按标签用python对xml进行排序

本文介绍了按标签用python对xml进行排序的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个 xml

<root>
 <node1>
  <B>text</B>
  <A>another_text</A>
  <C>one_more_text</C>
 </node1>
 <node2>
  <C>one_more_text</C>
  <B>text</B>
  <A>another_text</A>
 </node2>
</root>

我想得到如下输出:

<root>
 <node1>
  <A>another_text</A>
  <B>text</B>
  <C>one_more_text</C>
 </node1>
 <node2>
  <A>another_text</A>
  <B>text</B>
  <C>one_more_text</C>
 </node2>
</root>

我尝试了一些代码，例如:

I tried with some code like:

from xml.etree import ElementTree as et

tr = et.parse(path_in)
root = tr.getroot()
for children in root.getchildren():
    for child in children.getchildren():
        # sort it

tr.write(path_out)

我不能使用标准函数 sort 和 sorted 因为它排序错误(不是按标签).提前致谢.

I cannot use standard function sort and sorted because it sorted wrong way (not by tag).Thanks in advance.

推荐答案

您需要:

获取每个顶级节点"的子元素
按 tag 属性(节点名称)
重置每个顶级节点的子节点

示例工作代码:

from operator import attrgetter
from xml.etree import ElementTree as et

data = """  <root>
 <node1>
  <B>text</B>
  <A>another_text</A>
  <C>one_more_text</C>
 </node1>
 <node2>
  <C>one_more_text</C>
  <B>text</B>
  <A>another_text</A>
 </node2>
</root>"""


root = et.fromstring(data)
for node in root.findall("*"):  # searching top-level nodes only: node1, node2 ...
    node[:] = sorted(node, key=attrgetter("tag"))

print(et.tostring(root))

打印:

<root>
 <node1>
  <A>another_text</A>
  <B>text</B>
  <C>one_more_text</C>
 </node1>
 <node2>
  <A>another_text</A>
  <B>text</B>
  <C>one_more_text</C>
  </node2>
</root>

请注意，我们没有使用 getchildren() 方法在这里(它实际上是 deprecated 自 Python 2.7) - 使用每个 Element 实例是一个可迭代的子节点.

Note that we are not using getchildren() method here (it is actually deprecated since Python 2.7) - using the fact that each Element instance is an iterable over the child nodes.

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getChildren