UNIX Count Script

Question

I need a UNIX Kornshell (ksh) script that counts the number of files in directory. If the files exceed 20 files, then email results. I want the script to run every hour. I do not have access to cron. I am somewhat new to UNIX. Windows guy all my career. This is what I have so far..

#!/bin/ksh
# count.sh

while :
do
 for dir in /directory1/
 do
  echo "$dir `du $dir |wc -l`" 
 done > ./message
 mailx -s 'Dir Count' [email protected] < ./message
 sleep 3600
done

Any help is greatly appreciated.

Jonathan Leffler · Accepted Answer · 2014-01-12 19:18:12Z

1

The du command as shown will produce one line per directory inside the target directory, /directory1.

Your question isn't clear. It says 'if the files exceed 20 files', which I interpret as meaning 'if the number of files exceeds 20'. You're not dealing with files (as opposed to directories, FIFOs, and other types), and you don't test for "more than 20". I'm going to simplify things, and assume you mean 'if the number of names in the directory exceeds 20'.

dir=/directory1/
while :
do
    names=$(ls "$dir" | wc -l)
    if [ "$names" -gt 20 ]
    then echo "$dir $names" | mailx -s 'Dir Count' [email protected]
    fi
    sleep 3600
done

answered Jan 12, 2014 at 19:18

Jonathan Leffler

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4 Comments

tripleee Over a year ago

Maybe printf "%s\n" "$dir"/* | wc -l to avoid pesky ls parsing issues?

Jonathan Leffler Over a year ago

Maybe...in general, something like that might be necessary, but it is probably not a problem here.

tripleee Over a year ago

If the directory is big and/or load is high, you could be spending more than one hour between runs. Maybe consider a self-scheduling at job if precise timing is important.

Jonathan Leffler Over a year ago

If you use at you get into issues of knowing where your program is installed so you can arrange for it to be run again, and you need to reschedule at the top of the program to avoid time skew but it can still build up over time, but cron avoids that altogether.

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