regex - perl 正则表达式从字符串中删除引号

Question

我想从字符串中删除尾随和前导单引号，但如果引号从字符串中间开始，则不会。

示例：我想'text is single quoted'从text is 'partly single quoted'. 示例 2：在abc 'foo bar' another' baz中，不应删除引号，因为缺少字符串开头的引号。

这是我的代码：

use strict;
use warnings;

my @names = ("'text is single quoted'", "text is 'partly single quoted'");
map {$_=~ s/^'|'$//g} @names;
print $names[0] . "\n" . $names[1] . "\n";

|正则表达式中的 or ( )^'|'$显然也从第二个字符串中删除了第二个引号，这是不希望的。我认为^''$这意味着它仅在第一个和最后一个字符是单引号时才匹配，但这不会删除任何单引号或字符串。

Answer 1

您可以使用捕获组。

s/^'(.*)'$/$1/

^断言我们在开头并$断言我们在一行的结尾。.*贪婪地匹配任何字符零次或多次。

代码：

use strict;
use warnings;

my @names = ("'text is single quoted'", "text is 'partly single quoted'");
s/^'(.*)'$/$1/ for @names;

print $names[0], "\n", $names[1], "\n";

输出：

text is single quoted
text is 'partly single quoted'

Answer 2

你试过这个正则表达式吗？

/^'([^']*)'$/$1/

理由是：“用单引号替换任何字符串开头和结尾，并且不包含单引号，字符串本身（不包括开头和结尾引号）”......

你可以在这里测试它：regex101.com

完整的代码应该是：

my @names = ("'text is single quoted'", "text is 'partly single quoted'");
s/^'([^']*)'$/$1/ for @names;

print join "\n", @names;

哪个输出：

$ perl test.pl
text is single quoted
text is 'partly single quoted'