[Zope] Yet another newbie question (file upload and python scripts)

Janusz Zamecki janusz at pipi.ma.cx
Sat Jan 28 04:30:29 EST 2006


Hi, after googling and RTFM I have no other option but ask you for help.

I have the following DTML document:

<dtml-var standard_html_header>
<form action=3D"checkTheFile" method=3D"post" enctype=3D"multipart/form-d=
ata">
Select the file:
<input type=3D"file" name=3D"the_file">
<input type=3D"submit" name=3D"go" value=3D"Next >>">
</form>
<dtml-var standard_html_footer>


And here is the checkTheFile python script:
## Script (Python) "checkTheFile"
##bind container=3Dcontainer
##bind context=3Dcontext
##bind namespace=3D
##bind script=3Dscript
##bind subpath=3Dtraverse_subpath
##parameters=3D
##title=3D
##
# Import a standard function, and get the HTML request and response objec=
ts.
from Products.PythonScripts.standard import html_quote
request =3D container.REQUEST
RESPONSE =3D  request.RESPONS


filename=3Dhtml_quote(request.form['the_file'].filename)
context.fs.rpt.manage_addFile(filename, request.form['the_file'],
content_type=3D"text/xml")
# the next line does'n work:
file_content=3Drequest.form['the_file'].read()
# EOF

I need to temporary save uploaded file then I have to validate it. But
I've received the following error message:

Site error:
Error Type: AttributeError
Error Value: read

I've tested it on zope 2.7.7 and 2.9.0 on suse 10.0. (python 2.4.1).

So, It is obvious to me that I've made a mistake, but I have no idea
where it is?

I hope that someone could show me the right way....

Best regards to all,

Janusz

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