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Convert timestamp to epoch time and calculate hours plus conversion backwards

Posted on 2013-01-17
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Last Modified: 2013-01-18
Dear experts, after my last problem I stumbled upon another one.

I need to convert "dd.mm.yyyy:hh24:mi:ss" to epoch time and add some hours to it (e.g. 4h). This output has to be compared to the current time/date.

I need to do so in ksh or perl. Any hints are welcome.
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Question by:skahlert2010
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10 Comments
 

Expert Comment

by:sarathy_sri
ID: 38786611
#! /usr/bin/perl
my @time = localtime(time);
print "@time\n";

print scalar localtime(time);
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LVL 6

Expert Comment

by:deiaccord
ID: 38786638
You should be able to use the perl DateTime module to do what you want

Here is a quick example that takes a date string in the format you specify and prints the difference in hours from the current datetime.

use DateTime;

$datestring = "15.06.2005:15:30:30";
$datestring =~ m#(\d{2})\.(\d{2})\.(\d{4})\:(\d{2})\:(\d{2})\:(\d{2})#;
my $dt = DateTime -> new ( year => $3, month => $2, day => $1,
				hour => $4, minute => $5, second => $6);
				
$dt->add(hours => 4);

my $cmp = DateTime->now()-$dt;
print $cmp->in_units('hours');

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Author Comment

by:skahlert2010
ID: 38786642
Hi sarathy_sri,

I only receive an error using that code ==> syntax error: `(' unexpected
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Author Comment

by:skahlert2010
ID: 38786678
@deiaccord.

I fear perl is not the best soltion as the Perl DateTime Package is not installed on my Solaris machine. The solution needs to be universal in order to be applied on various hosts.

Any idea how to solve it differently? In Ksh maybe?

Otherwise you solution looks promising without being able to test it! Sorry!
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LVL 6

Accepted Solution

by:
deiaccord earned 500 total points
ID: 38787020
I believe Time::Local is standard with most perl distributions so you should be able to use that to get the difference in epoc seconds though it's not quite as flexible.

use Time::Local;

$datestring = "15.06.2005:15:30:30";
$datestring =~ m#(\d{2})\.(\d{2})\.(\d{4})\:(\d{2})\:(\d{2})\:(\d{2})#;
$time = timelocal ($6,$5,$4,$1,$2-1,$3-1900); # seconds, mins, hours, mday, month (0-11), year (difference from 1900)
print "Difference is " . (time - $time) . " seconds\n";

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0
 

Author Comment

by:skahlert2010
ID: 38792056
Hi deiaaccord,

thanks for your solution. I compiled a small perl script using your code and invoke it from within my shell script. However, I cannot pass a parameter to perl in order to bind $datestring.


In ksh:

/tmp/epoch.pl $LS

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while LS equals "15.06.2005:15:30:30"

In Perl:

$datestring = $LS;

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Do you have a hint?
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Author Comment

by:skahlert2010
ID: 38792098
It seems to work with:

$datestring = $ENV{'LS'};

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0
 

Author Comment

by:skahlert2010
ID: 38792243
Well unfortunately this works only if I export the Variable in ksh but not within the script.
Thus, the question is still open.
0
 
LVL 6

Expert Comment

by:deiaccord
ID: 38792256
Glad you got it working, you could have also set it with the first argument to the script if you want to make it a more re-usable script

$datestring = $ARGV[0];

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Author Closing Comment

by:skahlert2010
ID: 38792467
Thanks a lot! Just what I needed! I appreciate your hint regarding passing the variable from ksh to perl with ARGV[n]!
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