如何在Python中创建目录结构的zip存档?


当前回答

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。

其他回答

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',以创建新的存档mode='w'(上面的默认值)。因此,假设您希望将3个不同的目录树捆绑在同一归档文件下。

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')

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)

好吧,在阅读了这些建议之后,我想出了一个非常类似的方法,它可以在2.7.x中工作,而不会创建“有趣”的目录名(绝对类似的名称),并且只会在zip中创建指定的文件夹。

或者只是为了以防万一,您需要在zip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" )

这里有这么多答案,我希望我可以贡献我自己的版本,它基于原始答案(顺便提一下),但具有更图形化的视角,也为每个zipfile设置使用上下文并对os.walk()进行排序,以便获得有序的输出。

有了这些文件夹和文件(以及其他文件夹),我想为每个cap_文件夹创建一个.zip:

$ 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

以下是我应用的内容,并附有评论,以更好地理解流程。

$ 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(路径)上的每一次迭代,我都会打开一个用于zipfile设置的上下文,然后对文件进行迭代,这是根目录中的文件列表,根据当前根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。

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

假设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))