Unix: System date and time to a variable, and using that variable

Caveat: I am a Unix novice.  There may be better ways to do this, but this worked for me.

Show system date and time in the default format:

$ date
Thu Jun 27 21:23:36 CDT 2013

Show system date and time in yymmdd_hhmmss format (two digit year):

$ date "+%y%m%d_%H%M%S"
130627_212355

Show system date and time in yyyymmdd_hhmmss format (four digit year):

$ date "+%Y%m%d_%H%M%S"
20130627_212416

Save the system date and time in yyyymmdd_hhmmss format to a variable, then show it:

$ var="$(date '+%Y%m%d_%H%M%S')"
$ echo datetime is $var
datetime is 20130627_212437

My requirements: Schedule a SAS program (pd.sas) to run at 4am on June 28th.  Send the SAS log to a file whose name includes the system date and time so we can archive the logs.

Step 1: Create a (script) file called go

#!/bin/bash
ts="$(date '+%Y%m%d_%H%M%S')"
echo "sas pd.sas -log pd_$ts.log" | at 0400 Jun 28

Step 2: Execute a (script) file called go

$ . go