Question

我正在尝试创建一个可能很大的文件夹的zip存档。为此，我正在使用python zipfile模块，但据我所知，没有选择将创建的归档文件拆分为最大大小的多个块。

压缩的存档应该通过电报发送，每个文件的大小限制为1.5 GB。因此，我需要分割生成的zip存档。

我真的不希望使用子进程和shell命令来创建此存档。

我当前的代码如下：

def create_zip(archive_name, directory):
    """Create a zip file from given dir path."""
    with ZipFile(archive_name, "w", ZIP_LZMA) as target_zip_file:
        for root, _, files in os.walk(directory):
            for file_to_zip in files:
                absolute_path = os.path.join(root, file_to_zip)
                zip_file_name = absolute_path[len(directory) + len(os.sep):]
                target_zip_file.write(absolute_path, zip_file_name)

    return target_zip_file

预先感谢

Answer 1

如果找不到zipfile的更好的本机方式，您仍然可以自己编写文件拆分算法。像这样：

outfile = archive_name
packet_size = int(1.5 * 1024**3)   # bytes

with open(outfile, "rb") as output:
    filecount = 0
    while True:
        data = output.read(packet_size)
        print(len(data))
        if not data:
            break   # we're done
        with open("{}{:03}".format(outfile, filecount), "wb") as packet:
            packet.write(data)
        filecount += 1

类似于将其放回接收器一侧。

Answer 2

这是我用来通过电报bot将文件发送到电报通道的工具。电报bot上传文件的大小限制为50MB

#! /usr/bin/python3
# -*- coding:utf-8 -*-
# apt-get install p7zip-full

import subprocess
import os
import math
import logzero

SPLIT_SIZE = 49

def file_split_7z(filePath, split_size=SPLIT_SIZE, removeOrigin=False):
    CMD_7Z = '7z a -v' + str(split_size) + "m -y '{}' '{}' -mx0"
    originFilePath = ''
    if os.path.splitext(filePath)[1] == '.7z':
        originFilePath = filePath
        filePath = os.path.splitext(originFilePath)[0] + '.7zo'
        os.rename(originFilePath, filePath)

    fz = os.path.getsize(filePath) / 1024 / 1024
    pa = math.ceil(fz / split_size)
    fn = os.path.splitext(filePath)[0].replace('.', '_')
    subprocess.call(CMD_7Z.format(fn, filePath), shell=True)
    file_list = []
    for i in range(pa):
        file_list.append('{}.7z.{:03d}'.format(fn, i + 1))
    if originFilePath:
        os.rename(filePath, originFilePath)
        filePath = originFilePath
    if removeOrigin:
        os.remove(filePath)
    file_list = [os.path.abspath(i) for i in file_list]
    return file_list


def split7zFile(filePath, logger=logzero.logger):
    file_size = os.path.getsize(filePath) / 2**20
    if file_size > SPLIT_SIZE :
        split_part = math.ceil(file_size / SPLIT_SIZE)
        split_size = math.ceil(file_size / split_part)
        logger.info('file size | {} | split num | {} |split size | {}'.format(file_size, split_part, split_size))
        filePathList = file_split_7z(filePath, split_size=split_size, removeOrigin=False)
    else:
        filePathList = [filePath]
    return filePathList

将zip归档文件分成多个块

2 个答案: