[Zope] File Upload via python

[Jonathan Bryant]

Hi all,

I've been playing around with Zope for several days and really like it. I'm 
amazed at how simple so many things that take alot of effort in php, asp 
etc. are amazingly simple in Zope. One of those things is file upload, which 
works but still puzzles me.

I have a form in a dtml method that looks like this:

<table>
	<form method="post" action="testAction" enctype="multi-part/form-data">
	<tr><td><strong>FileTitle</strong></td><td><input type="text" 
name="item_title" maxlength="30"></td></tr>
	<tr><td><strong>Select File</strong></td><td><input type="file" 
name="item"></td></tr>
        <tr><td align=center colspan=2><input type="submit" Value="Add 
File"></td></tr>
	</form>
</table>


I figured out that using a python script I can add the file just by 
referencing it's name from the form, which is incredibly cool:

context.manage_addProduct['OFSP'].manage_addFile('newFile', item_title, 
item)

What I'd like to do is easily grab the filename:

context.manage_addProduct['OFSP'].manage_addFile(item.filename, item_title, 
item)


My question is, how do I get the filename or headers? I'm guessing that when 
I reference 'item' I'm getting a FileUpload object. What confuses me is that 
I can't reference item.filename or item.headers['Content-Type']. I've also 
tried using REQUEST['item'].filename like I've seen in several examples. 
What am I doing wrong here?

Thanks in advance,


Jonathan Bryant
jonathanbryant@hotmail.com


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