在与环视匹配的行首插入空格答案

【问题标题】：Insert space at the beginning of line matched with lookarounds在与环视匹配的行首插入空格
【发布时间】：2021-10-15 09:31:35
【问题描述】：

我正在尝试在与环视匹配的行的开头添加空格。这是一个示例文本：

This is my 
first sentence
with 2 lines
that have to be indented. 

This is my 
second sentence
with 2 lines
that have to be indented.

这是我想要得到的输出：

This is my 
    first sentence
    with 2 lines
that have to be indented. 

This is my 
    second sentence
    with 2 lines
that have to be indented.

我试过用这个环顾四周： (?<=This is my)(.*?)(?=\nthat)

但是，我无法将行首与^ 匹配。也就是说，(?<=This is my)(^)(?=\nthat) 不起作用。

regex101

TIA！

【问题讨论】：

您使用哪种编程语言？

标签： regex regex-lookarounds regexp-replace

【解决方案1】：

(.+?\n)(.+?\n)(.+?\n)(.+?\n)

https://regex101.com/r/Ryb3pR/1

似乎使用一些代码而不是使用正则表达式更容易理解。

【讨论】：

【解决方案2】：

如果后向断言（如 .NET、较新的 Javascript 引擎或 Python PyPi 正则表达式模块）支持量词，并且您想匹配任意数量的行，则可以使用：

(?<=^This is my.*\n(?!that\b)(?:.*\n(?!that\b))*).*

(?<= 肯定的后视断言左边是
- ^ 字符串开始
- This is my.*\n(?!that\b) 匹配第一行，下一行断言不是that
- (?:非捕获组
  - .*\n(?!that\b) 匹配包括换行在内的整行，并在下一个断言不是that
- )*关闭非捕获组
) 近距离观察
.*整行匹配

Regex demo

const regex = /(?<=^This is my.*\n(?!that\b)(?:.*\n(?!that\b))*).*/gm;
const str = `This is my 
first sentence
with 2 lines
that have to be indented. 

This is my 
second sentence
testing
with 3 lines
that have to be indented. 
test

This is my 
that have to be indented. `;
console.log(str.replace(regex, `    $&`));

没有列出语言，但另一种选择是使用 2 个捕获组，并将缩进添加到第二组中的所有行。

该模式捕获第 1 组中的起始行，然后捕获第 2 组中不以 that 开头的所有行

^(This is my.*)((?:\n(?!that\b).*)*)(?=\nthat\b)

Regex demo

例如使用 Javascript：

const regex = /^(This is my.*)((?:\n(?!that\b).*)*)(?=\nthat\b)/gm;
const str = `This is my 
first sentence
with 2 lines
that have to be indented. 

This is my 
second sentence
testing
with 3 lines
that have to be indented. 
test

This is my 
that have to be indented. `;
console.log(str
  .replace(regex, (m, g1, g2) => g1 + g2
    .split('\n').map(s => '    ' + s)
    .join('\n'))
);

【讨论】：