Shell Programming and Scripting

#! /usr/bin/ksh
/var/logs/
ls -ltr 
XSLog2012.03.28-09.20.00.txt XSLog2012.03.28-09.32.00.txt
XSLog2012.03.28-09.49.00.txt XSLog2012.03.28-10.01.00.txt 
#prompt user for start date and time
echo you are about to search filename based time range of an hour"
echo Time year month day hour start = " read stime 
echo  Time year month day hour stop  = " read etime
# have left out rest of the action on the file that strips out data

echo enter year: "
read YYYY
echo enter month: "
read MM
echo enter day: "
read DD
echo enter Hour: "
read HH
 sed -e '/./{H;$!d;}' -e 'x;/1234/!d'  XSLog$YYYY.$MM.$DD-$HH*

printf "Enter start YYYY/MM/DD HH "
# Should work for YYYY/MM/DD HH or YYYY MM DD HH
IFS="/ " read S_YYYY S_MM S_DD S_HH

printf "Enter end YYYY/MM/DD HH "
IFS="/ " read E_YYYY E_MM E_DD E_HH

EARLIEST="XSLog$S_YYYY.$S_MM.$S_DD-$S_HH.00.00.txt"
LATEST="XSLog$E_YYYY.$E_MM.$E_DD-$E_HH.59.59.txt"

ls -1 | awk -v E="$EARLIEST" -v L="$LATEST" '($0 >= E) && ($0 <= L)' | xargs sed -e '/./{H;$!d;}' -e 'x;/1234/!d'