Question

扫描目录树并生成xml。我尝试了很多东西，但它失败了。

对于Ex XML文件结构。

<dir name="dir_A">
    <dir name="dir_AA"> 
       <file name="abc.doc"/>
    </dir>
    <dir name="dir_BA">
       <dir name="dir_BAA"> 
           <file name="abc.doc"/>
       </dir>
       <file name="abc.doc"/>
    </dir>
</dir>

我尝试的代码但不完整。我在开发期间删除了一些代码，我现在还没有，抱歉。

import xml.etree.ElementTree as ET
import os

class XMLOperations:

    def list_files(self, startpath):

        parent = None
        prevLevel = None
        xmlRoot = ET.Element("root")

        xmlRoot.set('xml','http://www.google.com')
        xmlRoot.set('xmlns','http://www.w3.org/1999/xlink')

        directory = ET.Element("directory")
        elementFile = ET.Element("file")

        for root, dirs, files in os.walk(startpath):

            level = root.replace(startpath, '').count(os.sep)
            current = os.path.basename(root)

            try:
                dir_name = root.split(startpath+"/")[1]
            except:
                continue

            depth = dir_name.count(os.sep)
            fList = dir_name.split(os.sep)

            if level == 0:
                ET.SubElement(xmlRoot, directory, name = current)

            else:
                for tags in fList:
                    ET.SubElement(xmlRoot, directory)
            if depth > 3:
                break

        #with open("output.xml",'w') as file:
        #    file.write(xmlRoot)

        ET.dump(xmlRoot)

谢谢。

Answer 1

使用os.walk获取root（当前目录），当前目录的目录列表，文件列表当前目录。
通过lxml Parser创建xml根元素。
按os.walk
这很重要：通过xpath获取当前目录的Parent元素，因此从当前位置路径创建xpath。例如xapth = "/dir[@name='dir_9']"或xapth = "/dir[@name='dir_9']/dir[@name='dir_apache']"
按目录列表附加dir元素。
按文件列表追加'file'元素。

<强>输入：

>>> p = '/home/vivek/Desktop/9' 
>>> import os
>>> for root, dires,files in os.walk(p):
...   print root
...   print dires
...   print files
...   print "="*10
... 
/home/vivek/Desktop/9
['apache', 'i18n', 'templates', 'common']
['manage.py', 'urls.py', 'settings.py', '__init__.py']
==========
/home/vivek/Desktop/9/apache
[]
['readeradmin.wsgi']
==========
/home/vivek/Desktop/9/i18n
[]
['__init__.py', 'models.py']
==========
/home/vivek/Desktop/9/templates
['admin', 'registration']
[]
==========
/home/vivek/Desktop/9/templates/admin
[]
['base_site.html']
==========
/home/vivek/Desktop/9/templates/registration
[]
['logged_out.html']
==========
/home/vivek/Desktop/9/common
[]
['views.py', '__init__.py', 'idmaptoalpha.py', 'tests.py', 'models.py']
==========

<强>代码：

import os 
import lxml.etree as PARSER
xml_root = PARSER.Element("dir", {"name":"dir_"+os.path.basename(p)})

base_location = os.path.dirname(p) + "/"
for root, dires, files in os.walk(p):
    # Get Parent by xpth
    xpath_tmp = root.split(base_location)[1]
    xpath_p = ""
    for i in xpath_tmp.split("/"):
        xpath_p = "%s/dir[@name='dir_%s']"%(xpath_p, i)

    parent = xml_root.xpath(xpath_p)[0]
    #- Append directory to parent element.
    for i in dires:
        parent.append(PARSER.Element("dir", {"name":"dir_"+i}))
    #- Append files to parent element.
    for i in files:
        parent.append(PARSER.Element("file", {"name":i}))


print PARSER.tostring(xml_root, method="xml", pretty_print=True)

<强>输出：

<dir name="dir_9">
  <dir name="dir_apache">
    <file name="readeradmin.wsgi"/>
  </dir>
  <dir name="dir_i18n">
    <file name="__init__.py"/>
    <file name="models.py"/>
  </dir>
  <dir name="dir_templates">
    <dir name="dir_admin">
      <file name="base_site.html"/>
    </dir>
    <dir name="dir_registration">
      <file name="logged_out.html"/>
    </dir>
  </dir>
  <dir name="dir_common">
    <file name="views.py"/>
    <file name="__init__.py"/>
    <file name="idmaptoalpha.py"/>
    <file name="tests.py"/>
    <file name="models.py"/>
  </dir>
  <file name="manage.py"/>
  <file name="urls.py"/>
  <file name="settings.py"/>
  <file name="__init__.py"/>
</dir>

在python中是否有任何扫描目录树并生成xml的代码

1 个答案: