0

我是 XML 文件，根有 10 个子子项（相同层次结构），名称为“testCase”

我无法弄清楚以下内容：首先，我正在执行以下操作来获取所有子孩子：

for testCase in root.iter('testCase'):
  1. 我需要从最后一个子“testCase”中获取一些属性。但我怎么知道它是最后一个'testCase'。有没有办法计算它们？
  2. 另外，有没有一种方法可以访问第 n 个子子项而无需通过 iter()？
4

2 回答 2

3

尝试以下示例。请参阅下面的输出。它显示了用作my.xml. 该元素表现为一个子元素列表（即它也可以被迭代）。有一些函数和迭代器可以独立于它们的位置获取文档顺序中的所有想要的元素（即，不管它们有多深，谁的孩子等等）。element.attrib表现为属性字典。该标准xml.etree.ElementTree还支持 XPath 的子集——见最后：

import xml.etree.ElementTree as et

tree = et.parse('my.xml')
root = tree.getroot()     # the root element of the tree

et.dump(root)             # here is how the input file looks inside

# Any element behaves as a list of children. This way, the last child
# of the list can be accessed via negative index.
print '-------------------------------------------'
print root[-1]

# Here is the content.
print '-------------------------------------------'
et.dump(root[-1])

# If the elements could be not direct children, you can use findall('tag') to 
# get the list of the elements. Then you access it again as the last element
# of the list
print '-------------------------------------------'
lst = root.findall('testCase')
et.dump(lst[-1])

# The number of the 'testCase' elements is simply the length of the list.
print '-------------------------------------------'
print 'Num. of test cases:', len(lst)

# The elem.iter('tag') works similarly. But if you want the last element,
# you must know when the element is the last one. It means you have to 
# loop through all of them anyway.
print '-------------------------------------------'
last = None  # init
for e in root.iter('testCase'):
    last = e

et.dump(last)

# The attributes of the elements take the form of the dictinary .attrib.
print '-------------------------------------------'
print last.attrib
print last.attrib['name']

# The standard xml.etree.ElementTree supports a subset of XPath. You can use
# it if you are familiar with XPath.
print '-------------------------------------------'
third = root.find('.//testCase[3]')
et.dump(third)

# ... including the last() function. For more complex cases, use lxml
# as pointed out by Emmanuel.
print '-------------------------------------------'
last = root.find('.//testCase[last()]')
et.dump(last)

它在我的控制台上打印以下内容：

    c:\tmp\___python\Sunny\so12669404>python a.py
<root>
  <testCase name="a" />
  <testCase name="b" />
  <testCase name="c" />
  <testCase name="d" />
</root>
-------------------------------------------
<Element 'testCase' at 0x231a630>
-------------------------------------------
<testCase name="d" />
-------------------------------------------
<testCase name="d" />
-------------------------------------------
Num. of test cases: 4
-------------------------------------------
<testCase name="d" />
-------------------------------------------
{'name': 'd'}
d
-------------------------------------------
<testCase name="c" />

-------------------------------------------
<testCase name="d" />
于 2012-10-01T08:21:17.087 回答
2

关于这种类型的操作，您应该使用XPath，这是浏览 XML 树的一种常见且简单的方法。我不认为标准的 Python ElementTree 支持 XPath，但lxml支持（也很常用），这是一个示例：

获取最后一个孩子：

>>> text = """<Root>
    <Child name="child1" />
    <Child name="child2" />
    <Child name="child3" />
    <Child name="child4" />
    <Child name="child5" />
</Root>"""
>>> from lxml import etree
>>> root = etree.fromstring(text)
>>> last_tag = root.xpath('/Root/Child[last()]')[0]
>>> last_tag.attrib['name']
'child5'

直接访问元素编号#n：

>>> tag3 = root.xpath('/Root/Child[3]')[0]
>>> tag3.attrib['name']
'child3'
于 2012-10-01T08:23:03.997 回答