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用Python解压缩文件

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问题：用Python解压缩文件
回答 0
回答 1
回答 2
回答 3
回答 4
回答 5

问题：用Python解压缩文件

我通读了zipfile文档，但不明白如何解压缩文件，只能解压缩文件。如何将zip文件的所有内容解压缩到同一目录中？

点击查看英文原文

I read through the zipfile documentation, but couldn’t understand how to unzip a file, only how to zip a file. How do I unzip all the contents of a zip file into the same directory?

回答 0

import zipfile
with zipfile.ZipFile(path_to_zip_file, 'r') as zip_ref:
    zip_ref.extractall(directory_to_extract_to)

差不多了！

点击查看英文原文 
import zipfile
with zipfile.ZipFile(path_to_zip_file, 'r') as zip_ref:
    zip_ref.extractall(directory_to_extract_to)

That’s pretty much it!

回答 1

如果您使用的是Python 3.2或更高版本：

import zipfile
with zipfile.ZipFile("file.zip","r") as zip_ref:
    zip_ref.extractall("targetdir")

您不需要使用close或try / catch，因为它使用了 上下文管理器构造。

点击查看英文原文

If you are using Python 3.2 or later:

import zipfile
with zipfile.ZipFile("file.zip","r") as zip_ref:
    zip_ref.extractall("targetdir")

You dont need to use the close or try/catch with this as it uses the context manager construction.

回答 2

extractall如果您使用的是Python 2.6+，请使用该方法

zip = ZipFile('file.zip')
zip.extractall()
点击查看英文原文

Use the extractall method, if you’re using Python 2.6+

zip = ZipFile('file.zip')
zip.extractall()

回答 3

您也只能导入ZipFile：

from zipfile import ZipFile
zf = ZipFile('path_to_file/file.zip', 'r')
zf.extractall('path_to_extract_folder')
zf.close()

适用于Python 2和Python 3。

点击查看英文原文

You can also import only ZipFile:

from zipfile import ZipFile
zf = ZipFile('path_to_file/file.zip', 'r')
zf.extractall('path_to_extract_folder')
zf.close()

Works in Python 2 and Python 3.

回答 4

尝试这个 ： 


import zipfile
def un_zipFiles(path):
    files=os.listdir(path)
    for file in files:
        if file.endswith('.zip'):
            filePath=path+'/'+file
            zip_file = zipfile.ZipFile(filePath)
            for names in zip_file.namelist():
                zip_file.extract(names,path)
            zip_file.close()

path：解压缩文件的路径

点击查看英文原文

try this : 


import zipfile
def un_zipFiles(path):
    files=os.listdir(path)
    for file in files:
        if file.endswith('.zip'):
            filePath=path+'/'+file
            zip_file = zipfile.ZipFile(filePath)
            for names in zip_file.namelist():
                zip_file.extract(names,path)
            zip_file.close()

path : unzip file’s path

回答 5

import os 
zip_file_path = "C:\AA\BB"
file_list = os.listdir(path)
abs_path = []
for a in file_list:
    x = zip_file_path+'\\'+a
    print x
    abs_path.append(x)
for f in abs_path:
    zip=zipfile.ZipFile(f)
    zip.extractall(zip_file_path)

如果文件不是zip，则不包含对该文件的验证。如果文件夹包含非.zip文件，它将失败。

点击查看英文原文 
import os 
zip_file_path = "C:\AA\BB"
file_list = os.listdir(path)
abs_path = []
for a in file_list:
    x = zip_file_path+'\\'+a
    print x
    abs_path.append(x)
for f in abs_path:
    zip=zipfile.ZipFile(f)
    zip.extractall(zip_file_path)

This does not contain validation for the file if its not zip. If the folder contains non .zip file it will fail.