grep出力からxml文書を作成する方法

Question

だから私はPythonで試してみましたが、あなたが望むことをする簡単なスクリプトを思いついたようです。ここにいる：

#!/usr/bin/env python2
# -*- coding: ascii -*-
"""pathlist2xml.py

Takes a list of file-system paths and
generates an XML representation of the
corresponding file-system hierarchy.
"""

import sys
from lxml.etree import Element, SubElement, fromstring, tostring, XMLParser
from xml.sax.saxutils import escape, unescape
from os.path import join, isdir
from posix import lstat
import fileinput

def insert_path(xmlroot, path):
    """Updates an XML element `xmlroot` and adds the
    child elements that represent the path `path`."""

    # Initialize a node cursor to start at the root node
    xmlcursor = xmlroot

    # Keep track of the relative path
    fullpath = ''

    # Iterate through the components of the path
    for path_component in path.split('/'):

        # Update the path
        fullpath = join(fullpath, path_component)

        # UTF and XML encode the strings
        fullpath_encoded = escape(fullpath.encode('string-escape'))
        path_component_encoded = escape(path_component.encode('string-escape'))

        # Check to see if the component if already represented by a node
        xmlnodes = xmlcursor.xpath("./*[@name='%s']" % path_component_encoded)

        # If the node exists, update the cursor
        if xmlnodes:
            xmlcursor = xmlnodes[0]

        # If the node doesn't exists, create it
        else:

            # Create the node
            if isdir(fullpath):
                xmlcursor = SubElement(xmlcursor, "directory")
            else:
                xmlcursor = SubElement(xmlcursor, "file")

            # (Optional) Add some file-attributes
            # xmlcursor.set('name', path_component)
            xmlcursor.set('name', path_component_encoded)
            xmlcursor.set('path', fullpath_encoded)
            xmlcursor.set('inode', str(lstat(fullpath).st_ino))

    # Return the modified root element (for convenience - not necessary)
    return(xmlroot)

def paths_to_xml(pathlist):
    """ Takes a list of file-system paths and generates an XML
    representation of the corresponding file-system hierarchy.
    """

    xmlroot = Element('root')

    for path in pathlist:
        insert_path(xmlroot, path.strip().strip('/'))

    return(xmlroot)

# Read a list of file paths standard input or from a list of files
if __name__ == "__main__":

    # Get the XML document
    xmlroot = paths_to_xml(fileinput.input())

    # Display the generated XML document
    print(tostring(xmlroot, pretty_print=True))

以下は、実際にどのように動作するかを示す小さなサンプルセッションです。まず、いくつかのディレクトリとファイルを作成しました。

mkdir -p /tmp/xmltest
cd /tmp/xmltest
touch file1
touch file2
mkdir dir1
touch dir1/file3
touch dir1/file4
mkdir dir2
mkdir dir2/dir3
touch dir2/dir3/file5

このサブ階層は次のとおりですtree。

.
├── dir1
│   ├── file3
│   └── file4
├── dir2
│   └── dir3
│       └── file5
├── file1
└── file2

以下は、次の出力でスクリプトを呼び出す方法の例ですfind。

find . | pathlist2xml.py

結果のXML出力は次のとおりです。

<root>
  <directory name="." path="." inode="3587802">
    <directory name="dir1" path="./dir1" inode="3587817">
      <file name="file3" path="./dir1/file3" inode="3587818"/>
      <file name="file4" path="./dir1/file4" inode="3587819"/>
    </directory>
    <directory name="dir2" path="./dir2" inode="3587820">
      <directory name="dir3" path="./dir2/dir3" inode="3587821">
        <file name="file5" path="./dir2/dir3/file5" inode="3587822"/>
      </directory>
    </directory>
    <file name="file1" path="./file1" inode="3587815"/>
    <file name="file2" path="./file2" inode="3587816"/>
  </directory>
</root>

find以下は、次のものと組み合わせた2番目の例ですgrep。

find . | grep dir2 | pathlist2xml.py

これは2番目の例の出力です。

<root>
  <directory name="." path="." inode="3587802">
    <directory name="dir2" path="./dir2" inode="3587820">
      <directory name="dir3" path="./dir2/dir3" inode="3587821">
        <file name="file5" path="./dir2/dir3/file5" inode="3587822"/>
      </directory>
    </directory>
  </directory>
</root>

Answer 1