要删除目录及其子目录中的所有文件，而不删除文件夹本身，只需执行以下操作:

import os
mypath = "my_folder" #Enter your path here
for root, dirs, files in os.walk(mypath, topdown=False):
    for file in files:
        os.remove(os.path.join(root, file))

    # Add this block to remove folders
    for dir in dirs:
        os.rmdir(os.path.join(root, dir))

# Add this line to remove the root folder at the end
os.rmdir(mypath)

删除文件夹中所有文件/删除所有文件的最简单方法

import os
files = os.listdir(yourFilePath)
for f in files:
    os.remove(yourFilePath + f)

我觉得这个密码起作用了。它不会删除文件夹，您可以使用此代码删除具有特定扩展名的文件。

import os
import glob

files = glob.glob(r'path/*')
for items in files:
    os.remove(items)

注:以防有人对我的答案投了反对票，我在这里有一些事情要解释。

Everyone likes short 'n' simple answers. However, sometimes the reality is not so simple. Back to my answer. I know shutil.rmtree() could be used to delete a directory tree. I've used it many times in my own projects. But you must realize that the directory itself will also be deleted by shutil.rmtree(). While this might be acceptable for some, it's not a valid answer for deleting the contents of a folder (without side effects). I'll show you an example of the side effects. Suppose that you have a directory with customized owner and mode bits, where there are a lot of contents. Then you delete it with shutil.rmtree() and rebuild it with os.mkdir(). And you'll get an empty directory with default (inherited) owner and mode bits instead. While you might have the privilege to delete the contents and even the directory, you might not be able to set back the original owner and mode bits on the directory (e.g. you're not a superuser). Finally, be patient and read the code. It's long and ugly (in sight), but proven to be reliable and efficient (in use).

这里有一个冗长而丑陋，但可靠而有效的解决方案。

它解决了一些其他答案没有解决的问题:

它正确地处理符号链接，包括不对符号链接调用shutil.rmtree()(如果它链接到一个目录，它将通过os.path.isdir()测试;甚至os.walk()的结果也包含符号链接目录)。 它可以很好地处理只读文件。

下面是代码(唯一有用的函数是clear_dir()):

import os
import stat
import shutil


# http://stackoverflow.com/questions/1889597/deleting-directory-in-python
def _remove_readonly(fn, path_, excinfo):
    # Handle read-only files and directories
    if fn is os.rmdir:
        os.chmod(path_, stat.S_IWRITE)
        os.rmdir(path_)
    elif fn is os.remove:
        os.lchmod(path_, stat.S_IWRITE)
        os.remove(path_)


def force_remove_file_or_symlink(path_):
    try:
        os.remove(path_)
    except OSError:
        os.lchmod(path_, stat.S_IWRITE)
        os.remove(path_)


# Code from shutil.rmtree()
def is_regular_dir(path_):
    try:
        mode = os.lstat(path_).st_mode
    except os.error:
        mode = 0
    return stat.S_ISDIR(mode)


def clear_dir(path_):
    if is_regular_dir(path_):
        # Given path is a directory, clear its content
        for name in os.listdir(path_):
            fullpath = os.path.join(path_, name)
            if is_regular_dir(fullpath):
                shutil.rmtree(fullpath, onerror=_remove_readonly)
            else:
                force_remove_file_or_symlink(fullpath)
    else:
        # Given path is a file or a symlink.
        # Raise an exception here to avoid accidentally clearing the content
        # of a symbolic linked directory.
        raise OSError("Cannot call clear_dir() on a symbolic link")

回答一个有限的、具体的情况: 假设你想在维护子文件夹树的同时删除文件，你可以使用递归算法:

import os

def recursively_remove_files(f):
    if os.path.isfile(f):
        os.unlink(f)
    elif os.path.isdir(f):
        for fi in os.listdir(f):
            recursively_remove_files(os.path.join(f, fi))

recursively_remove_files(my_directory)

也许有点跑题，但我想很多人会觉得它很有用

扩展mhawke的回答，这是我所实现的。它删除文件夹的所有内容，但不删除文件夹本身。在Linux上测试了文件、文件夹和符号链接，应该也能在Windows上运行。

import os
import shutil

for root, dirs, files in os.walk('/path/to/folder'):
    for f in files:
        os.unlink(os.path.join(root, f))
    for d in dirs:
        shutil.rmtree(os.path.join(root, d))