How do I cut just the file name from from ls -l output? [closed]

Question

I am in the process of concatenating ogg audio files in bulk with the help of ffmpeg using a mylist.txt file

The format of the mylist.txt file is

file '/path/to/file1.wav' file '/path/to/file2.wav' file '/path/to/file3.wav' 

My 'ls -l' output sample of WhattsAPP ogg audio files is similar to

-rw-rw-r-- 1 work work 64112 Nov 14 18:43 'WhatsApp Ptt 2019-11-11 at 10.19.18 AM.ogg' -rw-rw-r-- 1 work work 24616 Nov 14 18:43 'WhatsApp Ptt 2019-11-11 at 10.19.50 AM.ogg' -rw-rw-r-- 1 work work 26166 Nov 14 18:43 'WhatsApp Ptt 2019-11-11 at 10.20.18 AM.ogg' -rw-rw-r-- 1 work work 69895 Nov 14 18:43 'WhatsApp Ptt 2019-11-11 at 10.21.05 AM.ogg' -rw-rw-r-- 1 work work 85416 Nov 14 18:43 'WhatsApp Ptt 2019-11-11 at 10.27.09 AM.ogg' 

1. How do I get only the 'WhatsApp Ptt 2019-11-11 at 10.19.18 AM.ogg' portions of each line in the 'ls -l' output using cut, sed, awk or any other tool into a file using a single command?

2. How do I add the keyword "file" before all file names in the file using a single command?

3. Can process 1. & 2. be combined into a single command?

The contents of the final file need to look like:

file 'WhatsApp Ptt 2019-11-12 at 10.21.59 AM.ogg' file 'WhatsApp Ptt 2019-11-12 at 10.29.45 AM.ogg' file 'WhatsApp Ptt 2019-11-12 at 10.31.52 AM.ogg' file 'WhatsApp Ptt 2019-11-12 at 9.31.38 AM.ogg' 

Answer 1

It's simpler to use a glob instead of ls to work with a list of file names

for f in *; do echo file \'$f\' done 

Answer 2

/bin/ls -1

That leaves out all the goo :)

The problem being that ls in most distributions is aliased to something like “ls -la”. You can check that with “alias ls” and see if anything comes up

Also using an absolute path (with /bin/) will avoid using the alias.