在 Python 中解压文件
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Unzipping files in Python
提问by John Howard
I read through the zipfiledocumentation, but couldn't understand how to unzipa file, only how to zip a file. How do I unzip all the contents of a zip file into the same directory?
我通读了zipfile文档，但无法理解如何解压缩文件，只能理解如何压缩文件。如何将 zip 文件的所有内容解压缩到同一目录中？
采纳答案by Rahul
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!
差不多就是这样！
回答by Dan Breen
Use the extractallmethod, if you're using Python 2.6+
使用该extractall方法，如果您使用的是 Python 2.6+
zip = ZipFile('file.zip')
zip.extractall()
回答by user1741137
If you are using Python 3.2or later:
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")
You dont need to use the closeor try/catchwith this as it uses the context managerconstruction.
您不需要使用close或try/catch，因为它使用 上下文管理器构造。
回答by user3911901
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.
如果文件不是 zip，这不包含对文件的验证。如果文件夹包含非 .zip 文件，它将失败。
回答by simhumileco
You can also import only ZipFile:
您也可以只导入ZipFile：
from zipfile import ZipFile
zf = ZipFile('path_to_file/file.zip', 'r')
zf.extractall('path_to_extract_folder')
zf.close()
Works in Python 2and Python 3.
适用于Python 2和Python 3。
回答by Done Jin
try this :
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 : unzip file's path
path : 解压文件的路径
