如何在Python中创建目录结构的zip存档?
当前回答
函数创建zip文件。
def CREATEZIPFILE(zipname, path):
#function to create a zip file
#Parameters: zipname - name of the zip file; path - name of folder/file to be put in zip file
zipf = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
zipf.setpassword(b"password") #if you want to set password to zipfile
#checks if the path is file or directory
if os.path.isdir(path):
for files in os.listdir(path):
zipf.write(os.path.join(path, files), files)
elif os.path.isfile(path):
zipf.write(os.path.join(path), path)
zipf.close()
其他回答
最简单的方法是使用shutil.make_archive。它支持zip和tar格式。
import shutil
shutil.make_archive(output_filename, 'zip', dir_name)
如果您需要做一些比压缩整个目录更复杂的事情(例如跳过某些文件),那么您需要像其他人建议的那样深入到zipfile模块。
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!')
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。
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)
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)
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