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用AjaxUpload上传python图像

ajax-upload image-uploading python-imaging-library pylons python

4

resopollution · 技术社区 · 15 年前

我正在尝试将AjaxUpload与python结合使用: http://valums.com/ajax-upload/

我想知道如何用python访问上传的文件。网站上写着:

* PHP: $_FILES['userfile']
* Rails: params[:userfile]

python的语法是什么?

request.params['userfile']似乎不起作用。

事先谢谢!这是我的当前代码(使用pil作为图像导入)

im = Image.open(request.params['myFile'].file)

3 回复 | 直到 14 年前

1

0

user149513 15 年前

在Django,您可以使用:

request.FILES['file']

而不是:

request.POST['file']

我不知道在塔架上怎么做……也许这是同一个概念……

2

1

sth Alien 15 年前

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.

3

1

Dario Zamuner 14 年前

我在和金字塔合作,我也试着做同样的事情。过了一段时间,我想出了这个解决办法。

from cStringIO import StringIO
from cgi import FieldStorage

fs = FieldStorage(fp=request['wsgi.input'], environ=request)
f = StringIO(fs.value)

im = Image.open(f)

我不确定它是不是“正确”的,但它似乎有效。