Date format again

[Troy May]

Hello, this guy finally emailed his script to me.  The problem he is having
is with "$year".  Here's the dating part of the code:

---
$date = `/bin/date`;
chop($date);
($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime;
$thisday = (Sun,Mon,Tue,Wed,Thu,Fri,Sat)[(localtime) [6]];
$s = (localtime)[0];
$m = (localtime)[1];
$m = ($m + 35) ;

$h = (localtime)[2];

if ($m >=60)
{
$m = ($m - 60);
$h = ($h + 1);
}
if ($h >=24)
{
$h = 1;
}
else
{
$h = ($h + 10) ;
}

$month = (Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec)[(localtime)[4]];
$year = (localtime)[5];
$year = ($year + 1900);

$day = (localtime)[3];


# this coding is creating problem

($sec,$min,$hour,$mday,$mon,$year,undef,undef,undef) = localtime();
$mon++; # adjust from 0-11 to 1-12
$year %= 100;
$theDate = sprintf("%02u%02u%02u", $mday, $mon, $year);

--

Now, when he prints "$theDate", everything is ok with it (format-030302).
But when he prints just "$year" it displays "2" instead of "2002"

Any ideas?



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Re: Date format again

[Marcelo E. Magallon]

[ Don't Cc: me, I read this mailing list, thank you ]

>> Troy May <[EMAIL PROTECTED]> writes:

 > ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime;
 > $thisday = (Sun,Mon,Tue,Wed,Thu,Fri,Sat)[(localtime) [6]];
 > $s = (localtime)[0];
 > $m = (localtime)[1];
 > $m = ($m + 35) ;

 Is adding 35 minutes to the current date what this is all about?

 How about:

($sec, $min, $hour, $mday, $mon, $year) = (localtime(time+35*60))[0..5];

 and save yourself from all that adding and carrying?

 By the way, in the original program, if you happen to run it at, say
 23:59:59 and the clock changes to 0:00:00 while the program is still
 running, between one of the calls to localtime, you'll get horribly
 incorrect results.

 > $year %= 100;

 > Now, when he prints "$theDate", everything is ok with it (format-030302).
 > But when he prints just "$year" it displays "2" instead of "2002"

 2002 % 100 == 2

-- 
Marcelo

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Re: Date format again

[Alfred Wheeler]

$year is a number. Numbers do not retain leading zeros. "$year %= 100;" does
not return the string "02"; it returns the number 2. Try something like
this -
$s_year = sprintf("%02u", $year);
print "$s_year";

The sprintf function basically converts the number into a string. 

"sprintf returns a string formatted by the usual printf conventions 
of the C library function sprintf." -- 
http://www.perldoc.com/perl5.6.1/pod/func/sprintf.html

It's just the same thing you did with the $theDate variable only applied to
a single item

of the list on the right side of the assignment.


- Original Message -
From: "Troy May" <[EMAIL PROTECTED]>
To: "Beginners CGI List" <[EMAIL PROTECTED]>
Sent: Sunday, March 03, 2002 12:52 AM
Subject: Date format again


> Hello, this guy finally emailed his script to me.  The problem he is
having
> is with "$year".  Here's the dating part of the code:
>
> ---
> $date = `/bin/date`;
> chop($date);
> ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime;
> $thisday = (Sun,Mon,Tue,Wed,Thu,Fri,Sat)[(localtime) [6]];
> $s = (localtime)[0];
> $m = (localtime)[1];
> $m = ($m + 35) ;
>
> $h = (localtime)[2];
>
> if ($m >=60)
> {
> $m = ($m - 60);
> $h = ($h + 1);
> }
> if ($h >=24)
> {
> $h = 1;
> }
> else
> {
> $h = ($h + 10) ;
> }
>
> $month =
(Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec)[(localtime)[4]];
> $year = (localtime)[5];
> $year = ($year + 1900);
>
> $day = (localtime)[3];
>
>
> # this coding is creating problem
>
> ($sec,$min,$hour,$mday,$mon,$year,undef,undef,undef) = localtime();
> $mon++; # adjust from 0-11 to 1-12
> $year %= 100;
> $theDate = sprintf("%02u%02u%02u", $mday, $mon, $year);
>
> --
>
> Now, when he prints "$theDate", everything is ok with it (format-030302).
> But when he prints just "$year" it displays "2" instead of "2002"
>
> Any ideas?
>
>
>
> --
> To unsubscribe, e-mail: [EMAIL PROTECTED]
> For additional commands, e-mail: [EMAIL PROTECTED]
>



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