根据正则表达式拆分字符串，但保留定界符 [英] Split string based on regex but keep delimiters
问题描述
我正在尝试使用各种字符作为分隔符来分割字符串，并将这些分隔符保留在自己的数组索引中。例如，说我想分割字符串：
I'm trying to split a string using a variety of characters as delimiters and also keep those delimiters in their own array index. For example say I want to split the string:
if（x> 1）return x * fact（x-1）;
if (x>1) return x * fact(x-1);
使用'（'，'>'，'）'，'*'，'-'，';'和'\s '作为分隔符。我希望输出为以下字符串数组：{ if，（， x，>， 1，）， return， x， *， fact ，（， x，-， 1，），;}
using '(', '>', ')', '*', '-', ';' and '\s' as delimiters. I want the output to be the following string array: {"if", "(", "x", ">", "1", ")", "return", "x", "*", "fact", "(", "x", "-", "1", ")", ";"}
我到目前为止使用的正则表达式是
split（（？=（\\w +（？= [\\s\\ + \\-\\ * /&（ < =）>（> =）（==）（！=）= ;, \\.\ \\（\\）\\ [\\] \\ {\\}]）））））））
The regex I'm using so far is
split("(?=(\\w+(?=[\\s\\+\\-\\*/<(<=)>(>=)(==)(!=)=;,\\.\"\\(\\)\\[\\]\\{\\}])))")
会在每个单词字符处拆分，无论是否跟随
which splits at each word character regardless of whether it is followed by one of the delimiters. For example
test + 1
test + 1
输出{ t， e， s， t +， 1}而不是{ test +， 1}
outputs {"t","e","s","t+","1"} instead of {"test+", "1"}
为什么即使每个字符后没有我的定界符，它也会在每个字符处分割呢？还有一个正则表达式，它甚至可以在Java中实现吗？
谢谢
Why does it split at each character even if that character is not followed by one of my delimiters? Also is a regex which does this even possible in Java? Thank you
推荐答案
好，您可以使用环视法在字符之间的点处分割，而无需使用分隔符：
Well, you can use lookaround to split at points between characters without consuming the delimiters:
(?<=[()>*-;\s])|(?=[()>*-;\s])
这将在每个定界符前后创建一个分割点字符。不过，您可能需要从结果数组中删除多余的空白元素。
This will create a split point before and after each delimiter character. You might need to remove superfluous whitespace elements from the resulting array, though.
快速PowerShell测试（ | 标记为拆分点）：
Quick PowerShell test (| marks the split points):
PS Home:\> 'if (x>1) return x * fact(x-1);' -split '(?<=[()>*-;\s])|(?=[()>*-;\s])' -join '|'
if| |(|x|>|1|)| |return| |x| |*| |fact|(|x|-|1|)|;|
这篇关于根据正则表达式拆分字符串，但保留定界符的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
