如何将文本解析为特定的数组列表？

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我有以下字符串：

const text = '01. is simply dummy text of the printing. 02. It is a long established fact that a reader will be. 03. and web page editors now use Lorem Ipsum.';

我需要将其解析为以下数组列表：

[
  '01. is simply dummy text of the printing.',
  '02. It is a long established fact that a reader will be.',
  '03. and web page editors now use Lorem Ipsum.'
]

我已经尝试过这种方法，但是没有得到想要的结果：

const list = text.match(/\d+.\s(.*)./g);

[
  '01. is simply dummy text of the printing. 02. It is a long established fact that a reader will be. 03. and web page editors now use Lorem Ipsum.'
]

请您帮我解决这个问题。

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安蓝儿
安蓝儿

如果我们可以假设句子中没有数字，那么您可以寻找不包含数字且以句号结尾的任何内容。

Also a . in regex means "match anything", if you want a proper full stop, you need to escape it \.

const text = '01. is simply dummy text of the printing. 02. It is a long established fact that a reader will be. 03. and web page editors now use Lorem Ipsum.';

const list = text.match(/(\d+\.[^\d]+\.)/g);

console.log(list)
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w无意
w无意

您可以匹配数字，后跟一个点，直到第一次出现一个点，然后是空格或字符串的末尾。

\d+\.\s.*?\.(?= |$)

Regex demo

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相思浓
相思浓

您可以使用

/\d+\..*?\.(?=\s*\d+\.|$)/g

细节

  • \d+ - 1+ digits
  • \. - a dot (must be escaped)
  • .*? - any 0 or more chars other than line break chars as few as possible
  • \. - a dot...
  • (?=\s*\d+\.|$) - that is followed with 0+ whitespaces, 1+ digits and a dot or end of string.

参见JS演示：

const text = '01. is simply dummy text of the printing. 02. It is a long established fact that a reader will be. 03. and web page editors now use Lorem Ipsum.';
console.log(text.match(/\d+\..*?\.(?=\s*\d+\.|$)/g));
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