Shell script - getting Time difference using awk

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# 1  
Old 10-23-2013
RedHat Shell script - getting Time difference using awk

Hi..I have the data in a file like in this format, and I need the output time difference in seconds by using awk command. Start date/time and end date/time given in column 2,3 & 4,5. Please assist how to write shell script.
JOB1 10/09/2013  17:42:16 10/09/2013  17:43:46 SU 6202685/1
JOB2 10/09/2013  18:05:17 10/09/2013  18:06:30 SU 6202787/1
JOB3 10/10/2013  00:05:19 10/10/2013  00:06:42 SU 6204211/1
JOB4 10/10/2013  03:05:15 10/10/2013  03:06:19 SU 6204847/1
JOB5 10/16/2013  06:05:13 10/15/2013  20:05:45 SU 3142148/1

Moderator's Comments:
Mod Comment Please use mext time code tags for your data and code, thanks
# 2  
Old 10-23-2013
#! /bin/bash

while read a d1 t1 d2 t2 b
    s=$(date -d "$d1 $t1" +%s)
    e=$(date -d "$d2 $t2" +%s)
    if [ $e -ge $s ]
        echo "$(( e - s ))s"
        echo "$(( s - e ))s"
done < file

# 3  
Old 10-23-2013
try perl:
perl -e '
use Time::Local;
while (<>) {
   print $_;
   @a=split(/  */, $_);
   @d1=split(/[\/:]/, @a[1]. ":" . @a[2]);
   @d2=split(/[\/:]/, @a[3]. ":" . @a[4]);
   $td=$t2 - $t1;
   print $td . "\n";
}' infile

# 4  
Old 10-23-2013
What's your system? awk has date features on a few systems but not on others.
# 5  
Old 10-24-2013
Using SunOS... Great thanks... the first script working fine.
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