基于XML的解决方案是为ElementTree编写一个helper类，该类：在解析为ElementTree之前在抓取XML声明行在写入时，如果不同时写入编码属性(我检查了源代码)，则无法写入XML声明行。

解析输入文件一次，获取根元素的命名空间。向ElementTree注册该命名空间，将空字符串作为前缀。完成后，使用ElementTree再次使用该新设置分析源文件。

它有一个主要缺点：XML注释丢失。我了解到，对于这种情况，这是不可接受的(我最初认为输入数据没有任何注释，但事实证明它有)。

我的助手类示例：from xml.etree import ElementTree as ET

import re

class ElementTreeHelper():

def __init__(self, xml_file_name):

xml_file = open(xml_file_name, "rb")

self.__parse_xml_declaration(xml_file)

self.element_tree = ET.parse(xml_file)

xml_file.seek(0)

root_tag_namespace = self.__root_tag_namespace(self.element_tree)

self.namespace = None

if root_tag_namespace is not None:

self.namespace = '{' + root_tag_namespace + '}'

# Register the root tag namespace as having an empty prefix, as

# this has to be done before parsing xml_file we re-parse.

ET.register_namespace('', root_tag_namespace)

self.element_tree = ET.parse(xml_file)

def find(self, xpath_query):

return self.element_tree.find(xpath_query)

def write(self, xml_file_name):

xml_file = open(xml_file_name, "wb")

if self.xml_declaration_line is not None:

xml_file.write(self.xml_declaration_line + '\n')

return self.element_tree.write(xml_file)

def __parse_xml_declaration(self, xml_file):

first_line = xml_file.readline().strip()

if first_line.startswith('<?xml ') and first_line.endswith('?>'):

self.xml_declaration_line = first_line

else:

self.xml_declaration_line = None

xml_file.seek(0)

def __root_tag_namespace(self, element_tree):

namespace_search = re.search('^{(\S+)}', element_tree.getroot().tag)

if namespace_search is not None:

return namespace_search.group(1)

else:

return None

def __main():

el_tree_hlp = ElementTreeHelper('houses.xml')

dogs_tag = el_tree_hlp.element_tree.getroot().find(

'{ns}house/{ns}dogs'.format(

ns=el_tree_hlp.namespace))

one_dog_added = int(dogs_tag.text.strip()) + 1

dogs_tag.text = str(one_dog_added)

el_tree_hlp.write('hejsan.xml')

if __name__ == '__main__':

__main()

输出：<?xml version="1.0"?>

2821

3

如果有人对此解决方案有改进，请毫不犹豫地获取代码并加以改进。