如何创建目录的zip存档

时间:2009-12-06 11:12:54

标签: python zipfile

如何在Python中创建目录结构的zip存档?

27 个答案:

答案 0 :(得分:689)

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)

如果你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),那么你需要像其他人建议的那样深入研究zipfile模块。

答案 1 :(得分:426)

正如其他人所指出的那样,你应该使用zipfile。文档告诉您哪些功能可用,但没有真正解释如何使用它们来压缩整个目录。我认为最简单的解释一些示例代码:

#!/usr/bin/env python
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))

if __name__ == '__main__':
    zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
    zipdir('tmp/', zipf)
    zipf.close()

改编自:http://www.devshed.com/c/a/Python/Python-UnZipped/

答案 2 :(得分:56)

mydirectory的内容添加到新的zip文件中,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()

答案 3 :(得分:44)

  

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具有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_dir距离root_dir更远,则会排除base_dir以外的文件,但仍将父目录中的文件归档到root_dir

我确实在使用2.7测试Cygwin上有一个问题 - 它想要一个root_dir参数,对于cwd:

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

从shell使用Python

您也可以使用zipfile模块从shell执行此操作:

$ python -m zipfile -c zipname sourcedir

其中zipname是您想要的目标文件的名称(如果需要,添加.zip,它将不会自动执行),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__init__.py__main__.py压缩python包,并且您不想要父目录,那么

$ python -m zipfile -c zipname sourcedir/*

$ python zipname

会运行包。 (请注意,您无法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 +,并且特别想要压缩Python包,请使用zipapp

$ python -m zipapp myapp
$ python myapp.pyz

答案 4 :(得分:29)

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zip -r output.zip source_dir生成的归档条目相同。

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)

答案 5 :(得分:15)

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

  • 第一个参数:结果zip / tar文件的文件名
  • 第二个arg: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')

答案 6 :(得分:11)

要将压缩添加到生成的zip文件中,请查看this link

您需要更改:

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

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

答案 7 :(得分:5)

我对code given by Mark Byers做了一些更改。如果你有它们,下面的函数也会添加空目录。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。你可以在我的Gist中找到更优雅的课程: https://gist.github.com/Eccenux/17526123107ca0ac28e6

答案 8 :(得分:2)

我有另一个可能有用的代码示例,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!')

答案 9 :(得分:2)

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

您可能还希望os.walk()索引目录结构。

答案 10 :(得分:1)

如果您想要任何常见图形文件管理器的压缩文件夹之类的功能,您可以使用以下代码,它使用zipfile模块。使用此代码,您将获得zip文件,其路径为根文件夹。

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()

答案 11 :(得分:1)

这里有这么多答案,我希望我可以贡献自己的版本,该版本基于原始答案(顺便说一句),但具有更多的图形化视角,也为每个zipfile设置都使用了上下文并排序os.walk(),以使输出有序。

具有这些文件夹及其文件(以及其他文件夹),我想为每个.zip文件夹创建一个cap_

$ 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

这就是我应用的内容,并带有注释以更好地了解该过程。

$ 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os.walk(path)上的每次迭代,我都会打开一个用于zipfile设置的上下文,然后再迭代files,这是一个list文件从root目录开始,根据当前root目录形成每个文件的相对路径,并附加到正在运行的zipfile上下文中。

输出显示如下:

$ 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

答案 12 :(得分:1)

提供更大的灵活性,例如通过名称选择目录/文件使用:

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

您可以例如仅选择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)

或者只是在脚本调用目录中listdir,然后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)

答案 13 :(得分: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)

要追加到现有归档文件,请传递mode='a',以创建新的归档文件mode='w'(上面的默认设置)。假设您要将3个不同的目录树捆绑在同一个档案中。

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')

答案 14 :(得分:1)

现代Python(3.6+)使用pathlib模块进行简洁的OOP处理路径,使用pathlib.Path.rglob()进行递归通配。据我所知,这相当于George V. Reilly的答案:压缩拉链,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:正如可选类型提示所示,zip_name不能是Path对象(would be fixed in 3.6.2+)。

答案 15 :(得分:1)

使用 pathlib.Path 的解决方案,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')

答案 16 :(得分:1)

尝试下面的一个。这对我有用

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()

答案 17 :(得分:1)

以下是Nux给出的答案的变体,对我有用:

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 )

答案 18 :(得分:0)

对于其他研究此问题并尝试将其程序存储在同一目录中并且由于zip文件本身具有压缩性而变得非常深的树形结构并最终以递归的人,请尝试此操作。

这是Mark's answer和一些额外检查的结合,以确保zipfile本身没有递归zip压缩,并且没有不必要的深层文件夹结构。

import os
import zipfile

def zipdir(path, ziph, ignored_directories, ignored_files):
    # ziph is zipfile handle
    for root, dirs, files in os.walk(path):
      for file in files:
        if not any(ignored_dir in root for ignored_dir in ignored_directories):
          if not any(ignored_fname in file for ignored_fname in ignored_files):
            ziph.write(os.path.join(root, file))

# current working directory
this_dir = os.path.dirname(os.path.abspath(__file__))

# the directory within the working directory the zip will be created in (build/archives).
zip_dest_dir = os.path.join('build', 'archives')

# verify zip_dest_dir exists: if not, create it
if not os.path.isdir(zip_dest_dir):
    os.makedirs(zip_dest_dir, exist_ok=True)

# leave zip_dest_dir blank (or set dist_dir = this_dir) if you want the zip file in the working directory (same directory as the script)
dest_dir = os.path.join(this_dir, zip_dest_dir)

# name the zip file: remember the file extension
zip_filename = 'zipped_directory.zip'

# zip file's path
zip_path = os.path.join(dest_dir, zip_filename)

# create the zipfile handle: you can change ZIP_STORED to any other compression algorithm of your choice, like ZIP_DEFLATED, if you need actual compression
zipf = zipfile.ZipFile(zip_path, 'w', zipfile.ZIP_STORED)

# ignored files and directories: I personally wanted to ignore the "build" directory, alongside with "node_modules", so those would be listed here.
ignored_dirs = []

# ignore any specific files: in my case, I was ignoring the script itself, so I'd include 'deploy.py' here
ignored_files = [zip_filename]

# zip directory contents
zipdir('.', zipf, ignored_dirs, ignored_files)
zipf.close()

生成的zip文件应仅包含从工作目录开始的目录:因此,不能包含Users / user / Desktop / code /.../ working_directory /.../ etc。一种文件结构。

答案 19 :(得分:0)

显而易见的方法是使用shutil,就像第二个顶级答案所说的那样,但是如果您出于某种原因仍然希望使用ZipFile,并且如果您在这样做时遇到一些麻烦(例如ERR 13 in Windows 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,并将它们写入一个 zip 文件,而不是尝试直接压缩文件夹。

答案 20 :(得分:0)

以简洁的方式将文件夹层次结构保留在要归档的父目录下:

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,可以将其更改为使用pathlib for file globbing

答案 21 :(得分:0)

说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))

答案 22 :(得分:0)

用于创建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()

答案 23 :(得分:0)

好吧,在阅读了建议之后,我想到了一种与2.7.x十分相似的方式,而不创建“有趣的”目录名称(类似绝对的名称),并且只会在zip内创建指定的文件夹。

或者,以防万一,您需要在zip文件中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" )

答案 24 :(得分: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))

答案 25 :(得分:0)

我通过整合Mark Byers'来准备一个功能。 Reimund和Morten Zilmer的评论(相对路径和包括空目录)的解决方案。作为最佳实践,with用于ZipFile的文件构建。

该功能还准备了一个默认的zip文件名,其中包含压缩的目录名和' .zip'延期。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)))

答案 26 :(得分:0)

这是一种使用pathlib和上下文管理器的现代方法。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))