I am looking to display the 8th file name in a directory using ls -l and pipe for example ls -l | wc -1 would give me the count but what i want to have returned is the 8th file name in the directory list whatever it is. i have looked at grep and wildcards but still do not see which command would give me the result i am looking for.
Thanks
try
ls -l | head -n 9 | tail -n 1
if you want only name you can use cut at the end
ls -l | head -n 9 | tail -n 1 | tr -s ' ' | cut -d' ' -f8
tr is to replace multiply spaces and tabs by one space
Another way is like this.
ls -l | sed -n '9p'
If you only want the file or directory name then use this.
ls -l | sed -n '9p' | awk '{print $9}'
jcubic are you sure this is what you wanted?
ls -l | head -n 9 | tail -n 1 | tr -s ' ' | cut -d' ' -f8
I would think he would want the file or directory name like this.
ls -l | head -n 9 | tail -n 1 | tr -s ' ' | cut -d' ' -f9
ls | awk 'NR==8'
ls -l to get 1-column output, or even ls -1 if you are throwing output to a pipe (see e.g. What is the magic separator between filenames in ls output?).NR==8 will only return true for line 8, and then awk default action is to print it.
Just don't use -l
ls | head -8 | tail -1
You get file name, ls orders files same as ls -l.
