Question

我需要帮助编辑以下脚本来压缩目录的内容。我的最终目标是创建一个脚本，它将查看C：\ Test（里面会有多个目录），并在C：\ Test中创建一个包含每个目录内容的新zip文件。棘手的部分是我需要路径为C：\，即使目录的真路径是C：\ Test。这可能还是我在做梦？

由于

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)

Answer 1

您可以按如下方式更改存档内文件的路径：

a.write(PATH_ON_FILESYSTEM, 
        DESIRED_PATH_IN_ARCHIVE
)

使用单个拉链路径压缩多个子目录

1 个答案: