scala 如何通过正则表达式拆分此字符串？
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How to split this string by regex?
提问by Freewind
I have some string, they looks like:
我有一些字符串，它们看起来像：
div#title.title.top
#main.main
a.bold#empty.red
They are similar to haml, and I want to split them by regex, but I don't know how to define it.
它们类似于haml，我想通过正则表达式拆分它们，但我不知道如何定义它。
val r = """???""".r // HELP
val items = "a.bold#empty.red".split(r)
items // -> "a", ".bold", "#empty", ".red"
How to do this?
这个怎么做？
UPDATE
更新
Sorry, everyone, but I need to make this question harder. I'm very interested in
对不起，大家，但我需要让这个问题更难。我很感兴趣
val r = """(?<=\w)\b"""
But it failed to parse the more complex ones:
但它未能解析更复杂的：
div#question-title.title-1.h-222_333
I hope it will be parsed to:
我希望它会被解析为：
div
#question-title
.title-1
.h-222_333
I wanna know how to improve that regex?
我想知道如何改进该正则表达式？
采纳答案by Josh M.
I'm not completely sure what you need here but this should help:
我不完全确定您在这里需要什么，但这应该会有所帮助：
(?:\.|#)?\w+
It means a "term" is defined as an optional dot or hash followed by some word characters.
这意味着“术语”被定义为可选的点或散列，后跟一些单词字符。
You will end up with:
你最终会得到：
div
#title
.title
.top
#main
.main
a
.bold
#empty
.red
回答by Daniel C. Sobral
val r = """(?<=\w)\b(?!-)"""
Note that split takes a Stringrepresenting a regular expression, not a Regex, so you must not convert rfrom Stringto Regex.
请注意， split 采用 aString表示正则表达式，而不是 a Regex，因此您不能r从转换String为Regex。
Brief explanation on the regex:
关于正则表达式的简要说明：
(?<=...)is a look-behind. It states that this match must be preceded by the pattern..., or, in your case\w, meaning you want the pattern to follow a digit, letter, or underline.\bmeans word boundary. It is a zero-length match that happen between a word character (digits, letters and underscore) and a non-word character, or vice versa. Because it is zero-length,splitwon't remove any character when splitting.(?!...)is a negative-lookahead. Here I use to say that I'm not interested in word boundaries from a letter to a dash.
(?<=...)是后视。它指出此匹配必须以模式开头，或者...在您的情况下\w，这意味着您希望模式跟在数字、字母或下划线之后。\b表示词边界。它是在单词字符（数字、字母和下划线）和非单词字符之间发生的零长度匹配，反之亦然。因为它是零长度，split所以拆分时不会删除任何字符。(?!...)是负前瞻。在这里我经常说我对从字母到破折号的单词边界不感兴趣。
回答by Ken Bloom
Starting with Josh M's answer, he has a good regular expression, but since splittakes a regular expression matching the "delimiter", you need to use findAllInas follows:
从 Josh M 的回答开始，他有一个很好的正则表达式，但由于split采用了匹配“分隔符”的正则表达式，因此您需要使用findAllIn如下：
val r = """(?:\.|#)?\w+""".r
val items = r findAllIn "a.bold#empty.red"
//maybe you want a toList on the end also
Then you get the results
然后你得到结果
div#title.title.top -> List(div, #title, .title, .top)
#main.main -> List(#main, .main)
a.bold#empty.red -> List(a, .bold, #empty, .red)
