如何使用单个正则表达式匹配以“/ 1”或“/ 2”结尾的行？(How can I match lines ending with “/1” or “/2” using a single regular

如何使用单个正则表达式匹配以“/ 1”或“/ 2”结尾的行？(How can I match lines ending with “/1” or “/2” using a single regular expression?)

我正在解析文本行，并尝试使用单个正则表达式从行中删除尾部的“/ 1”或“/ 2”。

现在我有这个，它正常工作：

$id =~ s/\/1$//; $id =~ s/\/2$//;

如何将这些组合成一行？ 我尝试使用大括号，但我没有让它工作。

I'm parsing lines of text and trying to remove the trailing "/1" or "/2" from the lines with a single regular expression.

Right now I have this and it works correctly:

$id =~ s/\/1$//; $id =~ s/\/2$//;

How can I combine these into a single line? I tried using braces but I haven't gotten it to work.

最满意答案

$id =~ s/\/[12]$//;

使用匹配1或2的字符类[12] 。

$id =~ s/\/[12]$//;

using the character class [12] which matches either a 1 or a 2.