blocks|key|1017365|text|据我所知你已经很接近了。您可以使用dirname和basename来确保您正在获取正确的路径名：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1017366|>>>+os.path.dirname("/path/to/dst")
'/path/to'
>>>+os.path.basename("/path/to/dst")
'dst'|code-block|syntax|javascript|1017367|然后，使用chdir，您可以确保您在父级中，因此路径是相对的。|1017368|def+zip(src,+dst):
++++parent+=+os.path.dirname(dst)
++++folder+=+os.path.basename(dst)

++++os.chdir(parent):
++++for+dirname,+subdirs,+filenames+in+os.walk(folder):
++++++++...|1017369|这就产生了：|1017370|dst/a.txt
dst/b
dst/b/c.txt
...etc...|1017371|如果不想包含名称"dst“，可以只执行os.chdir(dst)，然后执行os.walk('.')。|1017372|希望这能有所帮助。|1017373|entityMap^0|H|7|P|8|0|0|5|5|0|0|0|0|J|D|11|C|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|1F|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|X|$]]

From what I can tell you are close. You could use <code>dirname</code> and <code>basename</code> to make sure you are grabbing the right path name:

<pre><code>&gt;&gt;&gt; os.path.dirname("/path/to/dst")
'/path/to'
&gt;&gt;&gt; os.path.basename("/path/to/dst")
'dst'
</code></pre>

Then using <code>chdir</code> you can make sure you are in the parent so the paths are relative.

<pre><code>def zip(src, dst):
 parent = os.path.dirname(dst)
 folder = os.path.basename(dst)

 os.chdir(parent):
 for dirname, subdirs, filenames in os.walk(folder):
 ...
</code></pre>

This creates:

<pre><code>dst/a.txt
dst/b
dst/b/c.txt
...etc...
</code></pre>

If do not want to include the name "dst" you can just do <code>os.chdir(dst)</code> and then <code>os.walk('.')</code>.

Hope that helps.

blocks|key|1219598|text|zipfile.write()方法采用一个可选的arcname参数，该参数指定The文件中文件的名称。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1219599|您可以使用它在开始时去掉通往src的路径。在这里，我使用os.path.abspath()确保src和os.walk()返回的文件名都有一个共同的前缀。|1219600|#!/usr/bin/env+python2.7

import+os
import+zipfile

def+zip(src,+dst):
++++zf+=+zipfile.ZipFile("%25s.zip"+%25+(dst),+"w",+zipfile.ZIP_DEFLATED)
++++abs_src+=+os.path.abspath(src)
++++for+dirname,+subdirs,+files+in+os.walk(src):
++++++++for+filename+in+files:
++++++++++++absname+=+os.path.abspath(os.path.join(dirname,+filename))
++++++++++++arcname+=+absname[len(abs_src)+%2B+1:]
++++++++++++print+'zipping+%25s+as+%25s'+%25+(os.path.join(dirname,+filename),
++++++++++++++++++++++++++++++++++++++++arcname)
++++++++++++zf.write(absname,+arcname)
++++zf.close()

zip("src",+"dst")|code-block|syntax|javascript|1219601|具有如下目录结构：|1219602|src
└──+a
++++├──+b
++++│  +└──+bar
++++└──+foo|1219603|脚本打印：|1219604|zipping+src/a/foo+as+a/foo
zipping+src/a/b/bar+as+a/b/bar|1219605|生成的zip文件的内容如下：|1219606|Archive:++dst.zip
++Length+++++Date+++Time++++Name
+--------++++----+++----++++----
++++++++0++01-28-13+11:36+++a/foo
++++++++0++01-28-13+11:36+++a/b/bar
+--------+++++++++++++++++++-------
++++++++0+++++++++++++++++++2+files|1219607|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/zipfile#zipfile.ZipFile.write|1|http://docs.python.org/2/library/os.path.html#os.path.abspath^0|0|F|O|7|0|F|0|0|E|3|S|H|1B|3|1F|9|S|H|1|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|F|3|G|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]]|D|@$9|1P|A|1Q|1|1R]]|E|$]]|$1|H|3|I|5|J|7|1S|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1T|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1U|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1V|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1W|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1X|8|@]|D|@]|E|$]]|$1|W|3|X|5|J|7|1Y|8|@]|D|@]|E|$K|L]]|$1|Y|3|-4|5|6|7|1Z|8|@]|D|@]|E|$]]]|Z|$10|$5|11|12|13|E|$14|15]]|16|$5|11|12|13|E|$14|17]]]]

The <a href="http://docs.python.org/2/library/zipfile#zipfile.ZipFile.write"><code>zipfile.write()</code></a> method takes an optional <code>arcname</code> argument
that specifies what the name of the file should be inside the zipfile.

You can use this to strip off the path to <code>src</code> at the beginning. Here I
use <a href="http://docs.python.org/2/library/os.path.html#os.path.abspath"><code>os.path.abspath()</code></a> to make sure that both <code>src</code> and the
filename returned by <code>os.walk()</code> have a common prefix.

<pre><code>#!/usr/bin/env python2.7

import os
import zipfile

def zip(src, dst):
 zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED)
 abs_src = os.path.abspath(src)
 for dirname, subdirs, files in os.walk(src):
 for filename in files:
 absname = os.path.abspath(os.path.join(dirname, filename))
 arcname = absname[len(abs_src) + 1:]
 print 'zipping %s as %s' % (os.path.join(dirname, filename),
 arcname)
 zf.write(absname, arcname)
 zf.close()

zip("src", "dst")
</code></pre>

With a directory structure like this:

<pre><code>src
└── a
 ├── b
 │   └── bar
 └── foo
</code></pre>

The script prints:

<pre><code>zipping src/a/foo as a/foo
zipping src/a/b/bar as a/b/bar
</code></pre>

And the contents of the resulting zip file are:

<pre><code>Archive: dst.zip
 Length Date Time Name
 -------- ---- ---- ----
 0 01-28-13 11:36 a/foo
 0 01-28-13 11:36 a/b/bar
 -------- -------
 0 2 files
</code></pre>

I'm trying to create a function in my script that zips the contents of a given source directory (<code>src</code>) to a zip file (<code>dst</code>). For example, <code>zip('/path/to/dir', '/path/to/file.zip')</code>, where <code>/path/to/dir</code> is a directory, and <code>/path/to/file.zip</code> doesn't exist yet. I do not want to zip the directory itself, this makes all the difference in my case. I want to zip the files (and subdirs) in the directory. This is what I'm trying:

<pre><code>def zip(src, dst):
 zf = zipfile.ZipFile("%s.zip" % (dst), "w")
 for dirname, subdirs, files in os.walk(src):
 zf.write(dirname)
 for filename in files:
 zf.write(os.path.join(dirname, filename))
 zf.close()
</code></pre>

This creates a zip that is essentially <code>/</code>. For example, if I zipped <code>/path/to/dir</code>, extracting the zip creates a directory with "path" in it, with "to" in that directory, etc.

Does anyone have a function that doesn't cause this problem?

I can't stress this enough, it needs to zip the files in the directory, not the directoy itself.

Create .zip in Python?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我试图在脚本中创建一个函数，将给定源目录(src)的内容压缩到zip文件(dst)。例如，zip('/path/to/dir', '/path/to/file.zip')，其中/path/to/dir是一个目录，而/path/to/file.zip还不存在。我不想压缩目录本身，这在我的情况下是完全不同的。我希望压缩目录...

问用Python创建.zip？
EN

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问用Python创建.zip？EN