Python zip 文件并避免目录结构
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zip file and avoid directory structure
提问by user1189851
I have a Python script that zips a file (new.txt).
我有一个压缩文件 ( new.txt)的 Python 脚本。
tofile = "/root/files/result/"+file
targetzipfile = new.zip # This is how I want my zip to look like
zf = zipfile.ZipFile(targetzipfile, mode='w')
try:
#adding to archive
zf.write(tofile)
finally:
zf.close()
When I do this I get the zip file. But when I try to unzip the file I get the text file inside of a series of directories corresponding to the path of the file i.e I see a folder called rootin the resultdirectory and more directories within it, i.e. I have
当我这样做时，我得到了 zip 文件。但是，当我尝试解压缩文件，我得到了一系列对应于文件的路径，即我看到一个名为文件夹目录的文本文件，里面root的result内它的目录和多个目录，即我有
/root/files/result/new.zip
/root/files/result/new.zip
and when I unzip new.zipI have a directory structure that looks like
当我解压缩时，new.zip我的目录结构看起来像
/root/files/result/root/files/result/new.txt.
Is there a way I can zip such that when I unzip I only get new.txt.
有没有一种方法可以压缩，这样当我解压缩时，我只能得到new.txt.
In other words I have /root/files/result/new.zipand when I unzip new.zip, it should look like
换句话说，我有/root/files/result/new.zip，当我解压缩时new.zip，它应该看起来像
/root/files/results/new.txt
采纳答案by user 12321
The zipfile.write()method takes an optional arcnameargument that specifies what the name of the file should be inside the zipfile
该zipfile.write()方法采用一个可选arcname参数，该参数指定 zipfile 中文件的名称
I think you need to do a modification for the destination, otherwise it will duplicate the directory. Use :arcnameto avoid it. try like this:
我认为您需要对目的地进行修改，否则会重复目录。使用 :arcname来避免它。试试这样：
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")
回答by aquil.abdullah
Check out the documentation for Zipfile.write.
查看Zipfile.write的文档。
ZipFile.write(filename[, arcname[, compress_type]]) Write the file named filename to the archive, giving it the archive name arcname (by default, this will be the same as filename, but without a drive letter and with leading path separators removed)
ZipFile.write(filename[, arcname[, compress_type]]) 将名为 filename 的文件写入存档，为其指定存档名称 arcname（默认情况下，这将与 filename 相同，但没有驱动器号和前导路径删除了分隔符）
https://docs.python.org/2/library/zipfile.html#zipfile.ZipFile.write
https://docs.python.org/2/library/zipfile.html#zipfile.ZipFile.write
Try the following:
请尝试以下操作：
import zipfile
import os
filename = 'foo.txt'
# Using os.path.join is better than using '/' it is OS agnostic
path = os.path.join(os.path.sep, 'tmp', 'bar', 'baz', filename)
zip_filename = os.path.splitext(filename)[0] + '.zip'
zip_path = os.path.join(os.path.dirname(path), zip_filename)
# If you need exception handling wrap this in a try/except block
with zipfile.ZipFile(zip_path, 'w') as zf:
zf.write(path, zip_filename)
The bottom line is that if you do not supply an archive name then the filename is used as the archive name and it will contain the full path to the file.
最重要的是，如果您不提供存档名称，则文件名将用作存档名称，它将包含文件的完整路径。
回答by William A. Romero R.
To illustrate most clearly,
为了最清楚地说明，
directory structure:
目录结构：
/Users
└── /user
. ├── /pixmaps
. │ ├── pixmap_00.raw
. │ ├── pixmap_01.raw
│ ├── /jpeg
│ │ ├── pixmap_00.jpg
│ │ └── pixmap_01.jpg
│ └── /png
│ ├── pixmap_00.png
│ └── pixmap_01.png
├── /docs
├── /programs
├── /misc
.
.
.
Directory of interest: /Users/user/pixmaps
感兴趣的目录：/Users/user/pixmaps
First attemp
第一次尝试
import os
import zipfile
TARGET_DIRECTORY = "/Users/user/pixmaps"
ZIPFILE_NAME = "CompressedDir.zip"
def zip_dir(directory, zipname):
"""
Compress a directory (ZIP file).
"""
if os.path.exists(directory):
outZipFile = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
for dirpath, dirnames, filenames in os.walk(directory):
for filename in filenames:
filepath = os.path.join(dirpath, filename)
outZipFile.write(filepath)
outZipFile.close()
if __name__ == '__main__':
zip_dir(TARGET_DIRECTORY, ZIPFILE_NAME)
ZIP file structure:
ZIP文件结构：
CompressedDir.zip
.
└── /Users
└── /user
└── /pixmaps
├── pixmap_00.raw
├── pixmap_01.raw
├── /jpeg
│ ├── pixmap_00.jpg
│ └── pixmap_01.jpg
└── /png
├── pixmap_00.png
└── pixmap_01.png
Avoiding the full directory path
避免完整目录路径
def zip_dir(directory, zipname):
"""
Compress a directory (ZIP file).
"""
if os.path.exists(directory):
outZipFile = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
# The root directory within the ZIP file.
rootdir = os.path.basename(directory)
for dirpath, dirnames, filenames in os.walk(directory):
for filename in filenames:
# Write the file named filename to the archive,
# giving it the archive name 'arcname'.
filepath = os.path.join(dirpath, filename)
parentpath = os.path.relpath(filepath, directory)
arcname = os.path.join(rootdir, parentpath)
outZipFile.write(filepath, arcname)
outZipFile.close()
if __name__ == '__main__':
zip_dir(TARGET_DIRECTORY, ZIPFILE_NAME)
ZIP file structure:
ZIP文件结构：
CompressedDir.zip
.
└── /pixmaps
├── pixmap_00.raw
├── pixmap_01.raw
├── /jpeg
│ ├── pixmap_00.jpg
│ └── pixmap_01.jpg
└── /png
├── pixmap_00.png
└── pixmap_01.png
回答by jinzy
zf.write(tofile)
to change
改变
zf.write(tofile, zipfile_dir)
for example
例如
zf.write("/root/files/result/root/files/result/new.txt", "/root/files/results/new.txt")
回答by Helenus the Seer
You can isolate just the file name of your sources files using:
您可以使用以下方法仅隔离源文件的文件名：
name_file_only= name_full_path.split(os.sep)[-1]
For example, if name_full_pathis /root/files/results/myfile.txt, then name_file_onlywill be myfile.txt. To zip myfile.txt to the root of the archive zf, you can then use:
例如，如果name_full_path是/root/files/results/myfile.txt，那么name_file_only将会是myfile.txt。要将 myfile.txt 压缩到存档的根目录zf，您可以使用：
zf.write(name_full_path, name_file_only)
回答by hubert
I face the same problem and i solve it with writestr. You can use it like this:
我面临同样的问题，我用writestr. 你可以这样使用它：
zipObject.writestr(<filename> , <file data, bytes or string>)
zipObject.writestr(<filename> , <file data, bytes or string>)
回答by Simon Buus Jensen
The arcnameparameter in the write method specifies what will be the name of the file inside the zipfile:
arcnamewrite 方法中的参数指定 zipfile 中文件的名称：
import os
import zipfile
# 1. Create a zip file which we will write files to
zip_file = "/home/username/test.zip"
zipf = zipfile.ZipFile(zip_file, 'w', zipfile.ZIP_DEFLATED)
# 2. Write files found in "/home/username/files/" to the test.zip
files_to_zip = "/home/username/files/"
for file_to_zip in os.listdir(files_to_zip):
file_to_zip_full_path = os.path.join(files_to_zip, file_to_zip)
# arcname argument specifies what will be the name of the file inside the zipfile
zipf.write(filename=file_to_zip_full_path, arcname=file_to_zip)
zipf.close()
