I have number and need to add the suffix: 'st', 'nd', 'rd', 'th'. So for example: if the number is 42 the suffix is 'nd' , 521 is 'st' and 113 is 'th' and so on. I need to do this in perl. Any pointers.
Try this:
my $ordinal;
if ($foo =~ /(?<!1)1$/) {
$ordinal = 'st';
} elsif ($foo =~ /(?<!1)2$/) {
$ordinal = 'nd';
} elsif ($foo =~ /(?<!1)3$/) {
$ordinal = 'rd';
} else {
$ordinal = 'th';
}
Use Lingua::EN::Numbers::Ordinate. From the synopsis:
use Lingua::EN::Numbers::Ordinate;
print ordinate(4), "\n";
# prints 4th
print ordinate(-342), "\n";
# prints -342nd
# Example of actual use:
...
for(my $i = 0; $i < @records; $i++) {
unless(is_valid($record[$i]) {
warn "The ", ordinate($i), " record is invalid!\n";
next;
}
...
}
Try this brief subroutine
use strict;
use warnings;
sub ordinal {
return $_.(qw/th st nd rd/)[/(?<!1)([123])$/ ? $1 : 0] for int shift;
}
for (42, 521, 113) {
print ordinal($_), "\n";
}
output
42nd
521st
113th
Here's a solution which I originally wrote for a code golf challenge, slightly rewritten to conform to usual best practices for non-golf code:
$number =~ s/(1?\d)$/$1 . ((qw'th st nd rd')[$1] || 'th')/e;
The way it works is that the regexp (1?\d)$ matches the last digit of the number, plus the preceding digit if it is 1. The substitution then uses the matched digit(s) as an index to the list (qw'th st nd rd'), mapping 0 to th, 1 to st, 2 to nd, 3 to rd and any other value to undef. Finally, the || operator replaces undef with th.
If you don't like s///e, essentially the same solution could be written e.g. like this:
for ($number) {
/(1?\d)$/ or next;
$_ .= (qw'th st nd rd')[$1] || 'th';
}
or as a function:
sub ordinal ($) {
$_[0] =~ /(1?\d)$/ or return;
return $_[0] . ((qw'th st nd rd')[$1] || 'th');
}
Another solution (though I like the preexisting answers that are independent of using modules better):
use Date::Calc 'English_Ordinal';
print English_Ordinal $ARGV[0];
来源:https://stackoverflow.com/questions/11369907/how-do-i-retrieve-an-integers-ordinal-suffix-in-perl-like-st-nd-rd-th