ElementTree - findall以递归方式选择所有子元素

Question

Python代码:

import xml.etree.ElementTree as ET
root = ET.parse("h.xml")
print root.findall('saybye')

h.xml代码:

<hello>
  <saybye>
   <saybye>
   </saybye>
  </saybye>
  <saybye>
  </saybye>
</hello>

代码输出,

[<Element 'saybye' at 0x7fdbcbbec690>, <Element 'saybye' at 0x7fdbcbbec790>]

saybyesaybye这里没有选择另一个孩子.那么,如何指示findall递归地向下走DOM树并收集所有三个saybye元素？

Answer 1

如果您不怕一点 XPath，则可以使用//表示查找任何后代节点的语法：

import xml.etree.ElementTree as ET
root = ET.parse("h.xml")
print(root.findall('.//saybye'))

不支持完整的 XPath，但这里有一个列表：https : //docs.python.org/2/library/xml.etree.elementtree.html#supported-xpath-syntax

Answer 2

从2.7版开始,您可以使用xml.etree.ElementTree.Element.iter:

import xml.etree.ElementTree as ET
root = ET.parse("h.xml")
print root.iter('saybye')

见19.7.xml.etree.ElementTree - ElementTree XML API

Answer 3

引用findall,

Element.findall()只查找带有当前元素直接子元素的标签的元素。

由于它只找到直接子节点，因此我们需要递归地查找其他子节点，如下所示

>>> import xml.etree.ElementTree as ET
>>> 
>>> def find_rec(node, element, result):
...     for item in node.findall(element):
...         result.append(item)
...         find_rec(item, element, result)
...     return result
... 
>>> find_rec(ET.parse("h.xml"), 'saybye', [])
[<Element 'saybye' at 0x7f4fce206710>, <Element 'saybye' at 0x7f4fce206750>, <Element 'saybye' at 0x7f4fce2067d0>]

更好的是，将其设为生成器函数，如下所示

>>> def find_rec(node, element):
...     for item in node.findall(element):
...         yield item
...         for child in find_rec(item, element):
...             yield child
... 
>>> list(find_rec(ET.parse("h.xml"), 'saybye'))
[<Element 'saybye' at 0x7f4fce206a50>, <Element 'saybye' at 0x7f4fce206ad0>, <Element 'saybye' at 0x7f4fce206b10>]