正则表达式 - 用相同数量的另一个字符替换一个字符的序列 [英] Regex- replace sequence of one character with same number of another character

问题描述

假设我有一个这样的字符串:

Let's say I have a string like this:

=====

我想用这个替换它:

-----

如果该字符的数量超过一定数量(我们会说 > 3)，我只想替换它.

I only want to replace it if it has more than a certain number of that character (we'll say > 3).

所以，这些应该是替代品:

So, these should be the replacements:

=== -> ===
==== -> ----
===== -> -----

应用是我想用二级标记替换markdown中的所有一级标题标记，而不改变嵌入的代码块.

The application is I want to replace all level 1 heading marks in markdown with a level 2 mark, without changing embedded code blocks.

我知道我可以做到:

/=/-/g，但这会匹配任何带有等号 (if (x == y)) 的内容，这是不可取的.

/=/-/g, but this matches anything with an equals sign (if (x == y)), which is undesirable.

或者这个:

/====+/----/g，但这并没有考虑原始匹配字符串的长度.

/===+/----/g, but this doesn't account for the length of the original matched string.

这可能吗?

推荐答案

Perl 是可能的:

my $string = "===== Hello World ====";
$string =~ s/(====+)/"-" x length($1)/eg;
# $string contains ----- Hello World ----

标志/e 使 Perl 在 s///的第二部分执行表达式.你可以用 oneliner 试试这个:

Flag /e makes Perl execute expression in second part of s///. You may try this with oneliner:

perl -e '$ARGV[0] =~ s/(====+)/"-" x length($1)/eg; print $ARGV[0]' "===== Hello World ===="

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