如何在Python中创建目录结构的zip存档?


当前回答

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

第一个参数:结果zip/tar文件的文件名,第二个参数:zip/tar,第三个参数:目录名

代码:

import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')

其他回答

如果您想要一个类似于任何通用图形文件管理器的压缩文件夹的功能,可以使用以下代码,它使用zipfile模块。使用这段代码,您将得到以路径为根文件夹的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()

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))

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

第一个参数:结果zip/tar文件的文件名,第二个参数:zip/tar,第三个参数:目录名

代码:

import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')

我通过将Mark Byers的解决方案与Reimund和Morten Zilmer的评论(相对路径和包括空目录)合并,准备了一个函数。作为最佳实践,在ZipFile的文件构造中使用。

该函数还准备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)))

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。