为什么BadZipFile:文件不是一个zip文件?

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这是我的代码——我尝试执行此脚本时遇到错误:

 Error    raise BadZipFile("File is not a zip file")  
          BadZipFile: File is not a zip file

这是我的源目录路径 :

data_dir = r'L:\DataQA\Python Unzip Files\Source Zipped'

我在“源压缩”(未压缩)文件夹中有多个压缩文件夹。当我将源文件的所有子文件夹压缩到单个压缩文件夹时,相同的代码工作。代码:

import os
import zipfile
import shutil
import json
import logging
import logging.config
import time

def my_start_time():
    global start_time, cumulative_time, start_time_stamp
    start_time = time.time()
    this_time = time.localtime(start_time)
    start_time_stamp = '{:4d}{:02d}{:02d} {:02d}:{:02d}:{:02d}'.format(\
                    this_time.tm_year, this_time.tm_mon, this_time.tm_mday,\
                    this_time.tm_hour, this_time.tm_min, this_time.tm_sec)
    cumulative_time = start_time - start_time 
    logging.info('Initial Setup: {:s}'.format(start_time_stamp))

def my_time():
    global cumulative_time
    time_taken = time.time() - start_time
    incremental_time = time_taken - cumulative_time
    cumulative_time = time_taken
    logging.info("Started: %s  Complete:  Cumulative: %.4f s  Incremental: %.4f s\n" \
          % (start_time_stamp, cumulative_time, incremental_time) )

logging.basicConfig(filename='myunzip_task_log.txt',level=logging.DEBUG)
my_start_time()

logging.info('Initial Setup...')

def write_to_json(data, file):
    value = False
    with open(file, 'w') as f:
        json.dump(json.dumps(data, sort_keys=True),f)   
        f.close()
        value = True
    return value


data_dir = r'L:\DataQA\Python Unzip Files\Source Zipped'
temp_dir =  r'L:\DataQA\Python Unzip Files\temp1'
new_dir = r'L:\DataQA\Python Unzip Files\temp2'
final_dir = r'L:\DataQA\Python Unzip Files\Destination Unzipped files'





big_list = os.listdir(data_dir)

archive_count = 0
file_count = 152865
basename1 = os.path.join(final_dir,'GENERIC_ROUGHDRAFT')
basename2 = os.path.join(final_dir,'XACTDOC')

my_time()
archive_count = len(big_list)
logging.info('Unzipping {} archives...'.format(archive_count))
for folder in big_list:
    prior_count = file_count
    logging.info('Starting: {}'.format(folder))

    try:
        shutil.rmtree(temp_dir)
    except FileNotFoundError: 
        pass
    os.mkdir(temp_dir)
    with zipfile.ZipFile(os.path.join(data_dir,folder),mode='r') as a_zip:
        a_zip.extractall(path = temp_dir)
        archive_count += 1
        logging.info('Cumulative total of {} archive(s) unzipped'.format(archive_count))
        bigger_list = os.listdir(temp_dir)
        logging.info('Current archive contains {} subfolders'.format(len(bigger_list)))
        for sub_folder in bigger_list:
            with zipfile.ZipFile(os.path.join(temp_dir,sub_folder),mode='r') as b_zip:
                b_zip.extractall(path = new_dir)
            file1 = "%s (%d).%s" % (basename1, file_count, 'xml')
            file2 = "%s (%d).%s" % (basename2, file_count, 'xml')
            shutil.copy(os.path.join(new_dir, 'GENERIC_ROUGHDRAFT.xml'), file1)
            shutil.copy(os.path.join(new_dir, 'XACTDOC.xml'), file2)
            file_count += 1
        logging.info('{} subfolders unzipped'.format(file_count - prior_count))
    #os.remove(data_dir)
    shutil.rmtree(data_dir)
    os.mkdir(data_dir)
    #os.unlink(data_dir)
    my_time()
logging.info('Total of {0} files -- {1} pairs -- should be in {2}'.format(2*(file_count-1), file_count-1, final_dir))

time.sleep(1)

my_time()

提问于
用户回答回答于

在两个zip存档打开语句:

with zipfile.ZipFile(os.path.join(data_dir,folder),mode='r')

with zipfile.ZipFile(os.path.join(temp_dir,sub_folder),mode='r')

我建议你在提取之前检查文件扩展名,例如:

import fnmatch
zfn = os.path.join(temp_dir,sub_folder)
if fnmatch.fnmatch(zfn,"*.zip"):
    with zipfile.ZipFile(zfn,mode='r') as whatever:

一些.zip文件可能已损坏,但不太可能。另外,如果你想提取.jar等拉链结构的文件扩展名不同,更换fnmatch

if zfn.lower().endswith(('.zip','.jar','.docx')):

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