Bash : sed -n arguments

Question

While reading results of a ls instance, I'm trying to display a specific term from a specific line using sed :

Here is what I managed to do until now :

echo $(ls -lAtro | sed -n '3p' | cut -d' ' -f 6 )

This is reading the 6th term of the 3rd line. What I'm trying to do is replace the '3p' by a more simple variable which would permit to do something like this :

linetoread=3
echo $(ls -lAtro | sed -n "$linetoread" | cut -d' ' -f 6 )

The above example doesn't work of course, but you get the idea. What can I do to achieve this ? Thanks in advance.

Answer 1

Try awk instead-

line=3 ; 
ls -lAtro | awk -v var="$line" 'NR==var {print $6}' 

This takes the variable line=3 and feeds it into awk (the awk -v part).

The NR==var ensures that you get the line corresponding to your variable.

Then you can change the print $6 part to match whichever column you want. Note that the default delimiter is a space.

Of course, as pointed out in the comments, you shouldn't be parsing the output of ls .