我有以下代码:
tree = etree.ElementTree(new_xml)
for e in new_xml.iter():
print tree.getpath(e), e.text
这会给我类似以下内容:
/Item/Purchases
/Item/Purchases/Purchase[1]
/Item/Purchases/Purchase[1]/URL http://tvgo.xfinity.com/watch/x/6091165185315991112/movies
/Item/Purchases/Purchase[1]/Rating R
/Item/Purchases/Purchase[2]
/Item/Purchases/Purchase[2]/URL http://tvgo.xfinity.com/watch/x/6091165185315991112/movies
/Item/Purchases/Purchase[2]/Rating R
但是,我需要获取的不是列表元素的路径,而是属性的路径。 xml 如下所示:
<Item>
<Purchases>
<Purchase Country="US">
<URL>http://tvgo.xfinity.com/watch/x/6091165US</URL>
<Rating>R</Rating>
</Purchase>
<Purchase Country="CA">
<URL>http://tvgo.xfinity.com/watch/x/6091165CA</URL>
<Rating>R</Rating>
</Purchase>
</Item>
我如何获得以下路径?
/Item/Purchases
/Item/Purchases/Purchase[@Country="US"]
/Item/Purchases/Purchase[@Country="US"]/URL http://tvgo.xfinity.com/watch/x/6091165185315991112/movies
/Item/Purchases/Purchase[@Country="US"]/Rating R
/Item/Purchases/Purchase[@Country="CA"]
/Item/Purchases/Purchase[@Country="CA"]/URL http://tvgo.xfinity.com/watch/x/6091165185315991112/movies
/Item/Purchases/Purchase[@Country="CA"]/Rating R
最佳答案
不太漂亮,但它能完成工作。
replacements = {}
for e in tree.iter():
path = tree.getpath(e)
if re.search('/Purchase\[\d+\]$', path):
new_predicate = '[@Country="' + e.attrib['Country'] + '"]'
new_path = re.sub('\[\d+\]$', new_predicate, path)
replacements[path] = new_path
for key, replacement in replacements.iteritems():
path = path.replace(key, replacement)
print path, e.text.strip()
为我打印这个:
/Item
/Item/Purchases
/Item/Purchases/Purchase[@Country="US"]
/Item/Purchases/Purchase[@Country="US"]/URL http://tvgo.xfinity.com/watch/x/6091165US
/Item/Purchases/Purchase[@Country="US"]/Rating R
/Item/Purchases/Purchase[@Country="CA"]
/Item/Purchases/Purchase[@Country="CA"]/URL http://tvgo.xfinity.com/watch/x/6091165CA
/Item/Purchases/Purchase[@Country="CA"]/Rating R
关于python - 如何使用属性获取lxml中所有元素的路径,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38936531/