如何在Python中创建目录结构的zip存档?


当前回答

下面是Nux给出的答案的一个变体,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 )

其他回答

使用python3.9、pathlib&zipfile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、打字,代码更少。

下面是Nux给出的答案的一个变体,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 )

前面的答案完全忽略了一点,即当您在Windows上运行代码时,使用os.path.join()可以很容易地返回POSIX不兼容的路径。当使用Linux上的任何常用归档软件处理文件时,生成的归档文件将包含名称中带有反斜杠的文件,这不是您想要的。请改用path.as_posix()作为arcname参数!

import zipfile
from pathlib import Path
with zipfile.ZipFile("archive.zip", "w", zipfile.ZIP_DEFLATED) as zf:
    for path in Path("include_all_of_this_folder").rglob("*"):
        zf.write(path, path.as_posix())

现代Python(3.6+)使用pathlib模块对路径进行类似OOP的简洁处理,使用pathlib.Path.rglob()进行递归globing。据我所知,这相当于乔治·V·莱利的回答:压缩压缩,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: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)。

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。