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How can i delete a keyword containing XYZ in unix


 
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# 1  
Old 09-11-2009
How can i delete a keyword containing XYZ in unix

Hi all,

I am trying to remove the words which has XYZ as a prt of that.

My input file is something like this :

Code:
PHNDAZLF-UPS-XYZ' aaaaaaa bbbbb
ADFRTEJKS-XYZ cccccccc ddddddd rrrrrr
SGETHEHDJ-ABC-RXY' hhhhh ttttt' kkkk
FHJSKSJDKD-XXX-YYY

Output expected is :

Code:
aaaaaaa bbbbb
cccccccc ddddddd rrrrrr
SGETHEHDJ-ABC-RXY' hhhhh ttttt' kkkk
FHJSKSJDKD-XXX-YYY


Last edited by rdhanek; 09-11-2009 at 05:31 AM..
# 2  
Old 09-11-2009
Have you tried grep? Have a read of the man page.

Regards
# 3  
Old 09-11-2009
Hi,

You can use the below code:

Code:
grep xyz filename > temp_filename
if [ $? -eq 0 ] ; then
echo "Pattern exist in filename"
sed "/$value/d" filename > temp_filename
mv temp_filename filename
echo "Pattern deleted"
else
echo "Pattern does not exist in filename"
fi

Cheers,
Shazin
# 4  
Old 09-11-2009
Or Simply

Simply do this:
grep -v XYZ > file_without_XYZ
# 5  
Old 09-11-2009
I apologize for the confusion. But i want to delete the words having XYZ as a part of them and not the total lines.

I have also modified the example above now.
Not sure if this is still possible with grep??
# 6  
Old 09-11-2009
Its simple using sed

Code:
sed 's/xyz//' FILENAME

# 7  
Old 09-11-2009
something like:

Code:
sed -e "s/[a-zA-Z-][a-zA-Z-]*XYZ[' ]*//g" -e "s/XYZ[a-zA-Z-][a-zA-Z-]*//g" filename


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