I want to create a variable that will store hardcoded hours and minutes value. Rest stuff like min, day, date, year should be current date and time. How can I do this? I tried timelocal, localtime and many others, but they didn't work.
If you are interested in a complete date time object (like it seems according your description), use DateTime. Example:
use DateTime;
my $dt = DateTime->now; # initialize with current date time
$dt->set_hour(20); #set your own hour
$dt->set_minute(20); #set your own minute
Unless I'm missing something, you indeed use localtime, as in perldoc:
http://perldoc.perl.org/functions/localtime.html
($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(time);
Then, if you want to stick to using the broken-down components, you could overwrite whichever variables you want to "fix":
$hour = $my_hardcoded_hour;
$min = $my_hardcoded_minute;
If instead you need to translate this back to "seconds since the epoch" (Jan 1, 1970, 12:00AM GMT) format, you need the Time::Local package:
http://perldoc.perl.org/Time/Local.html
use Time::Local;
$epoch_current_time_with_fixed_hour_min =
timelocal($sec, $my_hardcoded_minutes, $my_hardcoded_hour, $mday, $mon, $year);
Related
I want to subtract two timestamps in Perl. I converted them to unix-time via the function below and convert the unix timestamp back to how it was. In the example below the result is 01:20:00 instead of 00:20:00
(I think it has sth to do with the start of the unix timestamp 1.1.1970 01:00:00 but not sure how to resolve it)
Any idea? Many thanks for your help in advance.
use POSIX qw( strftime );
use Time::Local qw( timelocal );
sub to_epoch {
$_ = shift;
my #a = split /\W+/, $_;
my $b = timelocal($a[5],$a[4],$a[3],$a[2],$a[1],$a[0]);
return $b;
}
my $h_end = "2018.11.12 00:50:00";
my $h_start = "2018.11.12 00:30:00";
my $duration = to_epoch($h_end) - to_epoch($h_start);
my $convert_back = POSIX::strftime("%H:%M:%S", localtime($duration));
print $convert_back , "\n";
Ouptut: 01:20:00
It works for me. But I think that's because I'm in GMT and you're in CET (GMT+1).
The flaw is in your final step. You are confusing two concepts - a point in time and a duration.
You correctly convert your two points in time to Unix epoch numbers and then you subtract those numbers to get the number of seconds between them. That number is a duration. And you want to convert that duration into a human-readable format. Using localtime() and POSIX::strtime() is not the way to do that. POSIX::strftime() and localtime() deal with points in time, not durations.
The number you get is 1,200. By passing that to localtime() you are saying "what is the epoch number 1,200 when converted to a date and time in my local timezone?" 1,200 is 20 past midnight on Jan 1st 1970 GMT. But in your local, Frankfurt, timezone, it's 20 past 1am. Which is why you're getting 1:20 and I'm getting 0:20.
There are a couple of ways to fix this. You can do the conversion manually.
my $duration = 1_200;
my $mins = int($duration/60);
my $secs = $duration % 60;
Or you can use a proper date/time handling module like DateTime (along with its associated module DateTime::Duration).
It might work if you use timegm() and gmtime() in place of timelocal() and localtime() - but I really don't recommend this approach as it perpetuates the confusion between points in time and durations.
Update: A version using DateTime.
#/usr/bin/perl
use strict;
use warnings;
use DateTime;
use DateTime::Format::Strptime;
my $h_end = '2018.11.12 00:50:00';
my $h_start = '2018.11.12 00:30:00';
my $date_p = DateTime::Format::Strptime->new(
pattern => '%Y.%m.%d %H:%M:%S'
);
my $duration = $date_p->parse_datetime($h_end)
- $date_p->parse_datetime($h_start);
printf '%02d:%02d:%02d', $duration->in_units('hours', 'minutes', 'seconds');
1200, the value of $duration, signifies the following when treated as a epoch timestamp
1970-01-01T01:20:00+01:00
^^^^^^^^
The solution is to replace
strftime("%H:%M:%S", localtime($duration));
with
strftime("%H:%M:%S", gmtime($duration));
This gives
1970-01-01T00:20:00Z
^^^^^^^^
Of course, this is still a hack. You're not suppose to be passing a duration to gmtime. Use an appropriate module instead.
use DateTime::Format::Strptime qw( );
my $format = DateTime::Format::Strptime->new(
pattern => '%Y.%m.%d %H:%M:%S',
on_error => 'croak',
);
my $h_end = $format->parse_datetime('2018.11.12 00:50:00');
my $h_start = $format->parse_datetime('2018.11.12 00:30:00');
my $dur = $h_end - $h_start;
printf "%02d:%02d:%02d\n", $dur->in_units(qw( hours minutes seconds ));
By the way,
timelocal($a[5],$a[4],$a[3],$a[2],$a[1],$a[0])
should be
timelocal($a[5],$a[4],$a[3],$a[2],$a[1]-1,$a[0])
I am trying to compare a file creation time which is in the format: 08-07-2016 08:16:26 GMT with the current time using time() in perl.
Since time() returns epoch time, I am not sure how to find the time difference between these two different time formats.
I tried something like below and for obvious reasons, I get an error saying: "Argument 08-07-2016 08:16:26 GMT" isn't numeric in subtraction".
my $current_time = time();
my $time_diff = $creation_time - $current_time;
if ($time_diff > 10) { #compare if the difference is greater than 10hours
# do something...
}
Some of the questions I have:
Since I want to compare only the hour difference, how can I extract just the hours from both these time formats?
I am unsure if the comparison of $time_diff > 10 is right. How to represent 10hours? 10*60?
OR is there a way to at least convert any given time format into epoch using DateTime or Time::Local?
How can I pass a a date parameter to a DateTime constructor?
my $dt1 = DateTime-> new (
year =>'1998',
month =>'4',
day =>'4',
hour =>'21',
time_zone =>'local'
);
Instead can we do something like
my $date = '08-07-2016 08:16:26 GMT';
my $dt1 = DateTime->new($date); # how can i pass a parameter to the constructor
print Dumper($dt1->epoch);
Thanks in advance for any help.
Time::Piece has been a standard part of Perl since 2007.
#!/usr/bin/perl
use strict;
use warnings;
use 5.010;
use Time::Piece;
use Time::Seconds;
my $creation_string = '08-07-2016 08:16:26 GMT';
my $creation_time = Time::Piece->strptime($creation_string, '%d-%m-%Y %H:%M:%S %Z');
my $current_time = gmtime;
my $diff = $current_time - $creation_time;
say $diff; # Difference in seconds
say $diff->pretty;
I have an epoch value of 1372252864.901871 which needs to be converted to localtime in perl. I know that localtime($val) where $val is set to 1372252864.901871 converts the value to local time. But I also need to add values of 100 msecs or say 86.7 msecs to this and determine the computed local time. How do I do this? Does the function localtime() take care of fractional epoch value that I have given?
If you just want to add to an epoch time and get another epoch time, just do:
1372252864.901871 + .0867
If you are doing something else, no, localtime does not preserve fractional seconds. You probably want to use DateTime to deal with fractional seconds:
use DateTime;
my $time = DateTime->from_epoch(epoch=>1372252864.901871, time_zone=>'local');
$time->add(nanoseconds=>86.7e6);
print $time->epoch + $time->nanosecond / 1e9;
(time_zone isn't needed for this example, but sounds like what you might want for other things you might be doing with this time)
You might want Time::HiRes:
The Time::HiRes module implements a Perl interface to the usleep,
nanosleep, ualarm, gettimeofday, and setitimer/getitimer system calls,
in other words, high resolution time and timers
The localtime function breaks down a raw time like 1372252864 into year, month, day, etc. Any fractional part of the raw time is quietly ignored.
I'm not aware of any function, either built-in or in a module, that supports broken-down times with a resolution finer than 1 second.
But since you already have the fractional seconds, I'd say you can already do what you need to do. Just use localtime (or gmtime) to break down the integer part of your raw timestamp, and then take the fractional part:
$timestamp = 1372252864.901871;
#parts = localtime($timestamp);
$fraction = $timestamp - int($timestamp);
$milliseconds = int($fraction * 1000.0);
It should be easy enough to write your own high-resolution replacement for localtime that does this for you.
If you need to add 86.7 milliseconds to a timestamp, you'll need to use the raw form anyway:
$timestamp = 1372252864.901871;
$timestamp += 86.7/1000.0;
The simplest way that occurs to me is to use Time::HiRes::Value to handle arithmetic on times to 1μs accuracy, and Time::Piece to format the result as a date/time string.
Time::HiRes::Value holds the time as a two-element array; the first field has the whole seconds while the second has the number of microseconds. That's the same format that Time::HiRes functions work with, but there's nothing that I know of that will natively convert to a date/time string so I've used Time::Piece to convert the whole seconds and just appended the fractional seconds.
It looks like this
use strict;
use warnings;
use Time::HiRes::Value;
use Time::Piece;
my $dt = Time::HiRes::Value->new(1372252864.901871);
my $delta = Time::HiRes::Value->new(86.7E-3); # 86.7ms
$dt += $delta;
print dt_to_string($dt), "\n";
sub dt_to_string {
my ($dt) = #_;
Time::Piece->new($dt->[0])->datetime =~ s{(\d+)\z}{$1 + $dt->[1] / 1E6}er;
}
output
2013-06-26T14:21:4.98857
I have following date in string format and I need to convert it to milliseconds using a Perl script. I have tried to convert it using DateTime::Format::Strptime and it returns 0 after convert to millisecond.
Date String : 01-13-15:14.16
#!/usr/bin/perl
use DateTime::Format::Strptime;
my $strp = DateTime::Format::Strptime->new(
pattern => '%m-%d-%y:%H.%M',
on_error => 'croak',
);
my $dt = $strp->parse_datetime("01-13-15:14.16");
print $dt->millisecond."\n";
How can I convert datetime to milliseconds
I think it's likely that what you want is not milliseconds but epoch, which is the number of seconds since January 1 1970. Your date-time format has neither seconds nor milliseconds, so both of these fields will be reported as zero, and if you really want milliseconds then you can simply multiply the epoch by 1000.
It looks like this
use strict;
use warnings;
use 5.010;
use DateTime::Format::Strptime;
my $strp = DateTime::Format::Strptime->new(
pattern => '%m-%d-%y:%H.%M',
on_error => 'croak',
);
my $dt = $strp->parse_datetime('01-13-15:14.16');
say $dt->epoch;
output
1421158560
However, the DateTime module is enormous and slow, and there is a better solution in the core module Time::Piece, which would look like this
use strict;
use warnings;
use 5.010;
use Time::Piece;
my $dt = Time::Piece->strptime('01-13-15:14.16', '%m-%d-%y:%H.%M');
say $dt->epoch;
output
1421158560
It is impossible to convert a date to milliseconds. A date is a point in time. Milliseconds measure a duration.
The only way your question makes sense is if we assume you mean "how many milliseconds are there between my given date/time and some other fixed date/time". Because of the way that Unix measures time, let's assume that the fixed date/time you're measuring from is the Unix Epoch (1970-01-01 00:00:00).
You get the number of seconds since the epoch from a DateTime object with the epoch() method.
say $dt->epoch;
To get the number of milliseconds, just multiply that by a thousand.
say 1000 * $dt->epoch;
Your current timestamp doesn't include milliseconds. If you change to parsing strings that include them, then you can add that value on too.
say $dt->milliseconds + (1000 * $dt->epoch);
Adding my comment as an answer:
If you want to convert the datetime value into seconds since the epoch, try one of these:
print sprintf("%f", $dt->epoch + ($dt->millisecond/1000));
print sprintf("%s.%s", $dt->epoch, $dt->millisecond);
(My original comment did not account for millisecond being an integer)
Similar for the values in milliseconds, rather than seconds.millis:
print sprintf("%d", ($dt->epoch * 1000) + $dt->millisecond);
print sprintf("%d%04d", $dt->epoch, $dt->millisecond);
Remember, that if you don't supply DateTime with a value containing milliseconds (or higher resolution), the value will simply be zero.
I am trying to convert a date/time GMT 0 to GMT -6 in Perl.
For example, a DHCP Server lease time is in the following format:
2010/02/18 23:48:37
I am trying to convert that time to the Localtime zone (GMT -6) but need it to honor Daylight savings time.
The script below may be overkill, but I am not sure how to proceed from here. (Any suggestions would be awsome).
my $TIMESTART;
$TIMESTART = "2010/02/18 23:48:37";
$TIMESTART =~ s/\//-/g;
use DateTime;
use DateTime::TimeZone;
use DateTime::Format::MySQL;
my $dt = DateTime::Format::MySQL->parse_datetime($TIMESTART);
my $tz = DateTime::TimeZone->new( name => 'America/Chicago' );
print $tz->offset_for_datetime($dt) . "\n";
It will output the following lines:
2010-02-18T23:48:37
-21600
I need to be able to add -21600 to the date to get the local time zone of GMT -6 but I am not sure how to approch this.
Call set_time_zone method 2 times:
my $dt = DateTime::Format::MySQL->parse_datetime($TIMESTART);
$dt->set_time_zone('UTC'); ## set timezone of parsed date time
$dt->set_time_zone('America/Chicago'); ## change timezone in safe way
print DateTime::Format::MySQL->format_datetime($dt),"\n"; ## check the result
How it works:
when you create DateTime object without time zone specified, "floating" time zone is set
first call of set_time_zone change time zone to UTC without conversion
second call of set_time_zone change UTC to America/Chicago
This will convert UTC time into ETC time.
You can also use date/time in any format using +FORMAT parameter of date.
date --date='TZ="ETC" 18:30'
Time::Piece is a very lightweight piece of code of good quality. Or you can just use built-ins and strftime and POSIX::strptime
Don't Use Timezones With `DateTime::Format::MySQL->parse_datetime`
There is a very good reason you shouldn't be using this when doing anything relating to timezones. Why? Because it has absolutely no support for timezones whatsoever. Please see the documentation:
All of the parsing methods set the returned DateTime object's time zone to the floating time zone, because MySQL does not provide time zone information. (Source: MetaCPAN: DateTime::Format::MySQL.)
If you DO use this, it will assume your system time zone default. This might not be the timezone you want to convert from! (Although this will work for many people, it only works as long as your servers don't change their default timezone — after that, everything breaks.)
Use `DateTime->new()`, Because It DOES Support TimeZones
Use the constructor that DateTime provides directly (Source: MetaCPAN: DateTime.):
my $dt = DateTime->new(
year => 1966,
month => 10,
day => 25,
hour => 7,
minute => 15,
second => 47,
nanosecond => 500000000,
time_zone => 'America/Chicago',
);
Convert TimeZones with DateTime
Your starting time zone is whatever you feed into the constructor (i.e., America/Chicago in the above example). To convert a time to an ending time zone, use set_time_zone().
Working Code
In the code below, a time is converted from one timezone to another, and it does this perfectly every time, even you are converting New York time to San Francisco time with a server whose Linux OS is in Singapore time.
use strict;
use DateTime;
sub convertTimeZonesForTime {
my ($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 "$dt";
}
print(convertTimeZonesForTime({
'totimezone'=>'America/Denver',
'fromtimezone'=>'US/Eastern',
'time'=>'12:30:00',
}));
Output:
10:30:00
#!/usr/bin/perl -w
($sec,$min,$hour,$mday,$mon,$year) = gmtime(time+21600);
$year = $year + 1900;
printf("%02d:%02d:%02d %02d.%02d.%04d\n", $hour,$min,$sec,$mday,$mon,$year);