UNIX for Dummies Questions & Answers

Hi

I am writing a script to find the list of files in dir1 and my script is place in dir2

while doing ls of files dir1 it is displaying with path. I would like to omit the path and display the only file name so that I can pass it to my script as arguments.


for filename in $datafiles/*.dat
#cat $filename |
do
echo "in for loop $filename "
grep -i "dummy file" $filename > /dev/null || echo $filename >> temp
if [ -s temp ]
then
touch checkfile
fi
done

This script is taking the path also .


Any help Appreciated


Thank you
Madan

Something like this?

Code:

echo "./somedir1/somedir2/somefile"| sed 's!.*\/\([^\/]*\).*!\1!g'
somefile
#or
echo "./somedir1/somedir2/somefile"| awk -F"/" '{print $NF}'
somefile

And please use [c o d e] and [/ c o d e] to display code, ty

Use the basename command. Checkout man basename.