Printing nth and n+1th line after a pattern match


 
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# 1  
Printing nth and n+1th line after a pattern match

Hi ,
I want to print the nth and n+1 lines from a file once it gets a pattern match.
For eg:

aaa
bbb
ccc
ddd
gh
jjjj

If I find a match for bbb then I need to print bbb as well as 3rd and 4th line from the match.. Please help..Is it possible to get a command using sed Smilie
# 2  
much easier with awk:
Code:
awk '/bbb/ { print ; for(n=0; n<2; n++) { getline ; print } }' inputfile

# 3  
Thanks a lot Corona688 ... But is it possible to get a command in sed ..
I tired the below one .. but looks like some issue is there with the command ..

sed -n '/bbb/,+2p' filename
# 4  
Right tool for the right job and all that. Why sed? Is this homework?
# 5  
Smilie No not home work Smilie I am learning sed commands ..
# 6  
Code:
$ sed -n '/bbb/ {p;n;p;n;p;}' infile

# 7  
Try this one
Code:
grep -A2 'bbb' inputfile

 

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