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sed print first line before regexp and all lines after


 
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# 1  
Old 01-15-2013
sed print first line before regexp and all lines after

Hi All

I'm trying to extract the line just above a regexp and all lines after this.

I'm currently doing this in two steps

Code:
sed -n -e "/^+---/{g;p;}" -e h oldfile.txt > modified.txt
sed -e "1,/^+---/d" -e "/^$/d" oldfile.txt >>modified.txt

Sample
Code:
sometext will be here
sometext will be here
sometext will be here
sometext will be here



 ACD                   (A)                   
+------------------------+------------------------
 (some info)                        19082
 (something else)                        19082

Result
Code:
 ACD                   (A)                   
 (some info)                        19082
 (something else)                        19082

Is there a way I can achieve this using a one liner?
# 2  
Old 01-15-2013
I'm not sure about sed. Here is a solution using awk:
Code:
awk '/^\+/{ print x; getline; f=1; } { x=$0 } f==1 { print $0 }' oldfile.txt

# 3  
Old 01-16-2013
Code:
awk '/^\+-/{$0=p; f=1}{p=$0}f' file

This User Gave Thanks to Scrutinizer For This Post:
# 4  
Old 01-21-2013
I tried this one and it works. Thank you

Quote:
Originally Posted by Scrutinizer
Code:
awk '/^\+-/{$0=p; f=1}{p=$0}f' file


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