python - 在python中压缩文件

Question

我的程序运行顺利，但我希望我的文件从 ftp 压缩到我的本地驱动器

问题是在调用我的 main() 函数后只有 1 个文件被压缩

这是我的代码:

import os
import upload
import download
import zipfile
import ConfigParser
import ftputil

def main():
    
    #create a folder Temp on d drive for later use
    path = r'D:\Temp'
    os.mkdir(path)
    
    #parse all the  values at config.ini file
    config = ConfigParser.ConfigParser()
    config.readfp(open('config.ini'))
    server = config.get('main', 'Server')
    username = config.get('main', 'Username')
    password = config.get('main', 'Password')
    uploads = config.get('main', 'Upload folder')
    downloads = config.get('main', 'Download folder')

    #connect to ftp
    ftp = ftputil.FTPHost(server, username, password)

    dirlist = ftp.listdir(downloads)
    
    for list in dirlist:
        ftp.chdir(downloads)
        target = os.path.join(path, list)
        ftp.download(list, target)
        
    
    #########################################################
    #   THis section is where algo fails but the program run#
    ########################################################
    
    #zipping files
    absolute_path = r'D:\Temp'
    dirlist = os.listdir(absolute_path)
    filepath = r'D:\Temp\project2.zip'
    for list in dirlist:
        get_file = os.path.join(absolute_path, list)
        zip_name = zipfile.ZipFile(filepath, 'w')
        zip_name.write(get_file, 'Project2b\\' + list)
        
                
        

if __name__ == '__main__':
    print "cannot be"

Answer 1

当你这样做时:

for list in dirlist:
        get_file = os.path.join(absolute_path, list)
        zip_name = zipfile.ZipFile(filepath, 'w')
        zip_name.write(get_file, 'Project2b\\' + list)

您为每个要压缩的文件重新创建一个 ZipFile，"w" 模式意味着您从头开始重新创建它。

试试这个(在循环之前创建 zip 文件):

zip_name = zipfile.ZipFile(filepath, 'w')
for list in dirlist:
        get_file = os.path.join(absolute_path, list)
        zip_name.write(get_file, 'Project2b\\' + list)

或者这样，它将以追加模式打开 zip 文件:

for list in dirlist:
        get_file = os.path.join(absolute_path, list)
        zip_name = zipfile.ZipFile(filepath, 'a')
        zip_name.write(get_file, 'Project2b\\' + list)