如何计算Perl中字符串开头的空格量？

Question

如何计算Perl中字符串开头的空格量？

我现在有:

  $temp = rtrim($line[0]);
  $count = ($temp =~ tr/^ //);

但这给了我所有空间的数量.

Answer 1

$str =~ /^(\s*)/;
my $count = length( $1 );

如果你只想要实际的空格(而不是空格),那就是:

$str =~ /^( *)/;

编辑:tr不起作用的原因是它不是正则表达式运算符.您正在做的$count = ( $temp =~ tr/^ // );是替换所有实例^和 with itself (see comment below by cjm), then counting up how many replacements you've done. tr并不^认为"嘿,这是字符串伪字符的开头",它将其视为"嘿,这是一个^".

Answer 2

您可以使用 获取匹配的偏移量@-。如果您搜索非空白字符，这将是字符串开头的空白字符数：

#!/usr/bin/perl

use strict;
use warnings;

for my $s ("foo bar", " foo bar", "  foo bar", "  ") {
        my $count = $s =~ /\S/ ? $-[0] : length $s;
        print "'$s' has $count whitespace characters at its start\n";
}

或者，更好的是，使用@+来查找空格的结尾：

#!/usr/bin/perl

use strict;
use warnings;

for my $s ("foo bar", " foo bar", "  foo bar", "  ") {
    $s =~ /^\s*/;
    print "$+[0] '$s'\n";
}