使用python提取zip文件

Question

[我正在尝试在特定目录名称中“下载”所有zip文件，并将其所有内容提取到名为“ extracted”的目录中。

我不知道为什么，仅迭代现有文件名后，我收到一个错误消息，即没有这样的文件...

allFilesList = os.listdir(os.getcwd()+"/downloaded")
print allFilesList #verify - correct expected list
from zipfile import ZipFile
os.chdir(os.getcwd()+"/extracted/")
print os.getcwd() #verify - correct expected dir

for fileName in allFilesList: 
    print fileName
    with ZipFile(fileName, 'r') as zipFileObject:
        if os.path.exists(fileName):
            print "Skipping extracting " + fileName
            continue
        zipFileObject.extractall(pwd='hello')                   
        print "Saving extracted file to extracted/",fileName
print "all files has been successfully extracted"

错误消息：

File "program.py", line 77, in <module>
with ZipFile(fileName, 'r') as zipFileObject:
File "/usr/lib/python2.7/zipfile.py", line 779, in __init__
    self.fp = open(file, modeDict[mode])
IOError: [Errno 2] No such file or directory: 'zipFile1.zip'

Answer 1

您将从一个目录中获取文件名列表，然后更改为另一个目录，并尝试从该目录中提取可能不存在的文件：

allFilesList = os.listdir(os.getcwd()+"/downloaded")
# ...
os.chdir(os.getcwd()+"/extracted/")
# ...
    with ZipFile(fileName, 'r') as zipFileObject:

如果将该文件ZipFile命令更改为类似这样的内容：

    with ZipFile(os.path.join("..", "downloaded", fileName), 'r') as zipFileObject:

您应该能够在找到它的目录中打开该文件。