Shell Programming and Scripting

I have a log file which contains data like this
This log file is updated twice a day at 7am and 6pm,
I want a script(which i will make run at 7:10am and 6:10pm) which should fetch only the last appended lines since last update..

I mean.. if i execute the script at 7.10am 3/3/2010
it should fetch only
5555521.log exist
345745.log exist

and if i execute the script at 6:10pm 3/3/2010
it should fetch only
1982028.log exist
234454.log exist
123.log exist
666666.log exist
234355.log exist
566556.log exist
4456778565.log exist
2345454.log exist

How can i achieve this.. obviously i cannot user "Tail" command since i don't know how much lines of output it's gonna give..

Please let me know how can i achieve this???
Any help is highly appreciated !! thank You

of course you can.
wc before, count the lines, or characters

tail -c
tail -num

read man tail
man wc

Code:

gawk 'BEGIN{RS="";FS="\n";now=systime()}
      {split($1,a,"[ /:]")};
      {t=mktime(a[8]" "a[6]" "a[7]" "a[9]" "a[10]" "a[11])};
      {if ((now-t)<1000) print $0}' urfile

1000 is second, means to get the logs which was generated less than 1000s, you can ajust it for your request.

You can get the date and time variables then extract what you want using sed

Code:

var_time=$(date '+%m/%d/20%y %k')
 sed -n "/^Starting.*$var_time/,/^$/p" file

var_time=$(date '+%m/%d/%Y %k')