我想在这里做的就是如果日期或月份是一个数字,添加零 它的前面。现在它打印出日期为201188,我正在寻找 20110808
#!/usr/bin/perl
use Date::Calc qw(Add_Delta_Days);
my (undef, undef, undef, $day, $month, $year) = localtime();
$year +=1900;
$month +=1;
($year, $month, $day ) = Add_Delta_Days($year, $month, $day, -3)
if ($month =~ /\d{1}/){
s/$month/0$month/
}
if ($day =~/\d{1}/){
s/$day/0$day/
}
print $year,$month,$day;
答案 0 :(得分:5)
如果您愿意使用Date::Calc
,为什么不使用DateTime
?
use DateTime;
my $date = DateTime->now;
$date->subtract(days => 3);
print $date->ymd;
事实上,您可以将其减少为:
print DateTime->now->subtract(days => 3)->ymd
答案 1 :(得分:4)
改为使用printf
:
printf "%d-%02d-%02d", $year, $month, $day;
提供如下输出:
C:\perl>perl -we "printf qq(%d-%02d-%02d), 2011,5,4"
2011-05-04
C:\perl>perl -we "printf qq(%d-%02d-%02d), 2011,5,12"
2011-05-12
C:\perl>perl -we "printf qq(%d-%02d-%02d), 2011,22,12"
2011-22-12
答案 2 :(得分:2)
if($ month< 10)
{
$月= “0 $月”;
}
if($ day< 10)
{
$天= “0 $日”;
}