如何在Python中创建目录结构的zip存档?
当前回答
正如其他人所指出的,您应该使用zipfile。文档告诉哪些函数可用,但并没有真正解释如何使用它们压缩整个目录。我认为用一些示例代码来解释是最简单的:
import os
import zipfile
def zipdir(path, ziph):
# ziph is zipfile handle
for root, dirs, files in os.walk(path):
for file in files:
ziph.write(os.path.join(root, file),
os.path.relpath(os.path.join(root, file),
os.path.join(path, '..')))
with zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED) as zipf:
zipdir('tmp/', zipf)
其他回答
试试下面的一个。它对我有用。
import zipfile, os
zipf = "compress.zip"
def main():
directory = r"Filepath"
toZip(directory)
def toZip(directory):
zippedHelp = zipfile.ZipFile(zipf, "w", compression=zipfile.ZIP_DEFLATED )
list = os.listdir(directory)
for file_list in list:
file_name = os.path.join(directory,file_list)
if os.path.isfile(file_name):
print file_name
zippedHelp.write(file_name)
else:
addFolderToZip(zippedHelp,file_list,directory)
print "---------------Directory Found-----------------------"
zippedHelp.close()
def addFolderToZip(zippedHelp,folder,directory):
path=os.path.join(directory,folder)
print path
file_list=os.listdir(path)
for file_name in file_list:
file_path=os.path.join(path,file_name)
if os.path.isfile(file_path):
zippedHelp.write(file_path)
elif os.path.isdir(file_name):
print "------------------sub directory found--------------------"
addFolderToZip(zippedHelp,file_name,path)
if __name__=="__main__":
main()
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))
要提供更大的灵活性,例如按名称选择目录/文件,请使用:
import os
import zipfile
def zipall(ob, path, rel=""):
basename = os.path.basename(path)
if os.path.isdir(path):
if rel == "":
rel = basename
ob.write(path, os.path.join(rel))
for root, dirs, files in os.walk(path):
for d in dirs:
zipall(ob, os.path.join(root, d), os.path.join(rel, d))
for f in files:
ob.write(os.path.join(root, f), os.path.join(rel, f))
break
elif os.path.isfile(path):
ob.write(path, os.path.join(rel, basename))
else:
pass
对于文件树:
.
├── dir
│ ├── dir2
│ │ └── file2.txt
│ ├── dir3
│ │ └── file3.txt
│ └── file.txt
├── dir4
│ ├── dir5
│ └── file4.txt
├── listdir.zip
├── main.py
├── root.txt
└── selective.zip
例如,您可以只选择dir4和root.txt:
cwd = os.getcwd()
files = [os.path.join(cwd, f) for f in ['dir4', 'root.txt']]
with zipfile.ZipFile("selective.zip", "w" ) as myzip:
for f in files:
zipall(myzip, f)
或者只需在脚本调用目录中列出目录,然后从中添加所有内容:
with zipfile.ZipFile("listdir.zip", "w" ) as myzip:
for f in os.listdir():
if f == "listdir.zip":
# Creating a listdir.zip in the same directory
# will include listdir.zip inside itself, beware of this
continue
zipall(myzip, f)
使用pathlib.Path的解决方案,它独立于所使用的操作系统:
import zipfile
from pathlib import Path
def zip_dir(path: Path, zip_file_path: Path):
"""Zip all contents of path to zip_file"""
files_to_zip = [
file for file in path.glob('*') if file.is_file()]
with zipfile.ZipFile(
zip_file_path, 'w', zipfile.ZIP_DEFLATED) as zip_f:
for file in files_to_zip:
print(file.name)
zip_f.write(file, file.name)
current_dir = Path.cwd()
zip_dir = current_dir / "test"
tools.zip_dir(
zip_dir, current_dir / 'Zipped_dir.zip')
# import required python modules
# You have to install zipfile package using pip install
import os,zipfile
# Change the directory where you want your new zip file to be
os.chdir('Type your destination')
# Create a new zipfile ( I called it myfile )
zf = zipfile.ZipFile('myfile.zip','w')
# os.walk gives a directory tree. Access the files using a for loop
for dirnames,folders,files in os.walk('Type your directory'):
zf.write('Type your Directory')
for file in files:
zf.write(os.path.join('Type your directory',file))