我正在尝试将 AjaxUpload 与 Python 一起使用：http: //valums.com/ajax-upload/
我想知道如何使用 Python 访问上传的文件。在网站上，它说：
* PHP: $_FILES['userfile']
* Rails: params[:userfile]
Python 的语法是什么？
request.params['userfile'] 似乎不起作用。
提前致谢！这是我当前的代码（使用作为图像导入的 PIL）
im = Image.open(request.params['myFile'].file)
import cgi
#This will give you the data of the file,
# but won't give you the filename, unfortunately.
# For that you have to do some other trick.
file_data = cgi.FieldStorage.getfirst('file')
#<IGNORE if you're not using mod_python>
#(If you're using mod_python you can also get the Request object
# by passing 'req' to the relevant function in 'index.py', like "def func(req):"
# Then you access it with req.form.getfirst('file') instead. NOTE that the
# first method will work even when using mod_python, but the first FieldStorage
# object called is the only one with relevant data, so if you pass 'req' to the
# function you have to use the method that uses 'req'.)
#</IGNORE>
#Then you can write it to a file like so...
file = open('example_filename.wtvr','w')#'w' is for 'write'
file.write(file_data)
file.close()
#Then access it like so...
file = open('example_filename.wtvr','r')#'r' is for 'read'
#And use file.read() or whatever else to do what you want.
我正在与 Pyramid 合作，我也在尝试做同样的事情。一段时间后，我想出了这个解决方案。
from cStringIO import StringIO
from cgi import FieldStorage
fs = FieldStorage(fp=request['wsgi.input'], environ=request)
f = StringIO(fs.value)
im = Image.open(f)
我不确定它是否是“正确的”，但它似乎有效。
在 django 中，您可以使用：
request.FILES['file']
代替：
request.POST['file']
我不知道如何在塔里做……也许是同一个概念……