如何在空白之间而不是引号之间进行拆分? [英] How to split on white spaces not between quotes?

问题描述

我正在尝试仅在空白( \ s )上分割字符串，但是不在带引号"部分之间.

I am trying to split a string on white spaces only (\s), but that are not between a "quoted" section.

我通过以下方式匹配这些引用部分之间的所有文本:

I am matching all text in between these quoted sections in the following manner:

(['"`]).*?\1

Regex101

但是，当我尝试将其添加为负前瞻性时，仅在这些引号之外的空白处进行拆分时，我无法使它起作用:

However, when I try to add this as a negative lookahead, to only split on white spaces outside of those quotes, I can't get it to work:

\s(?!(['"`]).*?\1)

Regex101

如何只分割不在引号"中的空白?

How can I only split on the white spaces that are not in "quotes"?

推荐答案

\s(?=(?:[^'"`]*(['"`])[^'"`]*\1)*[^'"`]*$)

您可以将此正则表达式与 lookahead 一起使用.请参见演示.

You can use this regex with lookahead to split upon.See demo.

https://regex101.com/r/5I209k/4

或混合的刻度线类型.

https://regex101.com/r/5I209k/7

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