Python无法处理zipfile.BadZipFile中的异常 [英] Python cant handle exceptions from zipfile.BadZipFile

问题描述

需要处理zip文件是否损坏，因此它只需传递此文件即可继续使用下一个文件.

Need to handle if a zip file is corrupt, so it just pass this file and can go on to the next.

在Im试图捕获异常的下面的代码示例中，我可以将其传递.但是，当zip文件损坏时，我的脚本会失败*，并给我正常"的追溯错误*，而不是打印我的错误"，但是如果zip文件正常，则脚本运行正常.

In the code example underneath Im trying to catch the exception, so I can pass it. But my script is failing when the zipfile is corrupt*, and give me the "normal" traceback errors* istead of printing "my error", but is running ok if the zipfile is ok.

这是我正在处理的代码的一个简约示例.

This i a minimalistic example of the code I'm dealing with.

path = "path to zipfile" 

from zipfile import ZipFile

with ZipFile(path) as zf:
    try:
        print "zipfile is OK"
    except BadZipfile:
        print "Does not work "
        pass

回溯的一部分告诉我:引发BadZipfile，文件不是zip文件"

part of the traceback is telling me: raise BadZipfile, "File is not a zip file"

推荐答案

您需要将上下文管理器放入 try-except块:

You need to put your context manager inside the try-except block:

try:
    with ZipFile(path) as zf:
        print "zipfile is OK"
except BadZipfile:
    print "Does not work "

错误由引发 ZipFile引发，因此将其放置在外部意味着找不到针对引发的异常的处理程序.另外，请确保从zipfile中适当导入了BadZipFile.

The error is raised by ZipFile so placing it outside means no handler can be found for the raised exception. In addition make sure you appropriately import BadZipFile from zipfile.

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