如何获取lxml元素的css属性?

问题描述：

我想找到一个快速的函数来获取lxml元素的所有样式属性，这些属性应考虑css样式表，style属性元素并解决继承问题.

I want to find a fast function to get all style properties of a lxml element that take into account the css stylesheet, the style attribute element and tackle the herit issue.

例如:

html:

<body>
  <p>A</p>
  <p id='b'>B</p>
  <p style='color:blue'>B</p>
</body>

css:

body {color:red;font-size:12px}
p.b {color:pink;}

python:

elements = document.xpath('//p')
print get_style(element[0]) 
>{color:red,font-size:12px}
print get_style(element[1]) 
>{color:pink,font-size:12px}
print get_style(element[2]) 
>{color:blue,font-size:12px}

谢谢

答

您可以结合使用lxml和 cssutils . 此 cssutils实用程序模块应该可以你在问什么与该模块一起安装cssutils，然后运行以下代码:

You can do this with a combination of lxml and cssutils. This cssutils utility module should be able to do what you're asking. Install cssutils along with that module, then run the following code:

from style import *

html = """<body>
    <p>A</p>
    <p id='b'>B</p>
    <p style='color:blue'>B</p>
</body>"""

css = """body {color:red;font-size:12px}
p {color:yellow;}
p.b {color:green;}"""


def get_style(element, view):
    if element != None:
        inline_style = [x[1] for x in element.items() if x[0] == 'style']
        outside_style =  []
        if view.has_key(element):
            outside_style = view[element].getCssText()
        r = [[inline_style, outside_style]]
        r.append(get_style(element.getparent(), view))
        return r
    else:
        return None

document = getDocument(html)
view = getView(document, css)

elements = document.xpath('//p')
print get_style(elements[0], view)

如何获取lxml元素的css属性?

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