Shell Programming and Scripting

Hi guys,

I'm new to the forums and putting my foot in the door with SED and AWK. I was wondering if someone could help me as I think I'm making this harder than it needs to be...

I have a list of folders named as urls, inside these are log files and possibly a 'status' file. I'm trying to get a list of folders that do not have the file called 'status'.

So if the folder url2 has status the list would be:
url1
url3

The current command I'm using is:

Code:

ls -R | grep -v "****-**-**.log.*" | sed -n '/.\/www.*.*/,+1p'

At the moment this is giving me the following:
./url1

./url2
status

./url3

Any help would be much appreciated, only I only used the grep -v command because I was spending so much time on the regex haha.

directly from the system

Code:

find /path/to/dirs -type d |
while read dir
do
  ls ${dir}/status >/dev/null 2&>1 || echo $dir
done

I think I have what you want.

Or from a file with directory names.

Code:

while read dir
do
  ls ${dir}/status >/dev/null 2&>1 || echo $dir 
done < list_of_directories.txt

Ok so I ran your command on the home directory and it's getting very close to what I need.

This is the output I get, but it seems it is still showing the url2 directory which has the status file inside:

/home
/home/url1
/home/url2
/home/url3

Try something like this:

Code:

find .|grep www|nawk -F/ '/status$/{a[$2]=1};{b[$2]=1};END{for(i in b){if(a[i]!=1){print i}}}'

You have to be in the directory where the url-directories are to run this command though

I had to install nawk, but that worked perfectly, thank you!
Also thank you to Jim for your help.

You can just use awk as well the same way.

So you can... Awesome Thanks again.