How to parse multi-part content
Dave Kuhlman
dkuhlman at rexx.com
Mon Sep 27 19:03:06 EDT 2004
Dave Kuhlman wrote:
> John J. Lee wrote:
>
>> Dave Kuhlman <dkuhlman at rexx.com> writes:
>> [...]
>>> In case you are curious, this is content posted to my Zope server
>>> when I include an element '<input type="file" .../>' in my form.
>> [...]
>>
>> *Surely* Zope has a standard way of doing this. Try a Zope list?
>>
>
> That's a good suggestion. Thanks. Zope people are Python people,
> so they would give me the kind of help I'd need. I'll ask on the
> Zope users list.
>
> However, there is nothing Zope-specific about this. The content
> was produced by my Web browser (actually two Web browsers that I
> test with: Opera and Firefox).
>
I was wrong. You were right. There is a Zope way to do this.
Thanks for pushing me to dig deeper. It's a much easier way, too.
If there are any Zopesters reading, here is how to do it:
def my_external_method(request, ...):
# Retrieve a stream-like object.
myStream = request['myFileData']
# Read the data from the stream object.
data = myStream.read()
My problem was that I was so sure that I had to retrieve and parse
the content in the body of the request.
Thanks for help.
Dave
--
Dave Kuhlman
http://www.rexx.com/~dkuhlman
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