如何包装所有片段（不在里面）or) with?

Posted 2021-03-26

我有一个代码：

$html = 
'<h1>Headline</h1>
That should be paragraph 1.

That should be paragraph 2.

<div>Something.</div>
That should be paragraph 3.
';

我想将其改为：

$html = 
'<h1>Headline</h1>
<p>That should be paragraph 1.</p>

<p>That should be paragraph 2.</p>

<div>Something.</div>
<p>That should be paragraph 3.</p>
';

但是如何在php中实现它？

我正考虑用<p>，</p>或</p><p>替换每个空行，基于打开/关闭段落的一些计数器。但这将迫使用户在每个</h1>之后制作这样的新线。

也许一些正则表达式可以完成这项工作，而不会限制用户“你必须在</h1>之后和<h1>之前制作新的线条”？

方法越普遍越好。

答案

下面的负向前瞻声明该行不包含<h1>或<div>标签。如果是，那么它将整行的内容捕获到组1中。

^(?!.*?(?:<h1>|<div>))(.+)$

替换字符串：

<p>$1</p>

DEMO

码：

<?php
$string = <<<EOT
<h1>Headline</h1>
That should be paragraph 1.

That should be paragraph 2.

<div>Something.</div>
That should be paragraph 3.
EOT;
echo preg_replace('~^(?!.*?(?:<h1>|<div>))(.+)$~m', '<p>$1</p>', $string)
?>

输出：

<h1>Headline</h1>
<p>That should be paragraph 1.</p>

<p>That should be paragraph 2.</p>

<div>Something.</div>
<p>That should be paragraph 3.</p>