How to suppress printing of a file with a particular pattern?

[Marko_3]

Dear all

I have a bunch of files which I want to use a shell script to filter out for billing. Only those files with line 79 (fixed line number) having the string 'John' will not be sent to the printer for billing. How can I use a combination of awk and other shell functions to do it?

Many thanks

[James A. Donovan]

Something like the following should work.

for FILE in `ls /dir/dir/files`; do
PRINT=`cat $FILE |head -n 79 |tail -n 1|grep John`

if [ "X$PRINT" = "XJohn" ]; then
lp -d myprinter $FILE
fi
done

[Bruce Regittko_1]

Hi,

Here is a simple script that should do what you want. It will process each file in the current directory. You may replace the * with a file list if you want.

for filename in *
do
if head -79 $filename | tail -1 | grep -q John
then
: # do nothing
else
lp $filename
fi
done

The -q option to grep will suppress grep's stdout.

--Bruce

[Andreas Voss]

Hi,

just another idea:

for file in *
do
awk 'NR==79 && /John/{exit 1}' $file && lp $file
done

Regards

[Printaporn_1]

Hi,
Please use dos2ux command to convert dos file to unix file

#dos2ux dosfile > unixfile

Thanks