How can I automate this UNIX script

I want to automate the following commands so that I could run it as a cronjob on my Solaris10 server:

I have a source directory :
/var/www/temp1/
The files are like
1356652700.file1.dat.bz2.cor
1356652700.file2.dat.bz2.cor
1356652700.file3.dat.bz2.cor

I know the file name part except for the timestamp part (1356652700). Each time the cron is run, the timestamp part of the set of files will be different. In the next run I will probably have these in the source: 1356658200.file1.dat.bz2.cor , 1356658200.file2.dat.bz2.cor and 1356658200.file1.dat.bz2.cor.

So I do:

#!/bin/sh
DIR1=/var/www/temp1/
DIR2=/var/www/process/

Now I manually run:
$cd /var/www/process/
$mv 1356652700.file1.dat.bz2.cor 1356652700.file1.dat.bz2
$mv 1356652700.file2.dat.bz2.cor 1356652700.file2.dat.bz2
$mv 1356652700.file3.dat.bz2.cor 1356652700.file3.dat.bz2
$bzip2 -dc 1356652700.file1.dat.bz2 > 1356652700.file1.dat
$bzip2 -dc 1356652700.file2.dat.bz2 > 1356652700.file2.dat
$bzip2 -dc 1356652700.file3.dat.bz2 > 1356652700.file3.dat
$rm *.bz2
$mv 1356652700.file1.dat tab1.dat
$mv 1356652700.file2.dat tab2.dat
$mv 1356652700.file2.dat tab3.dat
$./my_cron.cron
$sqlplus mylogin/mypwd@myserver.com/SOMETHING <<ENDOFSQL
create or replace procedure my_prod as
begin
  declare
    db_package.db_proc1('1356652700.file1.dat');
    db_package.db_proc2('1356652700.file2.dat');
    db_package.db_proc2('1356652700.file3.dat');
  end;
end load_con;
exec DBMS_SCHEDULER.RUN_JOB('MY_JOB', FALSE);
exit;
ENDOFSQL