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awk/sed match and extraction

 
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# 1  
Old 09-21-2012
awk/sed match and extraction
Hi,

I have a file like this-

Code:
aa
12
23
34
aa
21
34
56
aa
78
45
56

I want to print out only the lines after the last aa. How do I do this? I tried using grep -A and sed -n, but both didnt work as I wanted to.
Could someone help me out please..
# 2  
Old 09-21-2012
try this..
Code:
$awk '{if($0 ~ /aa/) {s="";getline;s=$0} else{s=s"\n"$0}}END{print s}' file
78
45
56

This User Gave Thanks to pamu For This Post:
jamie_123
# 3  
Old 09-21-2012
it works Smilie thanks!
# 4  
Old 09-21-2012
This should also work correctly if there are no lines with "aa":
Code:
awk 'f{p=p $0 RS} /^aa$/{p=x; f=1} END{if(f)print p}' ORS= infile

This User Gave Thanks to Scrutinizer For This Post:
jamie_123
 
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