python遍历一个目录,输出所有的文件名的实例

import os
dir = "e:\\"
for root, dirs, files in os.walk(dir):
  for file in files:
    print os.path.join(root,file)

import os
path = r'e:\case'
fns = [os.path.join(root,fn) for root, dirs, files in os.walk(path) for fn in files]
for f in fns:
  print(f)
print(len(fns))

#coding=utf-8
import os

def GetFileList(dir, fileList):
  newDir = dir
  if os.path.isfile(dir):
    fileList.append(dir.decode('gbk'))
  elif os.path.isdir(dir): 
    for s in os.listdir(dir):
      #如果需要忽略某些文件夹，使用以下代码
      #if s == "xxx":
        #continue
      newDir=os.path.join(dir,s)
      GetFileList(newDir, fileList) 
  return fileList

list = GetFileList('D:\\workspace\\PyDemo\\fas', [])
for e in list:
  print e