我很难弄清楚如何用2.4解压缩zip文件。 extract()
未包含在2.4中。我只能在我的服务器上使用2.4.4。
有人可以提供一个简单的代码示例吗?
答案 0 :(得分:51)
您必须使用namelist()
和extract()
。考虑目录的示例
import zipfile
import os.path
import os
zfile = zipfile.ZipFile("test.zip")
for name in zfile.namelist():
(dirname, filename) = os.path.split(name)
print "Decompressing " + filename + " on " + dirname
if not os.path.exists(dirname):
os.makedirs(dirname)
zfile.extract(name, dirname)
答案 1 :(得分:12)
Vinko的答案存在一些问题(至少在我运行时)。我得到了:
IOError: [Errno 13] Permission denied: '01org-webapps-countingbeads-422c4e1/'
以下是解决方法:
# unzip a file
def unzip(path):
zfile = zipfile.ZipFile(path)
for name in zfile.namelist():
(dirname, filename) = os.path.split(name)
if filename == '':
# directory
if not os.path.exists(dirname):
os.mkdir(dirname)
else:
# file
fd = open(name, 'w')
fd.write(zfile.read(name))
fd.close()
zfile.close()
答案 2 :(得分:3)
修改Ovilia's answer,以便您也可以指定目标目录:
def unzip(zipFilePath, destDir):
zfile = zipfile.ZipFile(zipFilePath)
for name in zfile.namelist():
(dirName, fileName) = os.path.split(name)
if fileName == '':
# directory
newDir = destDir + '/' + dirName
if not os.path.exists(newDir):
os.mkdir(newDir)
else:
# file
fd = open(destDir + '/' + name, 'wb')
fd.write(zfile.read(name))
fd.close()
zfile.close()
答案 3 :(得分:1)
未完全测试,但应该没问题:
import os
from zipfile import ZipFile, ZipInfo
class ZipCompat(ZipFile):
def __init__(self, *args, **kwargs):
ZipFile.__init__(self, *args, **kwargs)
def extract(self, member, path=None, pwd=None):
if not isinstance(member, ZipInfo):
member = self.getinfo(member)
if path is None:
path = os.getcwd()
return self._extract_member(member, path)
def extractall(self, path=None, members=None, pwd=None):
if members is None:
members = self.namelist()
for zipinfo in members:
self.extract(zipinfo, path)
def _extract_member(self, member, targetpath):
if (targetpath[-1:] in (os.path.sep, os.path.altsep)
and len(os.path.splitdrive(targetpath)[1]) > 1):
targetpath = targetpath[:-1]
if member.filename[0] == '/':
targetpath = os.path.join(targetpath, member.filename[1:])
else:
targetpath = os.path.join(targetpath, member.filename)
targetpath = os.path.normpath(targetpath)
upperdirs = os.path.dirname(targetpath)
if upperdirs and not os.path.exists(upperdirs):
os.makedirs(upperdirs)
if member.filename[-1] == '/':
if not os.path.isdir(targetpath):
os.mkdir(targetpath)
return targetpath
target = file(targetpath, "wb")
try:
target.write(self.read(member.filename))
finally:
target.close()
return targetpath
答案 4 :(得分:-1)
我在Python 2.7.3rc2中进行测试,而ZipFile.namelist()
没有返回仅包含用于创建子目录的子目录名的条目,而只返回具有子目录的文件名列表,如下所示:
['20130923104558/control.json', '20130923104558/test.csv']
因此检查
if fileName == '':
根本不评估为True
。
因此,我修改了代码以检查dirName
是否存在destDir
内,并创建dirName
(如果不存在)。仅当fileName
部分不为空时才提取文件。因此,这应该考虑目录名称可以出现在ZipFile.namelist()
def unzip(zipFilePath, destDir):
zfile = zipfile.ZipFile(zipFilePath)
for name in zfile.namelist():
(dirName, fileName) = os.path.split(name)
# Check if the directory exisits
newDir = destDir + '/' + dirName
if not os.path.exists(newDir):
os.mkdir(newDir)
if not fileName == '':
# file
fd = open(destDir + '/' + name, 'wb')
fd.write(zfile.read(name))
fd.close()
zfile.close()