[SOLVED] How to filter output lines from bash command based on directorys

Issue

I have a file called sftp_output with the output of command from a sftp connection that list the content of some folders and look like this:

sftp> ls -l dir1/
-rw-------   1 200      100          1352 Jul 01 14:20 file1
-rw-------   1 200      100          1352 Jul 10 14:20 file2
sftp> ls -l dir2/
-rw-------   1 200      100          1352 Jul 01 14:20 file1
-rw-------   1 200      100          1352 Jul 10 14:20 file2
sftp> bye

What I need to do is filter all the files from dir1 to a single file called "dir1_contents" and files from dir2 to a file called "dir2_contents" . What is the best approach to do something like that?

The expected result needs to be something like this.

File: dir1_contents

file1
file2

What I need to do is filter all the files from dir1 to a single file called "dir1_contents" and files from dir2 to a file called "dir2_contents" . What is the best approach to do something like that?

I tried doing something like this

cat sftp_output | grep -v 'sftp' | awk '{print $9}'| sed '/^$/d'

Solution

This single awk can handle this:

awk 'match($0,/ls +[^\/]+/) { # match a line that has ls command
   s=substr($0,RSTART,RLENGTH)# get the matched substring into var s
   sub(/^ls +/, "", s)        # remove ls and 1+ spaces from s
   close(fn)                  # close file handle, if open
   fn = s "_contents"         # populate variable fn 
   next                       # move to next line
}
fn {                          # if fn is not empty
   print > fn                 # redirect current line to file `fn`
}' sftp_output

Answered By - anubhava

Answer Checked By - David Goodson (WPSolving Volunteer)