xpath解析XML简单明了,但是XML有命名空间的话就会出错了。解决方法是节点前加命名空间的前缀,下例中x、y是变量可以任意定义。
例如XML文档如下:
Text 解析代码片段:tree = etree.parse(path)
root = tree.getroot()
for child in root:
r=child.xpath(‘x:a/y:b/text()’,namespaces={‘x’: ‘www.jingfengshuo.com’,
‘y’: ‘jingfengshuo.com’})[0]
如果XML文档如下:
tree = etree.parse(path)
root = tree.getroot()
for child in root:
r=child.xpath(‘x:a/x:b/text()’,namespaces={‘x’: ‘www.jingfengshuo.com’})[0]