Shell Programming and Scripting

#!/usr/bin/ksh

###############################################################################
# #
#-----------------------------------------------------------------------------#
# Shell Script Name : getlogs #
#
# Location (directory) : /usr/local/bin #
#-----------------------------------------------------------------------------#
# #
###############################################################################

#typeset -u export SID=$1
#typeset -l export sid=$1

Hostname=$(hostname)

usermail="abcd@pqrst.com"

if [ "`find . -name "/dir1/dir2/abcd*.log" -mmin 10 | wc -l`" -gt "0" ]
then
ls -1tr /dir1/dir2/abcd*.log | sed -n '1,/^lastlog.txt$/! p' > newlog.txt

mv newlog.txt lastlog.txt

sleep 30

/usr/bin/mailx -s "${HOSTNAME}: subject" $usermail < $/dir1/dir2/newlog.txt
fi

When I run this script it gives an error ./getISTlogsnew.ksh
find: bad option -mmin

Please advise ..

Hi Yogi,

you can get latest file as

ls -1tr abc* | tail -1

scripter

Please use code tags.

your find doesn't support mmin option.
you can make a ref timestamp with touch.

with the reference timestamp file, you can use test FILE1 -ot FILE2 (is older than) to make the determination.

Please read my message fully.
I had said at the bottom:

Code:

If you want in your code check for [[ -s oldlog.log ]]

Use an "if" command and check if the file size is not 0.