如何使用一行正则表达式获取匹配的内容 [英] how to use one line regular expression to get matched content

问题描述

我是 ruby​​ 的新手，我想知道我是否可以只使用一行代码来完成这项工作.

I'm a newbie to ruby, I want to know if I can use just one line to do the job.

以本网站的搜索"为例.当用户输入 [ruby] regex 时，我可以使用以下代码获取标签和关键字

Take the 'search' of this site for example. When user typed [ruby] regex, I can use following code to get the tag and keyword

'[ruby] regex' =~ /\[(.*?)\](.*)/
tag, keyword = $1, $2

我们可以一行写吗?

更新

非常感谢！我可以让它变得更难更有趣，输入可能包含多个标签，例如:

Thank you so much! May I make it harder and more interesting, that the input may contains more than one tags, like:

[ruby] [regex] [rails] one line

是否可以使用一行代码来获取标签数组和关键字?我试过了，但失败了.

Is it possible to use one line code to get the tags array and the keyword? I tried, but failed.

推荐答案

您需要 Regexp#match 方法.如果你写 /\[(.*?)\](.*)/.match('[ruby] regex')，这将返回一个 MatchData 对象.如果我们称该对象为matches，那么，除其他外:

You need the Regexp#match method. If you write /\[(.*?)\](.*)/.match('[ruby] regex'), this will return a MatchData object. If we call that object matches, then, among other things:

• matches[0] 返回整个匹配的字符串.
• matches[n] 返回第 n 个捕获组 ($n).
• matches.to_a 返回一个由 matches[0] 到 matches[N] 组成的数组.
• matches.captures 返回一个只包含捕获组的数组(matches[1] 到 matches[N]).立>
• matches.pre_match 返回匹配字符串之前的所有内容.
• matches.post_match 返回匹配字符串之后的所有内容.

• matches[0] returns the whole matched string.
• matches[n] returns the nth capturing group ($n).
• matches.to_a returns an array consisting of matches[0] through matches[N].
• matches.captures returns an array consisting of just the capturing group (matches[1] through matches[N]).
• matches.pre_match returns everything before the matched string.
• matches.post_match returns everything after the matched string.

方法比较多，对应其他特殊变量等；您可以查看 MatchData 的文档 了解更多信息.因此，在这种特定情况下，您只需要编写

There are more methods, which correspond to other special variables, etc.; you can check MatchData's docs for more. Thus, in this specific case, all you need to write is

tag, keyword = /\[(.*?)\](.*)/.match('[ruby] regex').captures

<小时>

编辑 1: 好的，对于更难的任务，您将需要 String#scan 方法，它 @Theo 使用；但是，我们将使用不同的正则表达式.以下代码应该可以工作:

---

Edit 1: Alright, for your harder task, you're going to instead want the String#scan method, which @Theo used; however, we're going to use a different regex. The following code should work:

# You could inline the regex, but comments would probably be nice.
tag_and_text = / \[([^\]]*)\] # Match a bracket-delimited tag,
                 \s*          # ignore spaces,
                 ([^\[]*) /x  # and match non-tag search text.
input        = '[ruby] [regex] [rails] one line [foo] [bar] baz'
tags, texts  = input.scan(tag_and_text).transpose

input.scan(tag_and_text) 将返回标签-搜索-文本对的列表:

The input.scan(tag_and_text) will return a list of tag–search-text pairs:

[ ["ruby", ""], ["regex", ""], ["rails", "one line "]
, ["foo", ""], ["bar", "baz"] ]

transpose 调用会翻转它，这样你就有了一个由标签列表和搜索文本列表组成的对:

The transpose call flips that, so that you have a pair consisting of a tag list and a search-text list:

[["ruby", "regex", "rails", "foo", "bar"], ["", "", "one line ", "", "baz"]]

然后你可以对结果做任何你想做的事情.我可能会建议，例如

You can then do whatever you want with the results. I might suggest, for instance

search_str = texts.join(' ').strip.gsub(/\s+/, ' ')

这会将搜索片段与单个空格连接起来，去除前导和尾随空格，并用单个空格替换多个空格.

This will concatenate the search snippets with single spaces, get rid of leading and trailing whitespace, and replace runs of multiple spaces with a single space.

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