如何在Python中创建目录结构的zip存档?
当前回答
假设要压缩当前目录中的所有文件夹(子目录)。
for root, dirs, files in os.walk("."):
for sub_dir in dirs:
zip_you_want = sub_dir+".zip"
zip_process = zipfile.ZipFile(zip_you_want, "w", zipfile.ZIP_DEFLATED)
zip_process.write(file_you_want_to_include)
zip_process.close()
print("Successfully zipped directory: {sub_dir}".format(sub_dir=sub_dir))
其他回答
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)
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。
假设要压缩当前目录中的所有文件夹(子目录)。
for root, dirs, files in os.walk("."):
for sub_dir in dirs:
zip_you_want = sub_dir+".zip"
zip_process = zipfile.ZipFile(zip_you_want, "w", zipfile.ZIP_DEFLATED)
zip_process.write(file_you_want_to_include)
zip_process.close()
print("Successfully zipped directory: {sub_dir}".format(sub_dir=sub_dir))
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)
我还有另一个代码示例可能会有所帮助,使用python3、pathlib和zipfile。它应该可以在任何操作系统中工作。
from pathlib import Path
import zipfile
from datetime import datetime
DATE_FORMAT = '%y%m%d'
def date_str():
"""returns the today string year, month, day"""
return '{}'.format(datetime.now().strftime(DATE_FORMAT))
def zip_name(path):
"""returns the zip filename as string"""
cur_dir = Path(path).resolve()
parent_dir = cur_dir.parents[0]
zip_filename = '{}/{}_{}.zip'.format(parent_dir, cur_dir.name, date_str())
p_zip = Path(zip_filename)
n = 1
while p_zip.exists():
zip_filename = ('{}/{}_{}_{}.zip'.format(parent_dir, cur_dir.name,
date_str(), n))
p_zip = Path(zip_filename)
n += 1
return zip_filename
def all_files(path):
"""iterator returns all files and folders from path as absolute path string
"""
for child in Path(path).iterdir():
yield str(child)
if child.is_dir():
for grand_child in all_files(str(child)):
yield str(Path(grand_child))
def zip_dir(path):
"""generate a zip"""
zip_filename = zip_name(path)
zip_file = zipfile.ZipFile(zip_filename, 'w')
print('create:', zip_filename)
for file in all_files(path):
print('adding... ', file)
zip_file.write(file)
zip_file.close()
if __name__ == '__main__':
zip_dir('.')
print('end!')
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