如何使用Perl删除变量中的内容

Question

#! /usr/bin/perl
use warnings;
print "Please enter the number";
chomp($inNum =<>);
if($inNum =~ /^[0]+/)
{

 print "The length is ",length($inNum),"\n";
 print  " Trailing Zero's  present","\n";
 $inNum =~ s/^[0]+/  /; 
 print  "The new output is" , $inNum ,"\n"; 
 print "The new length is ",length($inNum),"\n";

 }
 else
 {
  print "The input format vaild";
 }

输出

请输入号码：000010

长度是：6

尾随零的礼物

新的输出是10

新的长度是：4

问题是新长度值应该为（2），但显示为（4）如何解决此问题？

Answer 1

您要s/^0+// ，而不是s/^[0]+/ / 。

---

#!/usr/bin/env perl
use strict;
use warnings FATAL => 'all';

print 'Please enter the number: ';
chomp(my $inNum = <>);
if ($inNum =~ /^0+/) {  # has padding zeroes
    printf "The length is <%d>.\n", length($inNum);
    print "Padding zeroes present.\n";
    $inNum =~ s/^0+/  /; # replace any padding zeroes with two spaces
    printf "The new output is <%s>.\n", $inNum;
    printf "The new length is <%d>.\n", length($inNum);
} else {
    print "The input format was invalid.\n";
}

---

Please enter the number: 000010
The length is <6>.
Padding zeroes present.
The new output is <  10>.
The new length is <4>.

Answer 2

看起来您正在用2个空格字符替换四个0。 尝试这个。

$inNum =~ s/^[0]+//; 

Answer 3

是的，您正在用空格替换，但是如果您不想更改reg表达式，也可以添加一个

 sub trim($) { my $string = shift; $string =~ s/^\\s+//; $string =~ s/\\s+$//; return $string; 

}

和使用

print "The new length is ",length(trim($inNum)),"\n";

Answer 4

如果您打算去除前导零，则可以考虑使用sprintf而不是正则表达式。

use feature qw(say);
use strict;
use warnings;

print "Please enter the number: ";
my $num = sprintf "%d", scalar <>;
say "$num";

请注意，如果您不输入数字，则会收到警告。

Answer 5

#!/usr/bin/perl

use strict;
use warnings;

print "Please enter the number";
my $num = 0 + <>;
print "The number is '$num'\n";

__END__