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>]

saybye``saybye此处未选择是另一个孩子。那么，如何指示findall递归地遍历DOM树并收集所有三个saybye元素？

Answer 1

报价findall，

Element.findall() 仅查找带有标签的元素，这些标签是当前元素的直接子元素。

由于它只能找到直系孩子，因此我们需要递归地找到其他孩子，像这样

>>> 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，使其成为生成器函数

>>> 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>]