语法

rename()方法语法格式如下：

os.rename(src, dst)

参数

  • src -- 要修改的目录名

  • dst -- 修改后的目录名

返回值

该方法没有返回值

实例1

import os
try:
  import xml.etree.cElementTree as ET
except ImportError:
  import xml.etree.ElementTree as ET
path = input("请输入需要重命名的目录")
listdir = os.listdir(path)

for i in listdir:
	if ".xml" in i: #限定xml文件
		oldpath = os.path.join(path,i) #os.path.join(a,b)用于拼接路径
		tree = ET.parser(oldpath)
		filename = tree.find("filename").text
		newpath = os.path.join(path,filename)
		print("filename",filename)
		print("oldpath:"+oldpath,"newpath:"+newpath)
		os.rename(oldpath,newpath)

'''呈现的效果是把path目录下的子目录xml文件循环访问，并用其中的filename节点的值重命名文件。'''

实例2

'''用指定的规则给target目录下的子目录重命名可以用os.walk'''
import os

def walk(path,pj_str):

	for root, dirs, files in os.walk(path):
		for file in files:
			#获取完整路径
			all_path = os.path.join(root,file)
			print(all_path,"allpath")
			allpath_list = all_path.split("\\")
			filename = allpath_list[-1]

			pjlist = pj_str.split(" ")
			print(len(pjlist))
			new_filename = allpath_list[int(pjlist[0])]+"-"+allpath_list[int(pjlist[1])]+"-"+allpath_list[int(pjlist[2])]+"-"+allpath_list[int(pjlist[3])]+"-"+filename
			newall_path = os.path.join(root,new_filename)
			print(all_path,"allpath")
			print(newall_path,"newallpath")
			print(root,"root?")
			# os.rename(all_path,newall_path)

path = input("请输入你要重命名的目录 ")
pj_str = input("请输入拼接规则：")
walk(path,pj_str)