遍历目录并替换每个子目录中名为1.mp3的文件为其上一级目录的名称

November 22nd 2023

import os def rename_mp3_files(base_dir): for root, dirs, files in os.walk(base_dir): for file in files: if file == ‘111 (1).wav’: # 获取上一级目录的名称 parent_dir = os.path.basename(os.path.abspath(os.path.join(root, os.pardir))) # 构造新的文件名 new_filename = os.path.

import os

def rename_mp3_files(base_dir):
    for root, dirs, files in os.walk(base_dir):
        for file in files:
            if file == '111 (1).wav':
                # 获取上一级目录的名称
                parent_dir = os.path.basename(os.path.abspath(os.path.join(root, os.pardir)))
                
                # 构造新的文件名
                new_filename = os.path.join(root, f'{parent_dir}.wav')
                
                # 重命名文件
                os.rename(os.path.join(root, file), new_filename)
                print(f'Renamed: {file} to {parent_dir}.wav')

# 用法示例
base_directory = 'C:/Users/Administrator/Desktop/111/108010slogo/spleeter'
rename_mp3_files(base_directory)