函数来修剪字符串前导空格和后跟空格

是否有一个内置的功能来修剪前导和尾随空格，使 trim(" hello world ") eq "hello world"？

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$string =~ s/^\s+|\s+$//g;

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Here's one approach using a regular expression:

$string =~ s/^\s+|\s+$//g ;     # remove both leading and trailing whitespace

Perl 6 will include a trim function:

$string .= trim;

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There's no built-in trim function, but you can easily implement your own using a simple substitution:

sub trim {
(my $s = $_[0]) =~ s/^\s+|\s+$//g;
return $s;
}

or using non-destructive substitution in Perl 5.14 and later:

sub trim {
return $_[0] =~ s/^\s+|\s+$//rg;
}

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Apply: s/^\s*//; s/\s+$//; to it. Or use s/^\s+|\s+$//g if you want to be fancy.

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No, but you can use the s/// substitution operator and the \s whitespace assertion to get the same result.

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This is available in String::Util with the trim method:

^{Editor's note: String::Util is not a core module, but you can install it from CPAN with [sudo] cpan String::Util.}

use String::Util 'trim';
my $str = "  hello  ";
$str = trim($str);
print "string is now: '$str'\n";

prints:

string is now 'hello'

However it is easy enough to do yourself:

$str =~ s/^\s+//;
$str =~ s/\s+$//;

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I also use a positive lookahead to trim repeating spaces inside the text:

s/^\s+|\s(?=\s)|\s+$//g

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One option is Text::Trim:

use Text::Trim;
print trim("  example  ");

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For those that are using Text::CSV I found this thread and then noticed within the CSV module that you could strip it out via switch:

$csv = Text::CSV->new({allow_whitespace => 1});

The logic is backwards in that if you want to strip then you set to 1. Go figure. Hope this helps anyone.