Re: [Zope] Stupid File Upload Question

[jpenny]

On Fri, Dec 15, 2000 at 04:59:07PM -0800, Jonothan Farr wrote:
> What sort of problems were you seeing with LocalFS? Maybe I can fix them.
> 
> Thanks,
> --jfarr
> 
> - Original Message - 
> From: <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Friday, December 15, 2000 4:22 PM
> Subject: [Zope] Stupid File Upload Question
> 
> 
> > This is for Jos:
> > 
> > I ran into the same problem today.  I had previously used this method,
> > and it works better than LocalFS for my purposes.  Anyway, be absolutly
> > sure that your form says:

Fiest, LocalFS turns out to have the same problem.  MSIE does not
interact well with method=get and file input methods.

> > 
> > 
> > ...
> > 
> > 
> > Then you should be able to access REQUEST.form['attachment'].filename
> > and REQUEST.form['attachment'].read()  (Well, I hope).
> > 
> > If the method=post is omitted, Netscape (linux at least) will work.
> > MSIE will not.
> > 
> > Man I ain't got enough hair for these kinds of problems!
> > 
> > Jim Penny

I want/need to collect some metadata along with the file.  Say I want
a file, a title, some keywords.  (And I have some idiots with 80MB files).

I would like to enforce the following:

No two files from a given user can have the same title.
A given file may not be uploaded twice by the same user with
different titles.

With LocalFS, as it stands, it is hard to determine the file that 
was uploaded.  (It is not returned by manage_upload, I could create
an id, but that has its own problems (downloading creates unexpected
files on the client PC's)).  This can be modified, but...

More importantly, it is really hard to see how to handle both of
the conditios above cleanly.  With LocalFS, I found no easy way
at all to combine the error checking implied by the metadata and
the error checking desired for the file name.

> > 
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> 

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Re: [Zope] Stupid File Upload Question

[Fred Yankowski]

On Fri, Dec 15, 2000 at 07:22:43PM -0500, [EMAIL PROTECTED] wrote:
> Anyway, be absolutly sure that your form says:
> 
> 

Of course, that should be ENCTYPE instead of the second METHOD
attribute name: 

  

As an aside, I found the following page to be a useful example of Zope
file-upload handling:


-- 
Fred Yankowski   [EMAIL PROTECTED]  tel: +1.630.879.1312
Principal Consultant www.OntoSys.com   fax: +1.630.879.1370
OntoSys, Inc 38W242 Deerpath Rd, Batavia, IL 60510, USA

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Re: [Zope] Stupid File Upload Question

[Jonothan Farr]

What sort of problems were you seeing with LocalFS? Maybe I can fix them.

Thanks,
--jfarr

- Original Message - 
From: <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, December 15, 2000 4:22 PM
Subject: [Zope] Stupid File Upload Question


> This is for Jos:
> 
> I ran into the same problem today.  I had previously used this method,
> and it works better than LocalFS for my purposes.  Anyway, be absolutly
> sure that your form says:
> 
> 
> ...
> 
> 
> Then you should be able to access REQUEST.form['attachment'].filename
> and REQUEST.form['attachment'].read()  (Well, I hope).
> 
> If the method=post is omitted, Netscape (linux at least) will work.
> MSIE will not.
> 
> Man I ain't got enough hair for these kinds of problems!
> 
> Jim Penny
> 
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>  http://lists.zope.org/mailman/listinfo/zope-announce
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> 
> 


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Re: [Zope] stupid file upload question

[Joh Johannsen]

Dieter Maurer wrote:

> Joh Johannsen writes:
>  > What sort of object is this "REQUEST.form['attached_file']"?   Is there some
>  > way to find out this sort of thing when you have a Python object?  (I'm new
>  > to Python)
> It is a "ZPublisher.HTTPRequest.FileUpload" object.
> >From its (source) documentation:
>
> File upload objects are used to represent file-uploaded data.
>
> File upload objects can be used just like files.
>
> In addition, they have a 'headers' attribute that is a dictionary
> containing the file-upload headers, and a 'filename' attribute
> containing the name of the uploaded file.
>
>  > Is that even the place to look to get the name of the file on the server
>  > after it is uploaded?
> You should not rely on the fact that the data is stored somewhere
> on the server.
>
>  > That's why I mentioned the quote from that How-To (which the above is
>  > basically a copy of):  "In you python external method you can now reference
>  > REQUEST.form['attached_file'] as a normal file. You can perform things such
>  > as read() on the object. "
> "REQUEST.form['attached_file']" should be a "FileUpload" object
> and as such have a "read" method (beside many others).
>
> Thus, "REQUEST.form['attached_file'].read()" should return the
> file content.
>
> In an earlier message, you said, it did not.
> What happened?
>

What happens is that if I have this:


def get_file_name(self,REQUEST):
s = REQUEST.form['attached_file'].filename
contents = REQUEST.form['attached_file'].read()
return s

I get a Zope error for the External Method, when I reload it by clicking on "edit"
Here is traceback...

Traceback (innermost last):
  File /usr/local/Zope-2.2.0-src/lib/python/ZPublisher/Publish.py, line 222, in
publish_module
  File /usr/local/Zope-2.2.0-src/lib/python/ZPublisher/Publish.py, line 187, in
publish
  File /usr/local/Zope-2.2.0-src/lib/python/Zope/__init__.py, line 221, in
zpublisher_exception_hook
(Object: get_file_name)
  File /usr/local/Zope-2.2.0-src/lib/python/ZPublisher/Publish.py, line 171, in
publish
  File /usr/local/Zope-2.2.0-src/lib/python/ZPublisher/mapply.py, line 160, in
mapply
(Object: manage_edit)
  File /usr/local/Zope-2.2.0-src/lib/python/ZPublisher/Publish.py, line 112, in
call_object
(Object: manage_edit)
  File
/usr/local/Zope-2.2.0-src/lib/python/Products/ExternalMethod/ExternalMethod.py,
line 201, in manage_edit
(Object: get_file_name)
  File
/usr/local/Zope-2.2.0-src/lib/python/Products/ExternalMethod/ExternalMethod.py,
line 210, in getFunction
(Object: get_file_name)
  File /usr/local/Zope-2.2.0-src/lib/python/App/Extensions.py, line 217, in
getObject
(Info: ('/usr/local/Zope2/Extensions/get_file_name.py', 'get_file_name'))
SyntaxError: (see above)


JJ


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Re: [Zope] stupid file upload question

[Dieter Maurer]

Joh Johannsen writes:
 > What sort of object is this "REQUEST.form['attached_file']"?   Is there some
 > way to find out this sort of thing when you have a Python object?  (I'm new
 > to Python)
It is a "ZPublisher.HTTPRequest.FileUpload" object.
>From its (source) documentation:

File upload objects are used to represent file-uploaded data.

File upload objects can be used just like files.

In addition, they have a 'headers' attribute that is a dictionary
containing the file-upload headers, and a 'filename' attribute
containing the name of the uploaded file.

 > Is that even the place to look to get the name of the file on the server
 > after it is uploaded?
You should not rely on the fact that the data is stored somewhere
on the server.

 > That's why I mentioned the quote from that How-To (which the above is
 > basically a copy of):  "In you python external method you can now reference
 > REQUEST.form['attached_file'] as a normal file. You can perform things such
 > as read() on the object. "
"REQUEST.form['attached_file']" should be a "FileUpload" object
and as such have a "read" method (beside many others).

Thus, "REQUEST.form['attached_file'].read()" should return the
file content.

In an earlier message, you said, it did not.
What happened?



Dieter

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Re: [Zope] stupid file upload question

[Joh Johannsen]

Thanks for the responses.  My first message was a little unclear...

Here is what I am doing in more detail:

On the client side (me), I have a windows machine, and I am uploading a
file:  "E:\test\my_file"

On the server, I am running Zope on Linux.  My upload form looks like this:












So I use this form, and I think the file gets uploaded, though I'm not quite
sure where it goes.  The "Customize" DTML gets called when the form is
submitted, it looks like this:


 

filename: 




And the "get_file_name" is a Python External Method, which looks like this:

def get_file_name(self,REQUEST):
s = REQUEST.form['attached_file'].filename
return s


Everything works great so far.

Now I want to do something with that file I just uploaded.

What is its name on the server?

The "filename" in the field I am pretty sure is NOT the name of the file on
the local system, since when I look at it, it says "E:\test\myfile" and this
is on a Linux system, so there is no such path.  That is the name file had
on my Windows system.

What sort of object is this "REQUEST.form['attached_file']"?   Is there some
way to find out this sort of thing when you have a Python object?  (I'm new
to Python)

Is that even the place to look to get the name of the file on the server
after it is uploaded?

That's why I mentioned the quote from that How-To (which the above is
basically a copy of):  "In you python external method you can now reference
REQUEST.form['attached_file'] as a normal file. You can perform things such
as read() on the object. "

This makes it seem like whatever is necessary, it is very easy, but there's
some detail that I am missing.  Maybe I just don't know what a "normal file"
is...

Regards,

JJ


Dieter Maurer wrote:

> Joh Johannsen writes:
>  > But it says: "In you python external method you can now reference
>  > REQUEST.form['attached_file'] as a normal file. You can
>  >   perform things such as read() on the object. "
>  >
>  > Now in my Python external method, I can reference things like:
>  >
>  > REQUEST.form['attached_file'].filename
>  >
>  > and that is fine.
>  >
>  > But what is the syntax for actually reading the file?  From the above
>  > quote I thought it was as simple as
>  > x=REQUEST.form['attached_file'].read()  but that doesn't work...
> Please tell us, how it does not work (in fact, it should work):
>
>Did you get an attribute error (read) or was the result wrong?
>
> You must use the 'enctype="multipart/form-data"' and
> 'method=post' form parameter to upload files. Did you do that?
>
> Dieter


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Re: [Zope] stupid file upload question

[Dieter Maurer]

Joh Johannsen writes:
 > But it says: "In you python external method you can now reference
 > REQUEST.form['attached_file'] as a normal file. You can
 >   perform things such as read() on the object. "
 > 
 > Now in my Python external method, I can reference things like:
 > 
 > REQUEST.form['attached_file'].filename
 > 
 > and that is fine.
 > 
 > But what is the syntax for actually reading the file?  From the above
 > quote I thought it was as simple as
 > x=REQUEST.form['attached_file'].read()  but that doesn't work...
Please tell us, how it does not work (in fact, it should work):

   Did you get an attribute error (read) or was the result wrong?

You must use the 'enctype="multipart/form-data"' and
'method=post' form parameter to upload files. Did you do that?


Dieter

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