UNIX for Advanced & Expert Users

How do I get the number of seconds since 1970, within a script, for the previous day at 23:59? I need this value to pass into a sql statement to cleanup records older than the previous day at midnight. It will be automated via cron so no hard coding allowed.

Thanks!

This will return the number of seconds elapsed between January 1, 1970, and the specified time , ie yesterday at 23:59:00.

Code:

#!/usr/bin/perl
$now = time();
($sec,$min,$hour,@etc) = localtime($now);
$calc_secs = $now - (60*60*$hour) - (60*$min) - $sec - 60;
print "$calc_secs\n";

Thanks to all for the perl help. I don't have perl on my servers so found a program on the net, compiled it and did this:

#setup the datetime variables to get last night at 23:59:59
cc=`date +"%C"`
yy=`date +"%y"`
mm=`date +"%m"`
dd=`date +"%d"`
hh=23
mi=59
ss=59

#assign the parameters to the variable
newtime="$mm $dd $cc$yy $hh $mi $ss"

#get the number of seconds since 1970 for tonight at 23:59:59
newseconds=`./seconds $newtime`

#get last night's seconds at midnight by subtracting one day from #midnight
let midnight=$newseconds-86400

#append the delete statement, with the system time, to the sql #script
echo "delete from oas_pvt_empty_order_data where expiration_date < $midnight;">> delete_contracts.sql