如何在Python中创建目录结构的zip存档?
当前回答
此函数将递归地压缩目录树,压缩文件,并在存档中记录正确的相对文件名。存档条目与zip-r output.zip source_dir生成的条目相同。
import os
import zipfile
def make_zipfile(output_filename, source_dir):
relroot = os.path.abspath(os.path.join(source_dir, os.pardir))
with zipfile.ZipFile(output_filename, "w", zipfile.ZIP_DEFLATED) as zip:
for root, dirs, files in os.walk(source_dir):
# add directory (needed for empty dirs)
zip.write(root, os.path.relpath(root, relroot))
for file in files:
filename = os.path.join(root, file)
if os.path.isfile(filename): # regular files only
arcname = os.path.join(os.path.relpath(root, relroot), file)
zip.write(filename, arcname)
其他回答
要保留要归档的父目录下的文件夹层次结构,请执行以下操作:
import glob
import os
import zipfile
with zipfile.ZipFile(fp_zip, "w", zipfile.ZIP_DEFLATED) as zipf:
for fp in glob(os.path.join(parent, "**/*")):
base = os.path.commonpath([parent, fp])
zipf.write(fp, arcname=fp.replace(base, ""))
如果需要,可以将其更改为使用pathlib进行文件globbing。
您可能想看看zipfile模块;文档位于http://docs.python.org/library/zipfile.html.
您可能还需要os.walk()来索引目录结构。
这是一种现代方法,使用pathlib和上下文管理器。将文件直接放在zip文件中,而不是放在子文件夹中。
def zip_dir(filename: str, dir_to_zip: pathlib.Path):
with zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED) as zipf:
# Use glob instead of iterdir(), to cover all subdirectories.
for directory in dir_to_zip.glob('**'):
for file in directory.iterdir():
if not file.is_file():
continue
# Strip the first component, so we don't create an uneeded subdirectory
# containing everything.
zip_path = pathlib.Path(*file.parts[1:])
# Use a string, since zipfile doesn't support pathlib directly.
zipf.write(str(file), str(zip_path))
如果您想要一个类似于任何通用图形文件管理器的压缩文件夹的功能,可以使用以下代码,它使用zipfile模块。使用这段代码,您将得到以路径为根文件夹的zip文件。
import os
import zipfile
def zipdir(path, ziph):
# Iterate all the directories and files
for root, dirs, files in os.walk(path):
# Create a prefix variable with the folder structure inside the path folder.
# So if a file is at the path directory will be at the root directory of the zip file
# so the prefix will be empty. If the file belongs to a containing folder of path folder
# then the prefix will be that folder.
if root.replace(path,'') == '':
prefix = ''
else:
# Keep the folder structure after the path folder, append a '/' at the end
# and remome the first character, if it is a '/' in order to have a path like
# folder1/folder2/file.txt
prefix = root.replace(path, '') + '/'
if (prefix[0] == '/'):
prefix = prefix[1:]
for filename in files:
actual_file_path = root + '/' + filename
zipped_file_path = prefix + filename
zipf.write( actual_file_path, zipped_file_path)
zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('/tmp/justtest/', zipf)
zipf.close()
现代Python(3.6+)使用pathlib模块对路径进行类似OOP的简洁处理,使用pathlib.Path.rglob()进行递归globing。据我所知,这相当于乔治·V·莱利的回答:压缩压缩,最顶层的元素是目录,保持空目录,使用相对路径。
from pathlib import Path
from zipfile import ZIP_DEFLATED, ZipFile
from os import PathLike
from typing import Union
def zip_dir(zip_name: str, source_dir: Union[str, PathLike]):
src_path = Path(source_dir).expanduser().resolve(strict=True)
with ZipFile(zip_name, 'w', ZIP_DEFLATED) as zf:
for file in src_path.rglob('*'):
zf.write(file, file.relative_to(src_path.parent))
注意:如可选类型提示所示,zip_name不能是Path对象(将在3.6.2+中修复)。
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