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Parse a file to display lines containing a word

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# 1  
Old 03-04-2010
Parse a file to display lines containing a word
Hi!

I'm trying to create a shell script to parse a file which might have multiple lines matching a pattern (i.e. containing some word).
I need to return all lines matching the pattern, but stripping the contents of that line until the pattern is matched

For example, if my input file was
Code:
aaaaaaa bbbbbbbb ccccccc
ddddddd eeeeeeee fffffffffff
ggggggg WORD hhhhhh
iiiiiiiiiiiiiiiii jjjjjjjjjjjjjjjj  kkkkkkkkk WORD mmmmmmm

I need to return
Code:
WORD hhhhhhhh
WORD mmmmmmm

What would be the best way of doing this, using only sed/awk/grep?

Thanks!
-orno
# 2  
Old 03-04-2010
One way:
Code:
sed -n 's/.*\(WORD.*\)/\1/p' file

# 3  
Old 03-04-2010
Code:
nawk 'match($0, "WORD") {print substr($0, RSTART)}' myFile
nawk '/WORD/ {print substr($0, index($0, "WORD"))}' myFile

# 4  
Old 03-04-2010
Bug Is this what you are looking for?
Code:
>sed "s/WORD /WORD_/" < orno.txt | sed 's/ /\n/g' | grep "^WORD" | sed "s/WORD_/WORD /"
WORD hhhhhh
WORD mmmmmmm

# 5  
Old 03-04-2010
Assuming the word needs to be parameterized and is stored in a shell variable:

Code:
$ w=WORD
$ awk -v w="$w" 'i=index($0,w){print substr($0,i)}' data

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