awk, print $PWD

[ste2703]

Hi,

let's say I have a directory with a certain number of .txt files.
I want to create a file where I get:

myself:~/etc/etc_plus/01.txt
myself:~/etc/etc_plus/02.txt

and so fort for each .txt file.

I tried

ls -lar *.txt | awk -v pwd="$PWD" '{print "$pwd"/"$8"}'

but it doesn't work.. any suggestion? thanks a lot

[ste2703]

sorry, I actually tried:

ls -lar *.txt | awk -v pwd="$PWD" '{print "$pwd"/"$8"}' > list_files.txt

[nmset]

After some experimenting and web searching about awk, it seems to me you should write this way :

ls -lar *.txt | awk -v vpwd="$(pwd)" '{print $vpwd"/"$8}' > list_files.txt

The slash is interpreted litterally and not as an operator if it is quoted.

You would have the same result with :

for f in $(ls -ar *.txt); do echo "$(pwd)/$f";done

i.e; without resorting to awk.

[ste2703]

thanks a lot.

your first solution seems to work like mine... instead of pwd I get file information, like rights, dimension, date, etc etc..

the second solution make the job.
nevertheless if I write a file, the last stream overwrite the previous one and at the end my file has just one file...

for file in $(ls -ar *.txt); do
comp=${file%.*}
echo $comp $PWD/$file > comp.txt
done

[ste2703]

Done!

I didn't know the difference between > and >>.

Thanks a bunch!

[keikun2004]

find "$PWD" -iname "*.txt"
do the work XD