How to parse multi-part content

[Dave Kuhlman]

Suppose that I have content that looks like what I've included at
the end of this message. Is there something in the standard
Python library that will help me parse it, break into the parts
separated by the boundary strings, extract headers from each
sub-part, etc?

Do I need to add something like the following to the beginning?

Content-Type: multipart/related;
type="multipart/alternative";
boundary="-----------------------------164697015457031 3593966717980"

I've tried working with the email, mimetools, and multifile
modules in the standard library. But my understanding of these
things is dim, and I have not had success.

Is there a beginner's guide somewhere that I should read?

In case you are curious, this is content posted to my Zope server
when I include an element '<input type="file" .../>' in my form.

Here is the content that I need to parse:


-----------------------------164697015457031 3593966717980
Content-Disposition: form-data; name="xschemaCo ntent"


-----------------------------164697015457031 3593966717980
Content-Disposition: form-data; name="xschemaFi le"; filename="po.xs d"
Content-Type: application/octet-stream

<xs:schema targetNamespace ="http://openuri.org/easypo"
xmlns:po="http://openuri.org/easypo"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
elementFormDefa ult="qualified" >

<xs:element name="purchase-order">
<xs:complexType >
<xs:sequence>
<xs:element name="customer" type="po:custom er"/>
<xs:element name="date" type="xs:dateTi me"/>
<xs:element name="line-item" type="po:line-item"
minOccurs="0" maxOccurs="unbo unded"/>
<xs:element name="shipper" type="po:shippe r"
minOccurs="0"/>
</xs:sequence>
</xs:complexType>
</xs:element>
<xs:complexTy pe name="customer" >
<xs:sequence>
<xs:element name="name" type="xs:string "/>
<xs:element name="address" type="xs:string "/>
</xs:sequence>
</xs:complexType>
<xs:complexTy pe name="line-item">
<xs:sequence>
<xs:element name="descripti on" type="xs:string "/>
<xs:element name="per-unit-ounces" type="xs:decima l"/>
<xs:element name="price" type="xs:double "/>
<xs:element name="quantity" type="xs:intege r"/>
</xs:sequence>
</xs:complexType>
<xs:complexTy pe name="shipper">
<xs:sequence>
<xs:element name="name" type="xs:string "/>
<xs:element name="per-ounce-rate" type="xs:decima l"/>
</xs:sequence>
</xs:complexType>
</xs:schema>

-----------------------------164697015457031 3593966717980
Content-Disposition: form-data; name="which"

superclass
-----------------------------164697015457031 3593966717980
Content-Disposition: form-data; name="Submit"

Submit
-----------------------------164697015457031 3593966717980--



--
Dave Kuhlman

http://www.rexx.com/~dkuhlman

[John J. Lee]

Re: How to parse multi-part content

Dave Kuhlman <dkuhlman@rexx. com> writes:
[...][color=blue]
> In case you are curious, this is content posted to my Zope server
> when I include an element '<input type="file" .../>' in my form.[/color]
[...]

*Surely* Zope has a standard way of doing this. Try a Zope list?


John

[Tim Roberts]

Re: How to parse multi-part content

Dave Kuhlman <dkuhlman@rexx. com> wrote:[color=blue]
>
>Suppose that I have content that looks like what I've included at
>the end of this message. Is there something in the standard
>Python library that will help me parse it, break into the parts
>separated by the boundary strings, extract headers from each
>sub-part, etc?
>...
>In case you are curious, this is content posted to my Zope server
>when I include an element '<input type="file" .../>' in my form.[/color]

Actually, you get this because your <form> header has
enctype="multip art/form-data". It happens that file upload only works with
that enctype, but you can use it without a file upload.

That's why cgi.py knows how to parse this. Look at cgi.parse_multi part.
--
- Tim Roberts, timr@probo.com
Providenza & Boekelheide, Inc.

[Dave Kuhlman]

Re: How to parse multi-part content

John J. Lee wrote:
[color=blue]
> Dave Kuhlman <dkuhlman@rexx. com> writes:
> [...][color=green]
>> In case you are curious, this is content posted to my Zope server
>> when I include an element '<input type="file" .../>' in my form.[/color]
> [...]
>
> *Surely* Zope has a standard way of doing this. Try a Zope list?
>[/color]

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).

Dave

--
Dave Kuhlman

http://www.rexx.com/~dkuhlman

[Dave Kuhlman]

Re: How to parse multi-part content

Tim Roberts wrote:
[color=blue]
> Dave Kuhlman <dkuhlman@rexx. com> wrote:[color=green]
>>
>>Suppose that I have content that looks like what I've included at
>>the end of this message. Is there something in the standard
>>Python library that will help me parse it, break into the parts
>>separated by the boundary strings, extract headers from each
>>sub-part, etc?
>>...
>>In case you are curious, this is content posted to my Zope server
>>when I include an element '<input type="file" .../>' in my form.[/color]
>
> Actually, you get this because your <form> header has
> enctype="multip art/form-data". It happens that file upload only works
> with that enctype, but you can use it without a file upload.
>
> That's why cgi.py knows how to parse this. Look at cgi.parse_multi part.[/color]

Ah. A clue. I think you're telling me that it's the CGI
specification that I need to be reading, right? I'll read some of
that.

Per your suggestion, I tried cgi.parse_multi part() and also
class cgi.FieldStorag e. They don't work. Or more correctly, I
don't know how to use them.

I guess I'll have to concede defeat, which in Python-speak means:
"It was easier to write it myself."

Basically, I wrote a little parser class ContentParser which
exposes a method get_content_by_ name. This method returns the
body (what follows two carriage returns, up to the next
boundary line) for a given name, where name is the value of the
"name" field in the line:

Content-Disposition: form-data; name="xschemaFi le"

I was in a bit of a hurry, so my solution (class ContentParser) is
not very elegant. But if anyone needs it, let me know.

And, thanks for the suggestions.

Dave


--
Dave Kuhlman

http://www.rexx.com/~dkuhlman

[Michael Foord]

Re: How to parse multi-part content

Dave Kuhlman <dkuhlman@rexx. com> wrote in message news:<2rp5dmF1c gkbkU1@uni-berlin.de>...[color=blue]
> Tim Roberts wrote:
>[color=green]
> > Dave Kuhlman <dkuhlman@rexx. com> wrote:[color=darkred]
> >>
> >>Suppose that I have content that looks like what I've included at
> >>the end of this message. Is there something in the standard
> >>Python library that will help me parse it, break into the parts
> >>separated by the boundary strings, extract headers from each
> >>sub-part, etc?
> >>...
> >>In case you are curious, this is content posted to my Zope server
> >>when I include an element '<input type="file" .../>' in my form.[/color]
> >
> > Actually, you get this because your <form> header has
> > enctype="multip art/form-data". It happens that file upload only works
> > with that enctype, but you can use it without a file upload.
> >
> > That's why cgi.py knows how to parse this. Look at cgi.parse_multi part.[/color]
>
> Ah. A clue. I think you're telling me that it's the CGI
> specification that I need to be reading, right? I'll read some of
> that.
>
> Per your suggestion, I tried cgi.parse_multi part() and also
> class cgi.FieldStorag e. They don't work. Or more correctly, I
> don't know how to use them.
>
> I guess I'll have to concede defeat, which in Python-speak means:
> "It was easier to write it myself."
>
> Basically, I wrote a little parser class ContentParser which
> exposes a method get_content_by_ name. This method returns the
> body (what follows two carriage returns, up to the next
> boundary line) for a given name, where name is the value of the
> "name" field in the line:
>
> Content-Disposition: form-data; name="xschemaFi le"
>
> I was in a bit of a hurry, so my solution (class ContentParser) is
> not very elegant. But if anyone needs it, let me know.
>
> And, thanks for the suggestions.
>
> Dave[/color]

If you are receiving this data to a python script on a server from an
HTML form (i.e. a cgi) then it's striaghtforward to do.

import cgi
theform = cgi.FieldStorag e()

parses the contents of the form into a dictionary like object.
The HTML form that posted the information will assign each file (or
element of the form) a name.
You can access the saved data ausing :

thedata = theform['name].value

Look under the cgi documentation for other attributes that uploaded
files will have. (Potential pitfall with 'list values' as well, where
several values have the same name - again see the docs to see ways
round this).

Regards,

Fuzzyman
http://www.voidspace.org.uk/atlantib...thonutils.html

[Dave Kuhlman]

Re: How to parse multi-part content

Dave Kuhlman wrote:
[color=blue]
> John J. Lee wrote:
>[color=green]
>> Dave Kuhlman <dkuhlman@rexx. com> writes:
>> [...][color=darkred]
>>> In case you are curious, this is content posted to my Zope server
>>> when I include an element '<input type="file" .../>' in my form.[/color]
>> [...]
>>
>> *Surely* Zope has a standard way of doing this. Try a Zope list?
>>[/color]
>
> 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).
>[/color]

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_met hod(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