perl 正则表达式从字符串中删除引号

Question

我想删除字符串中的尾随单引号和前导单引号，但如果引号从字符串中间开始，我就不会。

示例：我想删除 'text is single quoted' 中的引号，但不删除 text is 'partly single quoted' 中的引号。示例 2：在 abc 'foo bar' another' baz 中，不应删除引号，因为字符串开头的引号丢失。

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 (|) 显然也会从第二个字符串中删除第二个引号，这是不需要的。我认为 ^''$ 意味着它仅在第一个和最后一个字符是单引号时才匹配，但这不会从两个字符串中删除任何单引号。

Answer 1

你可以使用 capturing group.

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

^ 断言我们在开头，$ 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/^'(.*)'$// 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了吗？

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

理由是："substitute a any string starting and ending with a single quote, and that does not contain a single quote, with the string itself (starting and ending quotes excluded)"...

你可以在这里测试：regex101.com

完整代码应该是：

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

print join "\n", @names;

输出：

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

perl 正则表达式从字符串中删除引号

perl regex remove quotes from string

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