Python提取Linux内核源代码的目录结构实现方法

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Release: 2016-07-06 13:29:48
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今天用Python提取了Linux内核源代码的目录树结构,没有怎么写过脚本程序,我居然折腾了2个小时,先是如何枚举出给定目录下的所有文件和文件夹,os.walk可以实现列举,但是os.walk是只给出目录名和文件名,而没有绝对路径。使用os.path.listdir可以达到这个目的,然后是创建目录,由于当目录存在是会提示创建失败的错误,所以我先想删除所有目录,然后再创建,但是发现还是有问题,最好还是使用判断如果不存在才创建目录,存在时就不创建,贴下代码:

#  @This script can be used to iterate the given directory,and create the 

#  empty directory structure without file in it,e.g,I want to have you directory

#  as the linux kernel source, but i don't want the files, then this script comes.

#  @This script is running under python 3.1

#  @author:zhangchao

#  @Time:2011年7月25日18:43:26

###########################################################################

 

 

import os

import re

 

#listmydirs is created to recursivly list all the entrys in the specified path.

#In fact, we have os.walk to handle this problem

 

#

#level:目录的层数,不要也可以,主要是为了显示目录在那一层

#srcpath:内核源代码所在的路路径

#destpath:将要生成的内核源代码的目录结构所在路径

#

 

def createkerneldirs(level,srcpath,destpath):

  for entrys in os.listdir(srcpath): #学习listdir函数的用法

    tmpsrcpath=srcpath+os.sep+entrys

    tmpdestpath = tmpsrcpath.replace(srcpath,destpath)#将源路径中的E:\linux-2.6替换为E:\tmp,学习字符串替换函数的用法

  

    print('in level:'+str(level))

    print(tmpsrcpath)

    print(tmpdestpath)

     

    if os.path.isdir(tmpsrcpath):

      listmydirs(level+1,tmpsrcpath,tmpdestpath)

      if os.path.exists(tmpdestpath)==False: #如果文件不存在才创建文件

        os.makedirs(tmpdestpath)

 

if __name__=='__main__':

#将E:\linux-2.6的内核源代码目录结构拷贝到E:\tmp目录下

  createkerneldirs(1,r'E:\linux-2.6',r'E:\tmp')
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