字符串中的替换
我有以下字符串 ./test 我想用 test 替换它 所以，我用 perl 写了以下内容： 我的 $t =~ s/^.//; 但是，这会将 ./test 替换为 /test 任何人都可以建议我如何修复它，以便我也摆脱 / 。谢谢！
I have the following string ./test
and I want to replace it with test
so, I wrote the following in perl:my $t =~ s/^.//;
however, that replaces ./test with /test
can anyone please suggest how I fix it so I get rid of the / too. thanks!
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您需要转义点和斜线。
替换是
s/match/replace/。如果删除，则为s/match//。您想要匹配“以点和斜杠开头”，那就是^\.\/。You need to escape the dot and the slash.
The substitution is
s/match/replace/. If you erase, it'ss/match//. You want to match "starts with a dot and a slash", and that's^\.\/.该点不符合您的预期 - 它不是匹配点字符，而是匹配任何字符，因为它的特殊处理。要匹配点和正斜杠，您可以按如下方式重写表达式：
请注意，您可以随意使用不同的字符作为分隔符，以免与正则表达式中的任何此类字符混淆。
The dot doesn't do what you expect - rather than matching a dot character, it matches any character because of its special treatment. To match a dot and a forward slash, you can rewrite your expression as follows:
Note that you are free to use a different character as a delimiter, in order not to confuse it with any such characters inside the regular expression.
如果您想摆脱
./那么您需要在正则表达式中包含这两个字符。.和/在此表达式中都有特殊含义（.是一个正则表达式元字符，表示“任何字符”，而/> 是s///运算符的分隔符），因此我们需要通过在它们前面放置\来转义它们。解决
/问题的另一种（在我看来更好）方法是更改用作s///分隔符的字符。这些都记录在 perldoc perlop 中。
If you want to get rid of
./then you need to include both of those characters in the regex.Both
.and/have special meanings in this expression (.is a regex metacharacter meaning "any character" and/is the delimiter for thes///operator) so we need to escape them both by putting a\in front of them.An alternative (and, in my opinion, better) approach to the
/issue is to change the character that you are using as thes///delimiter.This is all documented in perldoc perlop.
您必须在点和正斜杠之前使用反斜杠：s/\.\//;
反斜杠用于编写在正则表达式中具有不同含义的符号。
You have to use a backward slash before the dot and the forward slash: s/\.\//;
The backslash is used to write symbols that otherwise would have a different meaning in the regular expression.
您需要编写
my $t =~ s/^\.\///;（请注意，需要对句点进行转义，以便匹配文字句点而不是任何字符）。如果斜杠太多，您还可以更改分隔符，改为写入，例如my $t =~ s:^\./::;。You need to write
my $t =~ s/^\.\///;(Note that the period needs to be escaped in order to match a literal period rather than any character). If that's too many slashes, you can also change the delimiter, writing instead, e.g.,my $t =~ s:^\./::;.如果你想让它匹配一个点，你需要转义点。否则它匹配任何字符。您可以选择备用分隔符——最好是在处理正斜杠时，以免在您也需要转义这些分隔符时看起来像倾斜的牙签。
You need to escape the dot if you want it to match a dot. Otherwise it matches any character. You can choose alternate delimiters --- best when dealing with forward slashes lest you get the leaning-toothpick look when you otherwise need to escape those too.
请注意，问题中的点与任何字符匹配，而不是文字“.”。
Note that the dot in your question is matching any character, not a literal '.'.
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