I am trying to convert hex time(getting first output from kernel module) into nanosecond granularity,
580a9272.0a9ce167
and I am trying to convert it using perl into human readable format:
while (<>) {
s/^([a-fA-F0-9]+)(\.)([a-fA-F0-9]+)(\s+.*)/sprintf("%s%s%s%s",&$converter(hex($1)), $2, hex($3), $4)/oe;
} continue {
print;
}
output : Fri Oct 21 18:10:58 2016.178053479
Converter uses localtime() and gmtime() directly
I want time with nano granularity and then year.Any help is highly appreciated.
POSIX::strftime doesn't support fractional seconds, so you need to build the output in parts.
use POSIX qw( strftime );
my $opt_gmt = 1;
my $hex = '580a9272.0a9ce167';
my ($s, $ns) = map hex($_), split /\./, $hex;
my $formatted_ns = sprintf("%09d", $ns);
my $formatted = strftime("%a %b %d %H:%M:%S.$formatted_ns %Y",
defined($opt_gmt) ? gmtime($s) : localtime($s));
say $formatted; # Fri Oct 21 22:10:58.178053479 2016
DateTime has native support for nanoseconds, so that presents an alternative.
use DateTime qw( );
my $opt_gmt = 1;
my $hex = '580a9272.0a9ce167';
my ($s, $ns) = map hex($_), split /\./, $hex;
my $dt = DateTime->from_epoch( epoch => $s );
$dt->set_nanosecond( $ns );
$dt->set_time_zone( defined($opt_gmt) ? 'UTC' : 'local' );
say $dt->strftime("%a %b %d %H:%M:%S.%N %Y"); # Fri Oct 21 22:10:58.178053479 2016
Related
I want to get the system clock (time and date) and display it in a human-readable format in Perl.
The format like 2014-09-12 15:13:56
#!/usr/local/bin/perl
my %months = qw(Jan Feb Mar Apr May Jun Jul
Aug Sep Oct Nov Dec);
#weekDays = qw(Sun Mon Tue Wed Thu Fri Sat Sun);
($second, $minute, $hour, $dayOfMonth, $month, $yearOffset, $dayOfWeek, $dayOfYear,
$daylightSavings) = localtime();
$year = 1900 + $yearOffset;
$now = "$year-$months-$dayOfMonth $hour:$minute:$second";
print $now;
When you run the program, you should see a much more readable date and time like this:
2014--12 16:57:15
how to get convert the month to number ?
Using Time::Piece (core module since perl v5.9.5)
use Time::Piece;
my $dt = localtime;
print $dt->ymd, " ", $dt->hms, "\n";
using DateTime
use DateTime;
my $dt = DateTime->now();
print $dt->ymd, " ", $dt->hms, "\n";
It's easier using a Perl module (POSIX doesn't requires installation):
use POSIX qw/strftime/;
my $now_string = strftime "%Y-%m-%d %H:%M:%S", localtime;
print $now_string, "\n"; #<-- prints: 2014-09-12 11:09:45 (with my local time)
Regarding to your code, there is a typo:
$now = "$year-$months-$dayOfMonth $hour:$minute:$second";
should be:
$now = "$year-$month-$dayOfMonth $hour:$minute:$second";
Be sure to write use strict; and use warnings; in the top place of your script. It prevents you from errors like that.
I like to put these date and time tasks into functions for reuse.
Here is my approach:
use strict;
use warnings;
my $time_stamp = getTodaysDateTime();
print "Program Started: $time_stamp \n";
# do some processing
$time_stamp = getTodaysDateTime();
print "Program Ended: $time_stamp \n";
# return date in specific format
# ex: 2014-09-12 14:11:43
sub getTodaysDateTime {
my ($sec,$min,$hour,$mday,$mon,$year,
$wday,$yday,$isdst) = localtime(time);
$year += 1900;
$mon += 1;
return sprintf("%d-%02d-%02d %02d:%02d:%02d",
$year,$mon,$mday,$hour,$min,$sec);
}
I searched through all the possible questions but couldn't find the answer,
so can Perl experts help me on this one?
I have two timestamps like 05/25/2011 05:22:03 PM and 05/25/2011 05:34:08 PM. They are stored in string form.
my $str1 = '05/25/2011 05:22:03';
my $str2 = '05/25/2011 05:34:08';
The latter being the time of a job ending and former being the time it started.
How do I find out the difference in dates and time? The dates are the same in this case but they could differ as well.
I recommend that you use the Time::Piece module. It has been a core module since the release of version 9.5 of Perl 5, so it shouldn't need installing.
This code demonstrates
use strict;
use warnings;
use Time::Piece;
my $str1 = 'Execution started at 05/25/2011 05:22:03 PM';
my $str2 = 'Execution completed at 05/25/2011 05:34:08 PM';
my #times = map Time::Piece->strptime(/(\d.+M)/, '%m/%d/%Y %H:%M:%S %p'), $str1, $str2;
my $delta = $times[1] - $times[0];
print $delta->pretty;
output
12 minutes, 5 seconds
You can take advantage of DateTime and its subtract_datetime() method, which returns a DateTime::Duration object.
use Date::Parse;
use DateTime;
my $t1 = '05/25/2011 05:22:03';
my $t2 = '05/25/2011 05:34:08';
my $t1DateTime = DateTime->from_epoch( epoch => str2time( $t1 ) );
my $t2DateTime = DateTime->from_epoch( epoch => str2time( $t2 ) );
my $diff = $t2DateTime->subtract_datetime( $t1DateTime );
print "Diff in minutes: " . $diff->in_units('minutes') . "\n";
print "Diff in hours: " . $diff->in_units('hours') . "\n";
print "Diff in months: " . $diff->in_units('months') . "\n";
In perl, how do I convert date like
Thu Mar 06 02:59:39 +0000 2008
to
2008-03-06T02:59:39Z
Tried HTTP::Date, it works if the question did not have +0000 in the string :(
DateTime::Format::Strptime will do this conversion.
#!/usr/bin/perl
use strict;
use warnings;
use 5.012;
use DateTime::Format::Strptime;
my $date = 'Thu Mar 06 02:59:39 +0000 2008 ';
my( #strp ) = (
DateTime::Format::Strptime->new( pattern => "%a %b %d %T %z %Y", ),
DateTime::Format::Strptime->new( pattern => "%FY%T%Z", )
);
my $dt = $strp[0]->parse_datetime( $date );
print $strp[1]->format_datetime( $dt );
prints 2008-03-06T02:59:39UTC
Chris
If you're absolutely, positively sure that the date will ALWAYS be in that format, you can simply use regular expressions to reformat it. The only thing is that you have to have a way of converting the month to a number. That way, you don't have to download any extra modules to do the date conversion:
my $date = "Thu Mar 06 02:59:39 +0000 2008"; #Original String
#Create the Month Hash (you might want all twelve months).
my %monthHash (Jan => "01", Feb => 2, Mar => 3);
# Use RegEx Matching to parse your date.
# \S+ means one or more non-spaces
# \s+ means one or more spaces
# Parentheses save that part of the string in $1, $2, $3, etc.
$date =~ m/\S+\s+(\S+)\s+(\S+)\s+(\S+)\s+\S+\s(.*)/;
my $monthString = $1;
my $day = $2;
my $time = $3;
my $year = $4;
# Convert Month string to a number.
my $month = $monthHash{$monthString};
#Reformat the string
$fmtDate="$year-$month-$day" . "T" . "$time" . "Z";
Otherwise I was going to say you can also try DateTime::Format::Strptime, but Chris Charley beat me to it.
So, edit it with a regex and use HTTP::Date:
( my $new_date_string = $old_state_string ) =~ s/[+-]\d{4,}\s+//;
Here i based one problem.. i have two timestamps with same format like (Tue Dec 14 18:23:19 2010 & Tue Dec 14 17:23:19 2010). how can i get the difference of two timestamps in hours.
please help me
use Date::Parse;
my $t1 = 'Tue Dec 14 17:23:19 2010';
my $t2 = 'Tue Dec 14 18:23:19 2010';
my $s1 = str2time( $t1 );
my $s2 = str2time( $t2 );
print $s2 - $s1, " seconds\n";
I use the DateTime family of classes for pretty much all of my date/time handling.
#!/usr/bin/perl
use strict;
use warnings;
use DateTime::Format::Strptime;
my $dp = DateTime::Format::Strptime->new(
pattern => '%a %b %d %H:%M:%S %Y'
);
# Create two DateTime objects
my $t1 = $dp->parse_datetime('Tue Dec 14 17:23:19 2010');
my $t2 = $dp->parse_datetime('Tue Dec 14 18:23:19 2010');
# The difference is a DateTime::Duration object
my $diff = $t2 - $t1;
print $diff->hours;
You can take advantage of DateTime and its subtract_datetime() method, which returns a DateTime::Duration object.
use Date::Parse;
use DateTime;
my $t1 = 'Tue Dec 14 17:23:19 2010';
my $t2 = 'Tue Dec 14 18:23:19 2010';
my $t1DateTime = DateTime->from_epoch( epoch => str2time( $t1 ) );
my $t2DateTime = DateTime->from_epoch( epoch => str2time( $t2 ) );
my $diff = $t2DateTime->subtract_datetime( $t1DateTime );
print "Diff in hours: " . $diff->in_units('hours') . "\n";
print "Diff in months: " . $diff->in_units('months') . "\n";
Using this function:
perl -e 'use Time::Local; print timelocal("00","00","00","01","01","2000"),"\n";'
It will return an epochtime - but only in GMT - if i want the result in GMT+1 (which is the systems localtime(TZ)), what do i need to change?
Thanks in advance,
Anders
use DateTime;
my $dt = DateTime->now;
$dt->set_time_zone( 'Europe/Madrid' );
There is only one standard definition for epochtime, based on UTC, and not different epochtimes for different timezones.
If you want to find the offset between gmtime and localtime, use
use Time::Local;
#t = localtime(time);
$gmt_offset_in_seconds = timegm(#t) - timelocal(#t);
While Time::Local is a reasonable solution, you may be better off using the more modern DateTime object oriented module. Here's an example:
use strict;
use DateTime;
my $dt = DateTime->now;
print $dt->epoch, "\n";
For the timezones, you can use the DateTime::TimeZone module.
use strict;
use DateTime;
use DateTime::TimeZone;
my $dt = DateTime->now;
my $tz = DateTime::TimeZone->new(name => "local");
$dt->add(seconds => $tz->offset_for_datetime($dt));
print $dt->epoch, "\n";
CPAN Links:
DateTime
You just need to set the timezone. Try:
env TZ=UTC+1 perl -e 'use Time::Local; print timelocal("00","00","00","01","01","2000"),"\n";'
Time::Local::timelocal is the inverse of localtime. The result will be in your host's local time:
$ perl -MTime::Local -le \
'print scalar localtime timelocal "00","00","00","01","01","2000"'
Tue Feb 1 00:00:00 2000
Do you want the gmtime that corresponds to that localtime?
$ perl -MTime::Local' -le \
'print scalar gmtime timelocal "00","00","00","01","01","2000"'
Mon Jan 31 23:00:00 2000
Do you want it the other way around, the localtime that corresponds to that gmtime?
$ perl -MTime::Local -le \
'print scalar localtime timegm "00","00","00","01","01","2000"'
Tue Feb 1 01:00:00 2000
An other example based on DateTime::Format::Strptime
use strict;
use warnings;
use v5.10;
use DateTime::Format::Strptime;
my $s = "2016-12-22T06:16:29.798Z";
my $p = DateTime::Format::Strptime->new(
pattern => "%Y-%m-%dT%T.%NZ",
time_zone => "UTC"
);
my $dt = $p->parse_datetime($s);
$dt->set_time_zone("Europe/Berlin");
say join ' ', $dt->ymd, $dt->hms; # shows 2016-12-22 07:16:29
The Algorithm
If you want to change a time value from one timezone to another timezone, you must be able to indicate both timezones.
After all, if you set if you want to convert "12:30" to GMT or US/Eastern or Venezuelan time, which means adding/subtracting some amount of hours or hours and minutes, you need to know what timezone is the starting time zone, otherwise, the calculation won't know how much to add or subtract.
If you use DateTime->now;, the timezone is defaulted to the system-time, which may not be the timezone you want to convert from.
In the below code, I demonstrate how to initialize the datetime object to the right starting timezone (fromtimezone) and how to convert that time to the ending timezone (totimezone)...
Working Code
I could not find a Perl sandbox online with the DateTime CPAN module installed.
use strict;
use DateTime;
sub convertTimeZonesForTime {
my ($args) = #_;
my $time = $args->{time};
my $date = $args->{date};
my $totimezone = $args->{totimezone};
my $fromtimezone = $args->{fromtimezone};
my $format = $args->{format} || '%H:%M:%S';
my ($year, $month, $day) = map {int $_} split('-', $date);
my ($hour, $minute, $second) = map {int $_} split(':', $time);
$year ||= 1999 if !defined $year;
$month ||= 1 if !defined $month;
$day ||= 1 if !defined $day;
$hour ||= 12 if !defined $hour;
$minute ||= 30 if !defined $minute;
$second ||= 0 if !defined $second;
my $dt = DateTime->new(
year=>$year,
month=>$month,
day=>$day,
hour=>$hour,
minute=>$minute,
second=>$second,
time_zone => $fromtimezone,
);
my $formatter = new DateTime::Format::Strptime(pattern => $format);
$dt->set_formatter($formatter);
$dt->set_time_zone($totimezone);
return "$dt";
}
print(convertTimeZonesForTime({
'totimezone'=>'America/Denver',
'fromtimezone'=>'US/Eastern',
'time'=>'12:30:00',
}));
Output:
10:30:00