708

如何在 Python 中创建目录结构的 zip 存档?

4

27 回答 27

1748

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zipfile像其他人建议的那样深入研究模块。

于 2014-09-03T17:26:41.437 回答
658

正如其他人指出的那样,您应该使用zipfile。该文档告诉您哪些功能可用,但并没有真正解释如何使用它们来压缩整个目录。我认为用一些示例代码最容易解释:

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, '..')))
      
zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('tmp/', zipf)
zipf.close()
于 2009-12-06T11:23:55.163 回答
78

要将 的内容添加mydirectory到新的 zip 文件,包括所有文件和子目录:

import os
import zipfile

zf = zipfile.ZipFile("myzipfile.zip", "w")
for dirname, subdirs, files in os.walk("mydirectory"):
    zf.write(dirname)
    for filename in files:
        zf.write(os.path.join(dirname, filename))
zf.close()
于 2009-12-06T11:28:24.663 回答
58

如何在 Python 中创建目录结构的 zip 存档?

在 Python 脚本中

在 Python 2.7+ 中,shutil有一个make_archive函数。

from shutil import make_archive
make_archive(
  'zipfile_name', 
  'zip',           # the archive format - or tar, bztar, gztar 
  root_dir=None,   # root for archive - current working dir if None
  base_dir=None)   # start archiving from here - cwd if None too

此处压缩的存档将命名为zipfile_name.zip. 如果base_dirroot_dir它更远,它将排除不在的文件base_dir,但仍会将父目录中的文件存档到root_dir.

我在 Cygwin 上使用 2.7 进行测试时确实遇到了问题 - 它需要一个 root_dir 参数,用于 cwd:

make_archive('zipfile_name', 'zip', root_dir='.')

从 shell 中使用 Python

zipfile您也可以使用模块从 shell 中使用 Python 执行此操作:

$ python -m zipfile -c zipname sourcedir

zipname您想要的目标文件的名称在哪里(.zip如果需要,它不会自动添加)并且 sourcedir 是目录的路径。

压缩 Python(或者只是不想要父目录):

如果您尝试使用 and 压缩 python 包__init__.py__main__.py并且您不想要父目录,它是

$ python -m zipfile -c zipname sourcedir/*

$ python zipname

将运行包。(请注意,您不能将子包作为压缩存档的入口点运行。)

压缩 Python 应用程序:

如果您有 python3.5+,并且特别想压缩 Python 包,请使用zipapp

$ python -m zipapp myapp
$ python myapp.pyz
于 2016-03-31T18:57:20.383 回答
39

此函数将递归压缩目录树,压缩文件,并在存档中记录正确的相对文件名。归档条目与由 生成的相同zip -r output.zip source_dir

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)
于 2013-06-13T06:57:03.097 回答
24

使用shutil,它是python标准库集的一部分。使用 shutil 非常简单(见下面的代码):

  • 第一个参数:生成的 zip/tar 文件的文件名,
  • 第二个参数:zip/tar,
  • 第三个参数:dir_name

代码:

import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')
于 2018-10-12T09:46:46.227 回答
19

现代 Python (3.6+) 使用该pathlib模块进行简洁的类似 OOP 的路径处理以及pathlib.Path.rglob()递归通配符。据我所知,这相当于 George V. Reilly 的回答:压缩压缩,最顶层元素是目录,保留空目录,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)。

于 2017-03-31T13:01:51.090 回答
12

要向生成的 zip 文件添加压缩,请查看此链接

你需要改变:

zip = zipfile.ZipFile('Python.zip', 'w')

zip = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
于 2013-04-27T17:14:02.897 回答
8

使用 python 3.9pathlibzipfile模块,您可以从系统中的任何位置创建 zip 文件。

def zip_dir(dir: Union[Path, str], filename: Union[Path, str]):
    """Zip the provided directory without navigating to that directory using `pathlib` module"""

    # Convert to Path object
    dir = Path(dir)

    with zipfile.ZipFile(filename, "w", zipfile.ZIP_DEFLATED) as zip_file:
        for entry in dir.rglob("*"):
            zip_file.write(entry, entry.relative_to(dir))

它整洁、有类型,并且代码更少。

于 2021-08-17T12:05:41.167 回答
5

我对Mark Byers 给出的代码进行了一些更改。如果您有空目录,下面的函数还将添加空目录。示例应该更清楚地说明添加到 zip 的路径是什么。

#!/usr/bin/env python
import os
import zipfile

def addDirToZip(zipHandle, path, basePath=""):
    """
    Adding directory given by \a path to opened zip file \a zipHandle

    @param basePath path that will be removed from \a path when adding to archive

    Examples:
        # add whole "dir" to "test.zip" (when you open "test.zip" you will see only "dir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir')
        zipHandle.close()

        # add contents of "dir" to "test.zip" (when you open "test.zip" you will see only it's contents)
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir', 'dir')
        zipHandle.close()

        # add contents of "dir/subdir" to "test.zip" (when you open "test.zip" you will see only contents of "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir', 'dir/subdir')
        zipHandle.close()

        # add whole "dir/subdir" to "test.zip" (when you open "test.zip" you will see only "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir', 'dir')
        zipHandle.close()

        # add whole "dir/subdir" with full path to "test.zip" (when you open "test.zip" you will see only "dir" and inside it only "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir')
        zipHandle.close()

        # add whole "dir" and "otherDir" (with full path) to "test.zip" (when you open "test.zip" you will see only "dir" and "otherDir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir')
        addDirToZip(zipHandle, 'otherDir')
        zipHandle.close()
    """
    basePath = basePath.rstrip("\\/") + ""
    basePath = basePath.rstrip("\\/")
    for root, dirs, files in os.walk(path):
        # add dir itself (needed for empty dirs
        zipHandle.write(os.path.join(root, "."))
        # add files
        for file in files:
            filePath = os.path.join(root, file)
            inZipPath = filePath.replace(basePath, "", 1).lstrip("\\/")
            #print filePath + " , " + inZipPath
            zipHandle.write(filePath, inZipPath)

上面是一个简单的函数,应该适用于简单的情况。您可以在我的 Gist 中找到更优雅的课程: https ://gist.github.com/Eccenux/17526123107ca0ac28e6

于 2013-06-10T09:28:26.360 回答
3

我有另一个可能有帮助的代码示例,使用 python3、pathlib 和 zipfile。它应该适用于任何操作系统。

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!')
于 2015-10-02T10:03:14.890 回答
3

对于保留要归档的父目录下的文件夹层次结构的简洁方法:

import glob
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,您可以将其更改为pathlib 用于文件通配符

于 2019-07-09T08:48:30.300 回答
2

如果您想要任何通用图形文件管理器的压缩文件夹等功能,您可以使用以下代码,它使用zipfile模块。使用此代码,您将拥有以路径作为根文件夹的 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()
于 2016-03-21T13:40:10.027 回答
2

这里有很多答案,我希望我可以用我自己的版本做出贡献,该版本基于原始答案(顺便说一下),但具有更图形化的视角,也为每个zipfile设置和排序使用上下文os.walk(),以便有一个有序输出。

有了这些文件夹和它们的文件(在其他文件夹中),我想.zip为每个cap_文件夹创建一个:

$ tree -d
.
├── cap_01
|    ├── 0101000001.json
|    ├── 0101000002.json
|    ├── 0101000003.json
|
├── cap_02
|    ├── 0201000001.json
|    ├── 0201000002.json
|    ├── 0201001003.json
|
├── cap_03
|    ├── 0301000001.json
|    ├── 0301000002.json
|    ├── 0301000003.json
| 
├── docs
|    ├── map.txt
|    ├── main_data.xml
|
├── core_files
     ├── core_master
     ├── core_slave

这是我应用的内容,并附有评论以更好地理解该过程。

$ cat zip_cap_dirs.py 
""" Zip 'cap_*' directories. """           
import os                                                                       
import zipfile as zf                                                            


for root, dirs, files in sorted(os.walk('.')):                                                                                               
    if 'cap_' in root:                                                          
        print(f"Compressing: {root}")                                           
        # Defining .zip name, according to Capítulo.                            
        cap_dir_zip = '{}.zip'.format(root)                                     
        # Opening zipfile context for current root dir.                         
        with zf.ZipFile(cap_dir_zip, 'w', zf.ZIP_DEFLATED) as new_zip:          
            # Iterating over os.walk list of files for the current root dir.    
            for f in files:                                                     
                # Defining relative path to files from current root dir.        
                f_path = os.path.join(root, f)                                  
                # Writing the file on the .zip file of the context              
                new_zip.write(f_path) 

基本上,对于每次迭代os.walk(path),我打开一个上下文进行zipfile设置,然后迭代迭代files,这是目录中的一个list文件root,根据当前root目录形成每个文件的相对路径,附加到zipfile正在运行的上下文.

输出如下所示:

$ python3 zip_cap_dirs.py
Compressing: ./cap_01
Compressing: ./cap_02
Compressing: ./cap_03

要查看每个.zip目录的内容,可以使用less命令:

$ less cap_01.zip

Archive:  cap_01.zip
 Length   Method    Size  Cmpr    Date    Time   CRC-32   Name
--------  ------  ------- ---- ---------- ----- --------  ----
  22017  Defl:N     2471  89% 2019-09-05 08:05 7a3b5ec6  cap_01/0101000001.json
  21998  Defl:N     2471  89% 2019-09-05 08:05 155bece7  cap_01/0101000002.json
  23236  Defl:N     2573  89% 2019-09-05 08:05 55fced20  cap_01/0101000003.json
--------          ------- ---                           -------
  67251             7515  89%                            3 files
于 2019-09-05T13:10:01.160 回答
1

您可能想查看zipfile模块;在http://docs.python.org/library/zipfile.html有文档。

您可能还想os.walk()索引目录结构。

于 2009-12-06T11:17:50.930 回答
1

以下是对我有用的 Nux 给出的答案的变体:

def WriteDirectoryToZipFile( zipHandle, srcPath, zipLocalPath = "", zipOperation = zipfile.ZIP_DEFLATED ):
    basePath = os.path.split( srcPath )[ 0 ]
    for root, dirs, files in os.walk( srcPath ):
        p = os.path.join( zipLocalPath, root [ ( len( basePath ) + 1 ) : ] )
        # add dir
        zipHandle.write( root, p, zipOperation )
        # add files
        for f in files:
            filePath = os.path.join( root, f )
            fileInZipPath = os.path.join( p, f )
            zipHandle.write( filePath, fileInZipPath, zipOperation )
于 2014-07-30T23:13:20.930 回答
1

试试下面的。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()
于 2015-04-23T11:18:16.770 回答
1

为了提供更大的灵活性,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,您可以仅选择dir4root.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)

或者只是listdir在脚本调用目录中并从那里添加所有内容:

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)
于 2018-09-25T13:32:32.100 回答
1

假设您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))
于 2019-04-09T08:25:30.457 回答
1

压缩文件或树(目录及其子目录)。

from pathlib import Path
from zipfile import ZipFile, ZIP_DEFLATED

def make_zip(tree_path, zip_path, mode='w', skip_empty_dir=False):
    with ZipFile(zip_path, mode=mode, compression=ZIP_DEFLATED) as zf:
        paths = [Path(tree_path)]
        while paths:
            p = paths.pop()
            if p.is_dir():
                paths.extend(p.iterdir())
                if skip_empty_dir:
                    continue
            zf.write(p)

要附加到现有存档,请传递mode='a', 以创建新存档mode='w'(上面的默认值)。因此,假设您想在同一个存档下捆绑 3 个不同的目录树。

make_zip(path_to_tree1, path_to_arch, mode='w')
make_zip(path_to_tree2, path_to_arch, mode='a')
make_zip(path_to_file3, path_to_arch, mode='a')
于 2020-09-17T05:50:27.767 回答
1

使用 的解决方案pathlib.Path,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')
于 2021-01-05T09:09:17.613 回答
1

显而易见的方法是使用shutil,就像第二个最佳答案所说的那样,但是如果你出于某种原因仍然希望使用ZipFile,并且如果你在这样做时遇到了一些麻烦(比如Windows中的ERR 13等) , 您可以使用此修复程序:

import os
import zipfile

def retrieve_file_paths(dirName):
  filePaths = []
  for root, directories, files in os.walk(dirName):
    for filename in files:
        filePath = os.path.join(root, filename)
        filePaths.append(filePath)
  return filePaths
 
def main(dir_name, output_filename):
  filePaths = retrieve_file_paths(dir_name)
   
  zip_file = zipfile.ZipFile(output_filename+'.zip', 'w')
  with zip_file:
    for file in filePaths:
      zip_file.write(file)

main("my_dir", "my_dir_archived")

这个递归遍历给定文件夹中的每个子文件夹/文件,并将它们写入 zip 文件,而不是尝试直接压缩文件夹。

于 2021-03-10T05:09:11.803 回答
0

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 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))
于 2016-12-14T21:50:12.490 回答
0

我通过将 Mark Byers 的解决方案与 Reimund 和 Morten Zilmer 的评论(相对路径并包括空目录)合并来准备一个函数。作为最佳实践,with在 ZipFile 的文件构造中使用。

该函数还准备了一个带有压缩目录名和“.zip”扩展名的默认 zip 文件名。因此,它只使用一个参数:要压缩的源目录。

import os
import zipfile

def zip_dir(path_dir, path_file_zip=''):
if not path_file_zip:
    path_file_zip = os.path.join(
        os.path.dirname(path_dir), os.path.basename(path_dir)+'.zip')
with zipfile.ZipFile(path_file_zip, 'wb', zipfile.ZIP_DEFLATED) as zip_file:
    for root, dirs, files in os.walk(path_dir):
        for file_or_dir in files + dirs:
            zip_file.write(
                os.path.join(root, file_or_dir),
                os.path.relpath(os.path.join(root, file_or_dir),
                                os.path.join(path_dir, os.path.pardir)))
于 2016-12-24T20:32:56.153 回答
0
# 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))
于 2017-06-15T06:24:08.640 回答
0

好吧,在阅读了这些建议之后,我想出了一种非常相似的方法,它适用于 2.7.x,而无需创建“有趣”的目录名称(绝对类似的名称),并且只会在 zip 中创建指定的文件夹。

或者以防万一你需要你的 zip 来包含一个包含所选目录内容的文件夹。

def zipDir( path, ziph ) :
 """
 Inserts directory (path) into zipfile instance (ziph)
 """
 for root, dirs, files in os.walk( path ) :
  for file in files :
   ziph.write( os.path.join( root, file ) , os.path.basename( os.path.normpath( path ) ) + "\\" + file )

def makeZip( pathToFolder ) :
 """
 Creates a zip file with the specified folder
 """
 zipf = zipfile.ZipFile( pathToFolder + 'file.zip', 'w', zipfile.ZIP_DEFLATED )
 zipDir( pathToFolder, zipf )
 zipf.close()
 print( "Zip file saved to: " + pathToFolder)

makeZip( "c:\\path\\to\\folder\\to\\insert\\into\\zipfile" )
于 2018-08-15T21:27:48.643 回答
0

创建 zip 文件的功能。

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()
于 2018-11-19T03:55:37.670 回答