Shell Programming and Scripting

I have a directory location where in some logs gets generated whenever some application build is triggered.
I need to send the generated log as an email to end user.
I will get files like abcyyyy_mm_dd_time.log and next file will have different time for same day.
Need to run a cron and take the latest file and email.
Need help how will I compare and get the latest file each time and send it across.
Any help asap will be highly appreciated ..

Below will always give you the latest

Code:

ls -1t abc*.log | head -1

Rest is just little coding.

Hi,

Below command can be used to take 5 recently modified or created files

Code:

  ls -1t dir/*|head -5

Thanks for the Info but ls -lt | abc.log | head -1 will give the latest .
But say in last 10 minutes 3 logs got generated then I need to get those 3 and send too.
In a nutshell whatever gets created after I last sent the log file needs to be captured and sent...

In such a case if abc.123.log is the last file I have sent
I will list down all the files with ls -lt | abc.*.log
and find the position of abc.123.log in the listing
and now head command with little alteration in -count can give me names of all the files.
Try it out...

btw it was 1(one) and not l(ell)

Make file prevstamp = when you have done mail process. Find those files which are newer.

Code:

#!/bin/ksh
smail()
{
fn="$1"
/usr/lib/sendmail -t -i <<EOF
From: myemail
To: youremal
Subject: $fn

Here is file $fn
$(cat $fn)
EOF
}

##main##
[ ! -f prevstamp ] && touch -m 20010101120000 prevstamp
touch newstamp
find . -maxdepth 1  -name "*.log" -newer prevstamp | while read fname
do
        smail "$fname"
done
cp -fp newstamp prevstamp 2>/dev/null
rm -f newstamp 2>/dev/null

Thanks for the script but why are you doing this here

20010101120000 prevstamp

Please explain...