python 批量修改文件夹的后缀名

# -*- coding:utf8 -*-
import os
import random
def rename(path):
    oldpath = path
    pos1 = path.find('.luac')
    # pos2 = path.find('_', pos1+5)
    newpath = path[:pos1]+'.lua'
    os.rename(path, newpath)  # 对文件进行重命名
def test(path):
    files = os.listdir(path)  # 获取当前目录的所有文件及文件夹
    for file in files:
        try:
            file_path = os.path.join(path, file)  # 获取绝对路径
            if os.path.isdir(file_path):  # 判断是否是文件夹
                test(file_path)  # 如果是文件夹，就递归调用自己
            else:
                print(file_path)
                extension_name = os.path.splitext(
                    file_path)  # 将文件的绝对路径中的后缀名分离出来
                if extension_name[1] == '.luac':
                    rename(file_path)
        except:
            continue  # 可能会报错，所以用了try-except,如果要求比较严格，不需要报错，就删除异常处理，自己调试
test(r'/Users/jichonggula/Desktop/src')
# for old_name in os.listdir(address):
#     portion = os.path.splitext(old_name)
#     if portion[1] == ".luac":  # 旧的后缀
#         new_name = portion[0] + ".lua"  # 新的后缀
#         old_name = os.path.join(address, old_name)
# # os.rename的参数需要的是路径和文件名，所以这里要加上路径，要不然脚本执行失败。
#         new_name = os.path.join(address, new_name)
#         print(old_name)
#         print(new_name)
#         os.rename(old_name, new_name)