soltani
# 3  
Old 02-01-2013
it works perfectly, thank you
# 4  
Old 02-02-2013
Elaborating on the OP's original idea
Code:
awk '{sub(/^\n/,x); print $1,$2}' RS=# FS='\n' OFS='\n' file

sed variation of the idea of making an exception for the first record
Code:
sed -n '1,2p; /#/{n;N;p;}'

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