9

我的程序运行顺利，但我希望将来自 ftp 的文件压缩到本地驱动器中

我的问题：调用我的 main() 函数后只有 1 个文件被压缩

这是我的代码：

主要的

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"
4

4 回答 4

11

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)

您为要压缩的每个文件重新创建一个 ZipFile，该"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，它将以附加模式打开 zipfile：

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)
于 2012-06-28T08:31:24.523 回答
4

看看 shutil 模块。有一个使用 shutil.make_archive() 的例子：

http://docs.python.org/library/shutil.html

于 2012-06-28T08:28:30.397 回答
1

如果你有很多文件，你可以并行压缩它们：

import zipfile
from pathlib import Path, WindowsPath
from typing import List, Text
import logging
from time import time
from concurrent.futures import ThreadPoolExecutor

logging.basicConfig(
    format="%(asctime)s - %(message)s", datefmt="%H:%M:%S", level=logging.DEBUG
)

PATH = (r"\\some_directory\subdirectory\zipped")


def file_names() -> List[WindowsPath]:
    p = Path(PATH)
    file_names = list(p.glob("./*.csv"))
    logging.info("There are %d files", len(file_names))
    return file_names


def zip_file(file: WindowsPath) -> None:
    zip_file_name = Path(PATH, f"{file.stem}.zip")
    with zipfile.ZipFile(zip_file_name, "w") as zip:
        zip.write(file, arcname=file.name, compress_type=zipfile.ZIP_DEFLATED)


def main(files: List[Text]) -> None:
    t0 = time()
    number_of_files = len(files)
    with ThreadPoolExecutor() as executor:
        for counter, _ in enumerate(executor.map(zip_file, files), start=1):
            # update progress every 100 files
            if counter % 100 == 0:
                logging.info(
                    "Processed %d/%d. TT: %d:%d",
                    counter,
                    number_of_files,
                    *divmod(int(time() - t0), 60),
                )

    logging.info(
        "Finished zipping %d files. Total time: %d:%d",
        len(files),
        *divmod(int(time() - t0), 60),
    )


if __name__ == "__main__":
    files = file_names()
    main(files)
于 2018-10-24T15:59:46.413 回答
0

最好的方法是将调试语句放在你的 for 循环中，有两种可能性；

一个是第一个forloop只从ftp文件夹下载一个文件

二是第一个循环下载所有文件，但第二个循环只压缩其中一个

使用打印语句查看在循环中下载/压缩了哪些文件，祝你好运

于 2012-06-28T08:27:58.913 回答