从PHP获取CSS样式 [英] Get CSS style from PHP

问题描述

我正在使用Simple DOM解析器解析页面。它是整洁的，但我想要获得每个元素应用的CSS样式。不仅内联样式，而且适用于该元素的每种风格都是内联的，页内的或外部的。

有没有一个类？如果没有，你会怎么做？我不太在乎重写样式，级联或浏览器特定样式。拥有所有直接应用的样式就足够了。

解决方案

正如Martin所说，在这样做时，你几乎在PHP中编写一个浏览器 - 这是一个大问题！与任何大项目一样，关键是将其分解成更易于管理的步骤（尽管其中一些步骤不完全简单）。

您需要：

• 将外部CSS文件链接到（如果有的话）


• （回覆Gumbo ）：找到（或开发）一种阅读和解释外部CSS的方式，页内css和inline css


• 确定什么样式适用于每个元素（包括.class应用的样式，#id和元素类型），并且每个元素的父母 - 包括哪些css规则覆盖其他css规则等。

我不会说这是不可能的，像 MPDF 这几乎是一回事（可能提供一个好的起点），但我不认为有一个整洁的快速修复。

I am parsing pages using Simple DOM parser. It is neat, but I would like to get the applied css style for each element. Not only the inline styles, but every style it applies to that element, being it inline, in-page or external.

Is there a class that does that? If not, how would you do it? I do not really care about overriding styles, cascade, or browser specific styles. Having all the directly applied styles would suffice.

解决方案

As Martin says, in doing this you're almost writing a browser in PHP - it's a big ask! As with any big project, the key is to break it down into more manageable steps (although some of these aren't exactly straightforward).

You'll need to:

• work out which (if any) external css files are linked to
• (to echo Gumbo): find (or develop) a way of reading and interpreting the external css, in-page css and inline css
• work out what styles apply to each element (including styles applied by .class, #id and element type), and the parents of each element - including which css rules override which other css rules, etc.

I wouldn't say it's impossible, as things like MPDF do almost the same thing (and may provide a good starting point) but I don't think there's a neat quick-fix.

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