在python中编辑XML作为字典？

问题描述

我试图从python中的模板xml文件生成自定义的xml文件。

I'm trying to generate customized xml files from a template xml file in python.

从概念上说，我想在模板xml中读取，删除一些元素，更改一些文本属性，并将新的xml输出到一个文件。我想要这样做：

Conceptually, I want to read in the template xml, remove some elements, change some text attributes, and write the new xml out to a file. I wanted it to work something like this:

conf_base = ConvertXmlToDict('config-template.xml')
conf_base_dict = conf_base.UnWrap()
del conf_base_dict['root-name']['level1-name']['leaf1']
del conf_base_dict['root-name']['level1-name']['leaf2']

conf_new = ConvertDictToXml(conf_base_dict)

现在我想写文件，但我看不到如何获得
ElementTree.ElementTree.write（）

now I want to write to file, but I don't see how to get toElementTree.ElementTree.write()

conf_new.write('config-new.xml')

有没有办法还是有人建议以不同的方式做这个事情？

Is there some way to do this, or can someone suggest doing this a different way?

推荐答案

为了在python中轻松处理XML，我喜欢库。它的工作原理如下：

For easy manipulation of XML in python, I like the Beautiful Soup library. It works something like this:

示例XML文件：

<root>
  <level1>leaf1</level1>
  <level2>leaf2</level2>
</root>

Python代码：

from BeautifulSoup import BeautifulStoneSoup, Tag, NavigableString

soup = BeautifulStoneSoup('config-template.xml') # get the parser for the xml file
soup.contents[0].name
# u'root'

可以使用节点名称作为方法：

You can use the node names as methods:

soup.root.contents[0].name
# u'level1'

还可以使用正则表达式：

It is also possible to use regexes:

import re
tags_starting_with_level = soup.findAll(re.compile('^level'))
for tag in tags_starting_with_level: print tag.name
# level1
# level2

添加和插入新节点非常简单：

Adding and inserting new nodes is pretty straightforward:

# build and insert a new level with a new leaf
level3 = Tag(soup, 'level3')
level3.insert(0, NavigableString('leaf3')
soup.root.insert(2, level3)

print soup.prettify()
# <root>
#  <level1>
#   leaf1
#  </level1>
#  <level2>
#   leaf2
#  </level2>
#  <level3>
#   leaf3
#  </level3>
# </root>

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