如何在Python中一次性修改嵌套目录的权限？

我目前通过以下代码逐个修改嵌套目录（例如路径/apps/sys/utils/prod/sales/apparel/）的权限。请问能否一次性完成权限修改，达到与循环逐个修改相同的效果？

import os
import stat
def change_permissions(path, new_mode):
    """Change permissions of the directory to new_mode."""
    if not os.path.islink(path):
        original_mode = os.stat(path).st_mode
        os.chmod(path, new_mode)
        oct_perm = oct(new_mode)
        unix_perm = oct_perm[-3: (len(oct_perm))]
        print(f"Permission of {path} changed to {new_mode} = {unix_perm}")
        return original_mode
    return None


target_full_path = "/apps/sys/utils/prod/sales/apparel/shirts.csv"
# Determine the directory path up to which permissions need to be changed
target_dir = os.path.dirname(target_full_path)
print(f"The abstract file dir path upto which perm to be changed {target_dir}")
# Change permissions for the specific directory needed
original_permissions = []
current_dir = "/apps/sys/utils/prod"
for part in target_dir.split(os.sep)[len(dest_root.split(os.sep)):]:
    current_dir = os.path.join(current_dir, part)
    print(f"\n### part = {part} and current_dir = {current_dir}")
    if not os.path.exists(current_dir):
        os.makedirs(current_dir)
        print(f"Created the current_dir = {current_dir}")
    original_mode = change_permissions(current_dir, stat.S_IRWXU | stat.S_IRWXG | stat.S_IRWXO)
    original_permissions.append((current_dir, original_mode))
    print (f"Original mode for current_dir ({current_dir}) is {original_mode}\n")

可以一次性完成权限修改，以下是两种实现方式，效果和你当前的循环逻辑完全一致：

方式一：用os.walk反向遍历目录层级 #

os.walk可以遍历目录树，反向遍历就能从最深层的子目录往上处理，一次性完成所有目标目录的权限修改，同时保留原权限记录：

import os
import stat

def change_permissions(path, new_mode):
    """Change permissions of the directory to new_mode."""
    if not os.path.islink(path):
        original_mode = os.stat(path).st_mode
        os.chmod(path, new_mode)
        oct_perm = oct(new_mode)
        unix_perm = oct_perm[-3:]
        print(f"Permission of {path} changed to {new_mode} = {unix_perm}")
        return original_mode
    return None

target_full_path = "/apps/sys/utils/prod/sales/apparel/shirts.csv"
target_dir = os.path.dirname(target_full_path)
current_root = "/apps/sys/utils/prod"

original_permissions = []

# 反向遍历目录树，从最底层子目录开始处理
for root, dirs, files in os.walk(current_root, topdown=False):
    # 只处理current_root到target_dir之间的目录
    if root.startswith(target_dir) and root != current_root:
        print(f"\n### Processing directory: {root}")
        original_mode = change_permissions(root, stat.S_IRWXU | stat.S_IRWXG | stat.S_IRWXO)
        original_permissions.append((root, original_mode))
        print(f"Original mode for {root} is {original_mode}\n")

# 额外处理target_dir本身（确保覆盖到最终目录）
if target_dir not in [item[0] for item in original_permissions]:
    print(f"\n### Processing directory: {target_dir}")
    original_mode = change_permissions(target_dir, stat.S_IRWXU | stat.S_IRWXG | stat.S_IRWXO)
    original_permissions.append((target_dir, original_mode))
    print(f"Original mode for {target_dir} is {original_mode}\n")

方式二：创建目录时直接指定权限（针对新建目录场景） #

如果你的代码中很多目录是通过os.makedirs新建的，可以直接在创建时指定权限，避免后续逐个修改：

import os
import stat

def change_permissions(path, new_mode):
    """Change permissions of the directory to new_mode."""
    if not os.path.islink(path):
        original_mode = os.stat(path).st_mode
        os.chmod(path, new_mode)
        oct_perm = oct(new_mode)
        unix_perm = oct_perm[-3:]
        print(f"Permission of {path} changed to {new_mode} = {unix_perm}")
        return original_mode
    return None

target_full_path = "/apps/sys/utils/prod/sales/apparel/shirts.csv"
target_dir = os.path.dirname(target_full_path)
current_root = "/apps/sys/utils/prod"

original_permissions = []

# 新建目录时直接指定权限，0o777对应stat.S_IRWXU | stat.S_IRWXG | stat.S_IRWXO
if not os.path.exists(target_dir):
    os.makedirs(target_dir, mode=0o777, exist_ok=True)
    print(f"Created directory tree: {target_dir} with permissions 0o777")

# 处理已存在的目录（如果有的话）
for root, dirs, files in os.walk(current_root, topdown=False):
    if root.startswith(target_dir) and root != current_root:
        print(f"\n### Processing existing directory: {root}")
        original_mode = change_permissions(root, 0o777)
        original_permissions.append((root, original_mode))
        print(f"Original mode for {root} is {original_mode}\n")

注意事项 #

• 两种方式都会保留原权限的记录，和你原代码的original_permissions列表功能一致
• 如果目录是符号链接，都会跳过处理，和原逻辑对齐
• 方式一适合处理已存在的目录树，方式二更适合需要新建目录的场景

内容的提问来源于stack exchange，提问作者A.G.Progm.Enthusiast