解析 XML 文件以获取文件夹结构？

Question

我必须从我的 XML 文件中检索文件夹结构。

我的文件夹结构：

XML 文件（捕获上述文件夹结构）如下：

<?xml version="1.0" encoding="utf-8"?>
<serverfiles name="Test">
  <serverfiles name="Fail">
    <serverfiles name="Cam1">
      <serverfiles name="Mod1">
        <serverfiles name="2019-03-07" />
        <serverfiles name="2019-03-08" />
      </serverfiles>
      <serverfiles name="Mod2">
        <serverfiles name="2019-03-07" />
        <serverfiles name="2019-03-08" />
      </serverfiles>
    </serverfiles>
  </serverfiles>
  <serverfiles name="Pass">
    <serverfiles name="Cam1">
      <serverfiles name="Mod1">
        <serverfiles name="2019-03-07" />
        <serverfiles name="2019-03-08" />
      </serverfiles>
      <serverfiles name="Mod2">
        <serverfiles name="2019-03-07" />
        <serverfiles name="2019-03-08" />
      </serverfiles>
    </serverfiles>
  </serverfiles>
</serverfiles>

我的 Python 脚本：

import xml.etree.ElementTree as ET
path_file = ET.parse(r'../../folder_structure.xml')
r = path_file.getroot()
l = [i.attrib for i in r.iter() if i.tag=='serverfiles']
print(l)

以上代码产生以下结果：

[{'name': 'Test'}, {'name': 'Fail'}, {'name': 'Cam1'}, {'name': 'Mod1'}, {'name': '2019-03-07'}, {'name': '2019-03-08'}, {'name': 'Mod2'}, {'name': '2019-03-07'}, {'name': '2019-03-08'}, {'name': '通过'}, {'name': 'Cam1'}, {'name': 'Mod1'}, {'name': '2019-03-07'}, {'name': '2019-03-08'}, {'name': 'Mod2'}, {'name': '2019-03-07'}, {'name': '2019-03-08'}]

这里的问题是，整个文件夹结构丢失了（父子关系丢失了）。如何修改我的脚本，以便我可以将我的列表呈现为文件夹结构？

Answer 1

这是一个使用递归的可能解决方案：

from pprint import pprint
import xml.etree.ElementTree as ET

def walk(e):
    name = e.attrib['name']
    children = [walk(c) for c in e if e.tag == 'serverfiles']
    struct = {'name': name}
    if children:
        struct['children'] = children
    return struct

path_file = ET.parse(r'folder_structure.xml')
r = path_file.getroot()
s = walk(r)
pprint(s)

还有输出：

{'children': [{'children': [{'children': [{'children': [{'name': '2019-03-07'},
                                                        {'name': '2019-03-08'}],
                                           'name': 'Mod1'},
                                          {'children': [{'name': '2019-03-07'},
                                                        {'name': '2019-03-08'}],
                                           'name': 'Mod2'}],
                             'name': 'Cam1'}],
               'name': 'Fail'},
              {'children': [{'children': [{'children': [{'name': '2019-03-07'},
                                                        {'name': '2019-03-08'}],
                                           'name': 'Mod1'},
                                          {'children': [{'name': '2019-03-07'},
                                                        {'name': '2019-03-08'}],
                                           'name': 'Mod2'}],
                             'name': 'Cam1'}],
               'name': 'Pass'}],
 'name': 'Test'}

编辑：更新代码以简化输出（基于评论）：

from pprint import pprint
import xml.etree.ElementTree as ET

def walk(e):
    name = e.attrib['name']
    children = [walk(c) for c in e if e.tag == 'serverfiles']
    return {name: children} if children else name

path_file = ET.parse(r'folder_structure.xml')
r = path_file.getroot()
s = walk(r)
pprint(s)

输出：

{'Test': [{'Fail': [{'Cam1': [{'Mod1': ['2019-03-07', '2019-03-08']},
                              {'Mod2': ['2019-03-07', '2019-03-08']}]}]},
          {'Pass': [{'Cam1': [{'Mod1': ['2019-03-07', '2019-03-08']},
                              {'Mod2': ['2019-03-07', '2019-03-08']}]}]}]}

结构更简单，但现在您必须考虑两种可能的类型——dict，当文件夹有子文件夹时，str，如果它是叶节点（没有子文件夹）。