解析XML文件，并创建文件

Question

的名单有一个info.xml文件下的每个在/ var /包/ {很多文件夹} /info.xml哪里都是不同的目录，但在info.xml显示目录的信息

如果文件类型为“config”，可以通过检查“config”是类型标签内的类型来找到，我需要解析每个{许多文件夹}并创建一个Path标签内的文件路径列表。

的info.xml文件西港岛线是这样的，

<Files> 
    <File> 
     <Path>usr/share/doc/dialog/samples/form1</Path> 
     <Type>doc</Type> 
     <Size>1222</Size> 
     <Uid>0</Uid> 
     <Gid>0</Gid> 
     <Mode>0755</Mode> 
     <Hash>49744d73e8667d0e353923c0241891d46ebb9032</Hash> 
    </File> 
    <File> 
     <Path>usr/share/doc/dialog/samples/form3</Path> 
     <Type>config</Type> 
     <Size>1294</Size> 
     <Uid>0</Uid> 
     <Gid>0</Gid> 
     <Mode>0755</Mode> 
     <Hash>f30277f73e468232c59a526baf3a5ce49519b959</Hash> 
    </File> 
</Files> 

Answer 1

这里是没有错误的处理非常简单的例子，用非常严格的定义XML文件的工作，但你应该把它作为开始，并继续与以下链接：

• http://docs.python.org/library/xml.dom.html
• http://docs.python.org/library/xml.dom.minidom.html
• http://docs.python.org/library/os.path.html
• http://docs.python.org/library/os.html

代码：

import os 
import os.path 
from xml.dom.minidom import parse 


def parse_file(path): 
    files = [] 
    try: 
     dom = parse(path) 
     for filetag in dom.getElementsByTagName('File'): 
      type = filetag.getElementsByTagName('Type')[0].firstChild.data 
      if type == 'config': 
       path = tag.getElementsByTagName('Path')[0].firstChild.data 
       files.append(path) 
     dom.unlink() 
    except: 
     raise 
    return files 


def main(): 
    files = [] 
    for root, dirs, files in os.walk('/var/packs'): 
     if 'info.xml' in files: 
      files += parse_file(os.path.join(root, 'info.xml')) 
    print 'The list of desired files:', files 


if __name__ == '__main__': 
    main() 

Answer 2

写作这一关我的头顶部，但在这里不用。我们将利用os.path.walk递归地下降到您的目录和minidom进行解析。

import os 
from xml.dom import minidom 

# opens a given info.xml file and prints out "Path"'s contents 
def parseInfoXML(filename): 
    doc = minidom.parse(filename) 
    for fileNode in doc.getElementsByTagName("File"): 
     # warning: we assume the existence of a Path node, and that it contains a Text node 
     print fileNode.getElementsByTagName("Path")[0].childNodes[0].data 
    doc.unlink() 

def checkDirForInfoXML(arg, dirname, names): 
    if "info.xml" in names: 
     parseInfoXML(os.path.join(dirname, "info.xml")) 

# recursively walk the directory tree, calling our visitor function to check for info.xml in each dir 
# this will include packs as well, so be sure that there's no info.xml in there 
os.path.walk("/var/packs" , checkDirForInfoXML, None) 

不是最有效的方式来完成它，我敢肯定，但如果你不希望任何错误/不管它会做。

Answer 3

使用lxml.etree和XPath：

files = [] 
for root, dirnames, filenames in os.walk('/var/packs'): 
    for filename in filenames: 
     if filename != 'info.xml': 
      continue 
     tree = lxml.etree.parse(os.path.join(root, filename)) 
     files.extend(tree.getroot().xpath('//File[Type[text()="config"]]/Path/text()')) 

如果LXML不可用，则可以选择使用etree API标准库：

files = [] 
for root, dirnames, filenames in os.walk('/var/packs'): 
    for filename in filenames: 
     if filename != 'info.xml': 
      continue 
     tree = xml.etree.ElementTree.parse(os.path.join(root, filename)) 
     for file_node in tree.findall('File'): 
      type_node = file_node.find('Type') 
      if type_node is not None and type_node.text == 'config': 
       path_node = file_node.find('Path') 
       if path_node is not None: 
        files.append(path_node.text)