从 XML 标签获取 URL

Question

我的 XML 文件：

<xml 
xmlns="http://www.myweb.org/2003/instance"
xmlns:link="http://www.myweb.org/2003/linkbase"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xlink="http://www.w3.org/1999/xlink"
xmlns:iso4217="http://www.myweb.org/2003/iso4217"
xmlns:utr="http://www.myweb.org/2009/utr">

<link:schemaRef xlink:type="simple" xlink:href="http://www.myweb.com/form/2020-01-01/test.xsd"></link:schemaRef>

我想从 <link:schemaRef> 标签中获取 URL: http://www.myweb.com/folder/form/1/2020-01-01/test.xsd。

我下面的 python 代码找到了 <link:schemaRef> 标签。但是我无法检索 URL.

from lxml import etree
with open(filepath,'rb') as f:
     file = f.read()    
root = etree.XML(file)
print(root.nsmap["link"]) #http://www.myweb.org/2003/linkbase
print(root.find(".//{"+root.nsmap["link"]+"}"+"schemaRef"))

Answer 1

使用：

>>> child = root.getchildren()[0]
>>> child.attrib
{'{http://www.w3.org/1999/xlink}type': 'simple', '{http://www.w3.org/1999/xlink}href': 'http://www.myweb.com/form/2020-01-01/test.xsd'}
>>> url = child.attrib['{http://www.w3.org/1999/xlink}href']

但是，我认为挑战在于您是否知道要使用的正确密钥（即 {http://www.w3.org/1999/xlink}href）。如果这是问题所在，那么我们只需要：

>>> print(root.nsmap['xlink'])   # Notice that the requested url is a href to the xlink
'http://www.w3.org/1999/xlink'
>>> key_url = "{"+key_prefix+"}href"
>>> print(child.attrib[key_url])
'http://www.myweb.com/form/2020-01-01/test.xsd'

Answer 2

这样试一下，看看行不行：

for i in root.xpath('//*/node()'):
if isinstance(i,lxml.etree._Element):
     print(i.values()[1])

输出：

http://www.myweb.com/form/2020-01-01/test.xsd

从 XML 标签获取 URL

Get the URL from XML tag

python

lxml