blocks|key|2009047|text|import+zipfile
with+zipfile.ZipFile(path_to_zip_file,+'r')+as+zip_ref:
++++zip_ref.extractall(directory_to_extract_to)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2009048|差不多就是这样了！|unstyled|2009049|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>import zipfile
with zipfile.ZipFile(path_to_zip_file, 'r') as zip_ref:
 zip_ref.extractall(directory_to_extract_to)
</code></pre>

That's pretty much it!

blocks|key|2009148|text|如果您使用的是Python+3.2或更高版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2009149|import+zipfile
with+zipfile.ZipFile("file.zip","r")+as+zip_ref:
++++zip_ref.extractall("targetdir")|code-block|syntax|javascript|2009150|您不需要对它使用close或try/catch，因为它使用context+manager结构。|offset|length|2009151|entityMap|0|LINK|mutability|MUTABLE|url|http://eigenhombre.com/2013/04/20/introduction-to-context-managers/^0|0|0|T|F|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

If you are using Python 3.2 or later:

<pre><code>import zipfile
with zipfile.ZipFile("file.zip","r") as zip_ref:
 zip_ref.extractall("targetdir")
</code></pre>

You dont need to use the close or try/catch with this as it uses the
<a href="http://eigenhombre.com/2013/04/20/introduction-to-context-managers/" rel="noreferrer">context manager</a> construction.

blocks|key|1795930|text|zipfile是一个有点低级的库。除非您需要它所提供的细节，否则您可以使用shutil的高级函数make_archive和unpack_archive。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1795931|this+answer中已经介绍了make_archive。至于unpack_archive|1795932|import+shutil
shutil.unpack_archive(filename,+extract_dir)|code-block|syntax|javascript|1795933|unpack_archive从filename的“扩展名”(.zip、.tar.gz等)自动检测压缩格式，make_archive也是如此。此外，从Python3.7开始，filename和extract_dir可以是任何path-like对象(例如pathlib.Path实例)。|1795934|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/zipfile.html|1|https://docs.python.org/3/library/shutil.html|2|https://docs.python.org/3/library/shutil.html#shutil.make_archive|3|https://docs.python.org/3/library/shutil.html#shutil.unpack_archive|4|https://stackoverflow.com/a/25650295|5|https://www.python.org/dev/peps/pep-0519/|6|https://docs.python.org/3/library/pathlib.html^0|0|7|11|6|1C|C|1P|E|0|7|0|11|6|1|1C|C|2|1P|E|3|0|H|C|W|E|0|4|4|0|0|0|E|F|8|U|4|Z|7|1H|C|2E|8|2N|B|33|9|5|3H|C|6|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@$9|1H|A|1I|1|1J]|$9|1K|A|1L|1|1M]|$9|1N|A|1O|1|1P]|$9|1Q|A|1R|1|1S]]|E|$]]|$1|F|3|G|5|6|7|1T|8|@$9|1U|A|1V|B|C]|$9|1W|A|1X|B|C]]|D|@$9|1Y|A|1Z|1|20]]|E|$]]|$1|H|3|I|5|J|7|21|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|22|8|@$9|23|A|24|B|C]|$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|C]|$9|2B|A|2C|B|C]|$9|2D|A|2E|B|C]|$9|2F|A|2G|B|C]]|D|@$9|2H|A|2I|1|2J]|$9|2K|A|2L|1|2M]]|E|$]]|$1|O|3|-4|5|6|7|2N|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]|10|$5|R|S|T|E|$U|11]]|12|$5|R|S|T|E|$U|13]]|14|$5|R|S|T|E|$U|15]]|16|$5|R|S|T|E|$U|17]]]]

<a href="https://docs.python.org/3/library/zipfile.html" rel="noreferrer"><code>zipfile</code></a> is a somewhat low-level library. Unless you need the specifics that it provides, you can get away with <a href="https://docs.python.org/3/library/shutil.html" rel="noreferrer"><code>shutil</code></a>'s higher-level functions <a href="https://docs.python.org/3/library/shutil.html#shutil.make_archive" rel="noreferrer"><code>make_archive</code></a> and <a href="https://docs.python.org/3/library/shutil.html#shutil.unpack_archive" rel="noreferrer"><code>unpack_archive</code></a>.
<code>make_archive</code> is already described in <a href="https://stackoverflow.com/a/25650295">this</a> answer. As for <code>unpack_archive</code>:
<pre class="lang-py prettyprint-override"><code>import shutil
shutil.unpack_archive(filename, extract_dir)
</code></pre>
<code>unpack_archive</code> detects the compression format automatically from the &quot;extension&quot; of <code>filename</code> (<code>.zip</code>, <code>.tar.gz</code>, etc), and so does <code>make_archive</code>. Also, <code>filename</code> and <code>extract_dir</code> can be any <a href="https://www.python.org/dev/peps/pep-0519/" rel="noreferrer">path-like</a> objects (e.g. <a href="https://docs.python.org/3/library/pathlib.html" rel="noreferrer">pathlib.Path</a> instances) since Python 3.7.

blocks|key|2009106|text|如果您使用的是Python+2.6%2B，请使用extractall方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2009107|zip+=+ZipFile('file.zip')
zip.extractall()|code-block|syntax|javascript|2009108|entityMap^0|M|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Use the <code>extractall</code> method, if you're using Python 2.6+

<pre><code>zip = ZipFile('file.zip')
zip.extractall()
</code></pre>

blocks|key|1795681|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1795682|import+zipfile
def+un_zipFiles(path):
++++files=os.listdir(path)
++++for+file+in+files:
++++++++if+file.endswith('.zip'):
++++++++++++filePath=path%2B'/'%2Bfile
++++++++++++zip_file+=+zipfile.ZipFile(filePath)
++++++++++++for+names+in+zip_file.namelist():
++++++++++++++++zip_file.extract(names,path)
++++++++++++zip_file.close()+|code-block|syntax|javascript|1795683|​|1795684|path+:解压文件的路径|1795685|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

try this :
 <pre><code>
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() 
</code></pre>

path : unzip file's path

blocks|key|2009484|text|如果你想在shell中做这件事，而不是写代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2009485|+python3+-m+zipfile+-e+myfiles.zip+myfiles/|code-block|syntax|javascript|2009486|myfiles.zip是压缩归档文件，myfiles是解压文件的路径。|offset|length|style|CODE|2009487|entityMap^0|0|0|0|B|J|7|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

If you want to do it in shell, instead of writing code.
<pre><code> python3 -m zipfile -e myfiles.zip myfiles/
</code></pre>
<code>myfiles.zip</code> is the zip archive and <code>myfiles</code> is the path to extract the files.

blocks|key|1795732|text|from+zipfile+import+ZipFile
ZipFile("YOURZIP.zip").extractall("YOUR_DESTINATION_DIRECTORY")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1795733|您将解压文件的目录以前不需要存在，您可以在此时命名它|unstyled|1795734|如果您的项目位于同一目录中，则YOURZIP.zip是压缩文件的名称。如果没有，请使用路径，即:+C://....//YOURZIP.zip|1795735|考虑通过路径中的其他/来转义/如果您有一个permission+denied，尝试以管理员身份启动您的ide+(即:+Anaconda)|offset|length|style|CODE|1795736|将在与项目相同的目录中创建YOUR_DESTINATION_DIRECTORY|1795737|entityMap^0|0|0|0|A|1|E|1|L|H|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|V|8|@$K|W|L|X|M|N]|$K|Y|L|Z|M|N]|$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|F|7|13|8|@]|9|@]|A|$]]]|R|$]]

<pre><code>from zipfile import ZipFile
ZipFile(&quot;YOURZIP.zip&quot;).extractall(&quot;YOUR_DESTINATION_DIRECTORY&quot;)
</code></pre>
The directory where you will extract your files doesn't need to exist before, you name it at this moment
YOURZIP.zip is the name of the zip if your project is in the same directory.
If not, use the PATH i.e : C://....//YOURZIP.zip
Think to escape the <code>/</code> by an other <code>/</code> in the PATH
If you have a <code>permission denied</code> try to launch your ide (i.e: Anaconda) as administrator
YOUR_DESTINATION_DIRECTORY will be created in the same directory than your project

blocks|key|2009235|text|import+os+
zip_file_path+=+"C:\AA\BB"
file_list+=+os.listdir(path)
abs_path+=+[]
for+a+in+file_list:
++++x+=+zip_file_path%2B'\\'%2Ba
++++print+x
++++abs_path.append(x)
for+f+in+abs_path:
++++zip=zipfile.ZipFile(f)
++++zip.extractall(zip_file_path)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2009236|如果文件不是zip文件，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。|unstyled|2009237|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>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)
</code></pre>

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

I read through the <a href="https://docs.python.org/3/library/zipfile.html" rel="noreferrer"><code>zipfile</code> documentation</a>, 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?

Unzipping files in Python

Python

我通读了一遍，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文件中的所有内容解压缩到同一目录中？

问在Python中解压文件
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