阅读上传的zip文件

Question

How can i read the fileitems in zip file that uploaded using form in html and using Cgi, i tried as this: 我如何阅读zip文件中的文件项目，该文件在html中使用表格上传并使用Cgi，我试过这样：

form = cgi.FieldStorage() 
file_upload = form['file']
if zipfile.is_zipfile(file_upload.filename):
    print "%s is a valid pkzip file" % file_upload.filename
else:
    print "%s is not a valid pkzip file" % file_upload.filename
zfile=zipfile.ZipFile(file_upload.filename,"r")
files_zip=zfile.namelist()

For example when i upload (test.zip)the error is No such file or directory: 'test.zip',and if i run the code without this zfile=zipfile.ZipFile(file_upload.filename,"r") , i get that test.zip is not a valid pkzip file. 例如，当我上传（test.zip）错误是没有这样的文件或目录：'test.zip'，如果我运行没有这个zfile=zipfile.ZipFile(file_upload.filename,"r") ，我得到test.zip不是有效的pkzip文件。 Thanks in advance. 提前致谢。

Answer 1

You could try to pass file_upload.file to ZipFile instead of file_upload.filename . 您可以尝试将file_upload.file传递给ZipFile而不是file_upload.filename 。

Here's a script that prints the list of files in the zip file: 这是一个打印zip文件中文件列表的脚本：

import sys
sys.stderr = sys.stdout
print "Content-Type: text/plain"
print

import cgi
import zipfile

form = cgi.FieldStorage()

filefield = form['somefile']
print "Filename:", filefield.filename

if filefield.file is not None and zipfile.is_zipfile(filefield.file):
    zfile = zipfile.ZipFile(filefield.file)
    print "Name list:\n\t",
    print "\n\t".join(zfile.namelist())

And the corresponding html form: 和相应的html表单：

<!DOCTYPE html>
<form enctype="multipart/form-data" action="file-upload" method=post>
<p><label for=somefile>File: <input type=file name=somefile>
<p><input type=submit>
</form>

阅读上传的zip文件

问题描述

1 个解决方案

解决方案1
3 已采纳 2012-03-25 22:32:37

阅读上传的zip文件

问题描述

1 个解决方案

解决方案1 3 已采纳 2012-03-25 22:32:37

解决方案1
3 已采纳 2012-03-25 22:32:37