我需要帮助编辑以下压缩目录内容的脚本。我的最终目标是创建一个脚本,该脚本将查看 C:\Test(其中将有多个目录)并使用 C:\Test 中每个目录的内容创建一个新的 zip 文件。棘手的部分是我需要路径是 C:\ 即使目录的真实路径是 C:\Test。这是可能的还是我在做梦?
谢谢
import zipfile, os
def makeArchive(fileList, archive):
try:
a = zipfile.ZipFile(archive, 'w', zipfile.ZIP_DEFLATED)
for f in fileList:
print "archiving file %s" % (f)
a.write(f)
a.close()
return True
except: return False
def dirEntries(dir_name, subdir, *args):
fileList = []
for file in os.listdir(dir_name):
dirfile = os.path.join(dir_name, file)
if os.path.isfile(dirfile):
if not args:
fileList.append(dirfile)
else:
if os.path.splitext(dirfile)[1][1:] in args:
fileList.append(dirfile)
# recursively access file names in subdirectories
elif os.path.isdir(dirfile) and subdir:
print "Accessing directory:", dirfile
fileList.extend(dirEntries(dirfile, subdir, *args))
return fileList
if __name__ == '__main__':
folder = r'C:\test'
zipname = r'C:\test\test.zip'
makeArchive(dirEntries(folder, True), zipname)