如何将文本解析为特定的数组列表？

由佑佴玎发布于 2020-05-18 17:40:59 javascriptregex

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。

安蓝儿 2020-05-18 17:40:59 回复
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)
```
w无意 2020-05-18 17:40:59 回复
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.*?\.(?= |$)
```
Regex demo
相思浓 2020-05-18 17:40:59 回复
您可以使用
```
/\d+\..*?\.(?=\s*\d+\.|$)/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));
```