如何将XML转换为Python对象？

我需要加载XML文件并将内容转换为面向对象的Python结构.我想接受这个:

    content

把它变成这样的东西:

main
main.object1 = "content"
main.object1.attr = "name"

XML数据将具有比这更复杂的结构,我不能硬编码元素名称.解析时需要收集属性名称并将其用作对象属性.

如何将XML数据转换为Python对象？

值得一看lxml.objectify.

xml = """

content
contenbar
me
"""

from lxml import objectify

main = objectify.fromstring(xml)
main.object1[0]             # content
main.object1[1]             # contenbar
main.object1[0].get("attr") # name
main.test                   # me

或者反过来构建xml结构:

item = objectify.Element("item")
item.title = "Best of python"
item.price = 17.98
item.price.set("currency", "EUR")

order = objectify.Element("order")
order.append(item)
order.item.quantity = 3
order.price = sum(item.price * item.quantity for item in order.item)

import lxml.etree
print(lxml.etree.tostring(order, pretty_print=True))

输出:

  
    
    17.98
    3
  
  53.94

我今天不止一次推荐这个,但尝试Beautiful Soup(easy_install BeautifulSoup).

from BeautifulSoup import BeautifulSoup

xml = """

    

"""

soup = BeautifulSoup(xml)
# look in the main node for object's with attr=name, optionally look up attrs with regex
my_objects = soup.main.findAll("object", attrs={'attr':'name'})
for my_object in my_objects:
    # this will print a list of the contents of the tag
    print my_object.contents
    # if only text is inside the tag you can use this
    # print tag.string