Increment a date variable in perl script


 
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# 1  
Old 01-06-2010
Question Increment a date variable in perl script

Hi,

I have a perl script which prints epoch value of date in milliseconds as the output.
My reuirement is that once the output is printed,the day variable shld increment by 1 and when i execute the script for the second time the output shld be for the new day value.

My script looks as below:
======================================================
Code:
#!/usr/bin/perl
use Time::Local ;
$sec=00;
$min=01;
$hours=00;
$day=14;
$month=7;
$year=109;
#
$Beg_time = (timelocal($sec,$min,$hours,$day,$month,$year) * 1000);
print "Epoch Beg Value: $Beg_time\n";

======================================================

1st execution output will be for "14th of August"

When i execute second time the output shld be for 15th August.

Can someone pls suggest.

Last edited by Franklin52; 01-06-2010 at 04:49 AM.. Reason: Please use code tags!
jyothi_wipro
# 2  
Old 01-06-2010
Store result in file .
Next time read the file and increment .
"File" can be file , DB , whatever persistent .
# 3  
Old 01-06-2010
hi,

I would like to increment $day value to 15 after first execution of the script.If i understood correctly the solution u gave is to store the result ie output in a file.
How to increment $day=14 to $day=15 ?
jyothi_wipro
# 4  
Old 01-06-2010
Debian

Code:
my $somefile = "/tmp/whatever" ;

sub store  { 
  my $what = shift ; 
  open my $file , ">$somefile" or die "$!" ; 
  print $file "$what"  ;
  close $file;
}

sub read_it { 
  open my $file , "<$somefile" or die "$!" ; 
  my $result = <$file> ; 
  close $file ;
  return $result ;  
}

if  ( -f  $somefile ) { 
   my $result = read_it();
   $result += 24 * 3600 ; # increment 1 day , date should be in epoch format 
   store($result);
   # do whatever you want with new result
} else {
  # original code  
  # 
  store($starting_date);
}

Store data in file in epoch format .
There is almost no exception handling in this code and it will not work concurrent .
# 5  
Old 01-06-2010
You can check the date/timestamp on the file and then just increment the date/day by 1.
# 6  
Old 01-06-2010
Try:

Code:
#!/usr/bin/perl -w
use Time::Local ;
$sec=00;
$min=01;
$hours=00;
$day=14;
$month=7;
$year=109;
#
$Beg_time = (timelocal($sec,$min,$hours,$day,$month,$year) * 1000);
$i=0;
while (++$i<10) {
$val=$Beg_time*86400000;
printf "Epoch Beg Value %d:%f\n",$i,$val;
}

# 7  
Old 01-06-2010
Thanks for the reply but this does not give me what is required.I hope my query was clear.
when i execute the code which i posted it prints me the epoch value for "Aug 14 00:01:00".
Once it has printed the output,i want that the day value is inremented by 1 ie "Aug 15 00:01:00" so that i can get the epoch value of Aug 15 when i execute the script the second time.


Hope I get some reply which suits my requirement.
jyothi_wipro
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