Question

这是我的XML文件

<?xml version="1.0"?>
<data>
    <country name="Liechtenstein">
        <rank>1</rank>
        <year>2008</year>
        <gdp>141100</gdp>
        <neighbor name="Austria" direction="E"/>
        <neighbor name="Switzerland" direction="W"/>
    </country>
</data>

如何拉取country的所有子节点？

例如，我需要输出为['rank','year','gdp','neighbor']

Answer 1

使用ElementTree lib拉出子节点。这可能会对你有所帮助。

import xml.etree.ElementTree as ET
tree = ET.parse("file.xml")
root = tree.getroot()
for child in root:
  print({x.tag for x in root.findall(child.tag+"/*")})

Answer 2

使用xml.etree.ElementTree模块的解决方案：

import xml.etree.ElementTree as ET

tree = ET.parse("yourxml.xml")
root = tree.getroot()
tag_names = {t.tag for t in root.findall('.//country/*')}

print(tag_names)  # print a set of unique tag names

输出：

{'gdp', 'rank', 'neighbor', 'year'}

'.//country/*' - xpath表达式，用于提取节点country

Answer 3

Have a look up to python documentation。它确实使用这个xml树作为示例。

import xml.etree.ElementTree as ET
tree = ET.parse('country_data.xml')
root = tree.getroot()

country = root[0].getchildren()
map(lambda e: e.tag, r)
# ['rank', 'year', 'gdp', 'neighbor', 'neighbor']

是的，当你被困住时，打开repl并一步一步地走。我不记得以上所有这些。最后在2年或3年前使用过xml解析器。但我知道，＆＃34;试着看看＆＃34;是最好的老师。

这些是步骤，我是如何提出解决方案的。

# imports and other stuff.
>>> tree = ET.parse('data.xml')
>>> root = tree.getroot()
>>> country = root[0]
>>> dir(country)
['__class__', '__delattr__', '__delitem__', '__dict__', '__doc__', '__format__', '__getattribute__', '__getitem__', '__hash__', '__init__', '__len__', '__module__', '__new__', '__nonzero__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__setitem__', '__sizeof__', '__str__', '__subclasshook__', '__weakref__', '_children', 'append', 'attrib', 'clear', 'copy', 'extend', 'find', 'findall', 'findtext', 'get', 'getchildren', 'getiterator', 'insert', 'items', 'iter', 'iterfind', 'itertext', 'keys', 'makeelement', 'remove', 'set', 'tag', 'tail', 'text']
>>> country.keys()
['name']
>>> country.getchildren()
[<Element 'rank' at 0x7f873cf53910>, <Element 'year' at 0x7f873cf539d0>, <Element 'gdp' at 0x7f873cf53a90>, <Element 'neighbor' at 0x7f873cf53c10>, <Element 'neighbor' at 0x7f873cf53c50>]
>>> country.getchildren()[0]
<Element 'rank' at 0x7f873cf53910>
>>> r = country.getchildren()[0]
>>> dir(r)
['__class__', '__delattr__', '__delitem__', '__dict__', '__doc__', '__format__', '__getattribute__', '__getitem__', '__hash__', '__init__', '__len__', '__module__', '__new__', '__nonzero__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__setitem__', '__sizeof__', '__str__', '__subclasshook__', '__weakref__', '_children', 'append', 'attrib', 'clear', 'copy', 'extend', 'find', 'findall', 'findtext', 'get', 'getchildren', 'getiterator', 'insert', 'items', 'iter', 'iterfind', 'itertext', 'keys', 'makeelement', 'remove', 'set', 'tag', 'tail', 'text']
>>> r.tag
'rank'
>>> r = country.getchildren()[0]
>>> r
<Element 'rank' at 0x7f873cf53910>
>>> r = country.getchildren()
>>> r
[<Element 'rank' at 0x7f873cf53910>, <Element 'year' at 0x7f873cf539d0>, <Element 'gdp' at 0x7f873cf53a90>, <Element 'neighbor' at 0x7f873cf53c10>, <Element 'neighbor' at 0x7f873cf53c50>]
>>> map(lambda e: e.tag, r)
['rank', 'year', 'gdp', 'neighbor', 'neighbor']

Answer 4

看起来Python可以使用一些API来解析XML文件：

http://python-guide-pt-br.readthedocs.io/en/latest/scenarios/xml/

https://www.tutorialspoint.com/python/python_xml_processing.htm

Answer 5

此代码在python 3.6

下测试

 import xml.etree.ElementTree as ET

 name = '4.xml'
 tree = ET.parse(name)
 root = tree.getroot()
 ditresult =[]

 for child in root:
     for child1 in child:         
          ditresult.append(child1.tag)

  print (ditresult)

=============

['rank', 'year', 'gdp', 'neighbor', 'neighbor']

如何使用python获取XML的所有子节点？

5 个答案: