[英]How can I use a regular expression to detect a list or enumeration in PHP
I'm getting data from an XML feed. 我从XML提要中获取数据。 I can't control the feed nor it's content.
我无法控制供稿,也不可以满足其内容。
Sometimes, the data contains a list / enumeration. 有时,数据包含一个列表/枚举。 I want to parse this as a clean HTML unordered list.
我想将其解析为干净的HTML无序列表。
The data I receive will be in a format like this: 我收到的数据将采用以下格式:
<p>Some text in a paragraph tag</p>
<p>
- List item one <br>
- List-item-two<br>
-List item three <br>
- Listitem four<br>
</p>
<p>Another paragraph with text, and maybe even more paragraphs after this one!
They might even contain - dashes - - - or <br><br> breaks!</p>
Note that not every list item is neatly formatted. 请注意,并非每个列表项都经过整齐的格式化。 Some contain trailing paces between the
<br>
tag or between the dash and the text. 有些包含在
<br>
标记之间或破折号与文本之间的尾随节奏。
How can I postprocess this in PHP to get this result: 我如何在PHP中对此进行后处理以获得结果:
<p>Some text in a paragraph tag</p>
<p><ul>
<li>List item one</li>
<li>List-item-two</li>
<li>List item three</li>
<li>Listitem four</li>
</ul></p>
<p>Another paragraph with text, and maybe even more paragraphs after this one!
They might even contain - dashes - - - or <br><br> breaks!</p>
Can I do it with a regular expression? 我可以使用正则表达式吗? If so, what would it look like?
如果是这样,它将是什么样?
Yes, I think regex are a good start point. 是的,我认为正则表达式是一个很好的起点。 Have a look to preg_replace
看看preg_replace
The regex could be something like this (not tested) : 正则表达式可能是这样的(未经测试):
$li = preg_replace('/^-([a-z]+)(<br>)?$/i', '<li>$1</li>', $entry);
Of course this is not working (you need support for whitespace and so on), but I think you get the idea. 当然这是行不通的(您需要对空格的支持等等),但是我想您已经明白了。
You can get started by replacing ^-\\s*\\b(.+)\\b\\s*<br>$
with <li>$1</li>
. 您可以通过将
^-\\s*\\b(.+)\\b\\s*<br>$
替换为<li>$1</li>
。 I'll leave the hard part of wrapping it all in a <ul/>
up to you. 我将把所有内容包装在
<ul/>
的困难部分留给您。
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