Beefy Boxes and Bandwidth Generously Provided by pair Networks
Perl-Sensitive Sunglasses
 
PerlMonks  

Re^3: A question to iterator in Datetime::Event::Recurrence

by Anonymous Monk
on Apr 11, 2012 at 10:25 UTC ( #964496=note: print w/ replies, xml ) Need Help??


in reply to Re^2: A question to iterator in Datetime::Event::Recurrence
in thread A question to iterator in Datetime::Event::Recurrence

Well I do remember that from your previous question ( Calculation of the time overlapping ), but it doesn't change my impressions, ditch DateTime::Event::Recurrence, learn your way around DateTime::Set


Comment on Re^3: A question to iterator in Datetime::Event::Recurrence
Re^4: A question to iterator in Datetime::Event::Recurrence
by vagabonding electron (Hermit) on Apr 11, 2012 at 12:05 UTC
    Aye-aye, sir! :-)
    #!/usr/bin/perl use strict; use warnings; use DateTime; use DateTime::Format::Strptime; use Datetime::Set; # use Data::Dumper; my $strp = DateTime::Format::Strptime->new( pattern => '%Y-%m-%d %T', ); my $datf = qq{2012-01-01 04:00:00}; my $datt = qq{2012-01-02 23:00:00}; my $start = $strp->parse_datetime($datf); my $end = $strp->parse_datetime($datt); while( $start < $end ){ print "$start\n" if $start->hour > 6 and $start->hour < 23; $start->add( hours => 1 ); }
    I had bad luck with an attempt over $set = DateTime::Set->from_recurrence however.
    This one:
    $set = DateTime::Set->from_recurrence( after => $start, before => $end, recurrence => sub { return $_[0]->truncate( to => 'day' )->add( hours => 1 ) }, ); my $iter = $set->iterator; while ( my $dt = $iter->next ) { print $dt->ymd; };
    produced an infinite loop with the error message "iterator can't find a previous value". I intended to add later
    my $set2 = $set->grep( sub { return ( $_->hour > 7 or $_->hour < 23); } ); my $iter = $set2->iterator; while ( my $dt = $iter->next ) { print $dt->ymd; };

    Where is my mistake?
    Many thanks again - also for answering my previous question!
    VE

      The "recurrence" function should return the "next" recurrence. This is one way to implement it:

      #!/usr/bin/perl use strict; use warnings; use DateTime; use DateTime::Format::Strptime; use Datetime::Set; # use Data::Dumper; my $strp = DateTime::Format::Strptime->new( pattern => '%Y-%m-%d %T', ); my $datf = qq{2012-01-01 04:00:00}; my $datt = qq{2012-01-02 23:00:00}; my $start = $strp->parse_datetime($datf); my $end = $strp->parse_datetime($datt); my $iter; my $set = DateTime::Set->from_recurrence( after => $start, before => $end, recurrence => sub { return $_[0]->add( hours => 1 )->truncate( to => 'hour' ) }, ); print "first set:\n"; $iter = $set->iterator; while ( my $dt = $iter->next ) { print $dt->datetime, "\n"; }; my $set2 = $set->grep( sub { return ( $_->hour > 7 && $_->hour < 23); } ); print "second set:\n"; $iter = $set2->iterator; while ( my $dt = $iter->next ) { print $dt->datetime, "\n"; };

Log In?
Username:
Password:

What's my password?
Create A New User
Node Status?
node history
Node Type: note [id://964496]
help
Chatterbox?
and the web crawler heard nothing...

How do I use this? | Other CB clients
Other Users?
Others musing on the Monastery: (12)
As of 2015-07-01 20:30 GMT
Sections?
Information?
Find Nodes?
Leftovers?
    Voting Booth?

    The top three priorities of my open tasks are (in descending order of likelihood to be worked on) ...









    Results (19 votes), past polls