Re: [Zope] Re: Sending XML straight down to a zope

2006-10-13 Thread Chris Withers 

Jonathan wrote:



As an alternative strategy, how about:

i) set up a url just for this problem user
ii) us apache rewrite to redirect to a non-zope cgi script
iii) use the cgi script to 'fix' the xml
iv) do a redirect with the fixed xml to a 'zope' url


Being all Nu Skool about this, I wonder if there's a wsgi thingy that 
could help here?


Chris

--
Simplistix - Content Management, Zope  Python Consulting
   - http://www.simplistix.co.uk

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Re: [Zope] Re: Sending XML straight down to a zope

2006-10-12 Thread Peter Bengtsson 



Tres Seaver wrote:

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Peter Bengtsson wrote:

I'm trying to send an XML straight into Zope without specifying it as
a parameter and with a Content-Length. It seems that Zope's mapply
function or whatever it's called digests the raw http body and tries
to turn it into parameters?


Here's the code on the Zope server (uploadExpenseXML()):
def uploadExpenseXML(self):
   return str(self.REQUEST.form.keys())


Here's the dummy code that sends the XML into my Zope:

xml_content = open('validxmlfile.xml').read()
http = httplib.HTTP(localhost, 8080)
http.putrequest(POST, /uploadExpenseXML)
http.putheader(User-Agent, Simple)
http.putheader(Host, localhost)
http.putheader(Content-Length, %d % len(xml_content))
http.endheaders()
http.send(xml_content)
reply, message, headers = http.getreply()
print http.getfile().read()


The result I get is:
['?xml version']

If I debug the value of that single REQUEST.form variable, it starts
like this:
'1.0 encoding=ISO-8859-1 standalone=yes ?\nXMLExpense
Version=1.0

Obviously, one solution would be to ask the XML sending company to not
post it like this but instead post it by parameter which I know will
work.
But, what if I can't change their minds?



PS. When faced with the same problem a long time ago I ended up
writing a mod_python app running one a different port that converted
the http post from mod_python into a parameter based http post. I
don't want to have to do that again.


Stock Zope can only handle POST request with XML payloads if they
conform to the XMLRPC spec:  they must have the 'Content-type' header of
'text/xml', and the document must be encoded as an xmlrpc request.

If you can get the vendor to use a PUT request instead of a POST, you
could process the result inside a custom object'c 'PUT' method.
Otherwise, you are going to need to create a custom derivative of
ZPublisher.HTTPRequest, and override its 'processInputs' to handle the
vendor's non-standard dump.  You could make that server listen on a
different port, for instance (I'm guessing you can tell them what URL to
POST to, right?)



Cool. Thanks for the suggestion. I've tried to make them change the way 
they post it to me and if that's not possible I'll consider your advice 
about either PUT or digging into ZPublisher which I was hoping not to 
have to do.




Tres.
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--
Peter Bengtsson,
work www.fry-it.com
home www.peterbe.com
hobby www.issuetrackerproduct.com
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Re: [Zope] Re: Sending XML straight down to a zope

2006-10-12 Thread Jonathan 


- Original Message - 
From: Peter Bengtsson [EMAIL PROTECTED]

To: Tres Seaver [EMAIL PROTECTED]
Cc: zope@zope.org
Sent: Thursday, October 12, 2006 1:09 PM
Subject: Re: [Zope] Re: Sending XML straight down to a zope





Tres Seaver wrote:

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Peter Bengtsson wrote:

I'm trying to send an XML straight into Zope without specifying it as
a parameter and with a Content-Length. It seems that Zope's mapply
function or whatever it's called digests the raw http body and tries
to turn it into parameters?


Here's the code on the Zope server (uploadExpenseXML()):
def uploadExpenseXML(self):
   return str(self.REQUEST.form.keys())


Here's the dummy code that sends the XML into my Zope:

xml_content = open('validxmlfile.xml').read()
http = httplib.HTTP(localhost, 8080)
http.putrequest(POST, /uploadExpenseXML)
http.putheader(User-Agent, Simple)
http.putheader(Host, localhost)
http.putheader(Content-Length, %d % len(xml_content))
http.endheaders()
http.send(xml_content)
reply, message, headers = http.getreply()
print http.getfile().read()


The result I get is:
['?xml version']

If I debug the value of that single REQUEST.form variable, it starts
like this:
'1.0 encoding=ISO-8859-1 standalone=yes ?\nXMLExpense
Version=1.0

Obviously, one solution would be to ask the XML sending company to not
post it like this but instead post it by parameter which I know will
work.
But, what if I can't change their minds?



PS. When faced with the same problem a long time ago I ended up
writing a mod_python app running one a different port that converted
the http post from mod_python into a parameter based http post. I
don't want to have to do that again.


Stock Zope can only handle POST request with XML payloads if they
conform to the XMLRPC spec:  they must have the 'Content-type' header of
'text/xml', and the document must be encoded as an xmlrpc request.

If you can get the vendor to use a PUT request instead of a POST, you
could process the result inside a custom object'c 'PUT' method.
Otherwise, you are going to need to create a custom derivative of
ZPublisher.HTTPRequest, and override its 'processInputs' to handle the
vendor's non-standard dump.  You could make that server listen on a
different port, for instance (I'm guessing you can tell them what URL to
POST to, right?)



Cool. Thanks for the suggestion. I've tried to make them change the way 
they post it to me and if that's not possible I'll consider your advice 
about either PUT or digging into ZPublisher which I was hoping not to have 
to do.


As an alternative strategy, how about:

i) set up a url just for this problem user
ii) us apache rewrite to redirect to a non-zope cgi script
iii) use the cgi script to 'fix' the xml
iv) do a redirect with the fixed xml to a 'zope' url

This would allow you to run a standard zope installation (running a hacked 
installation is a maintenance nightmare), and if the problem user ever gets 
their act together, you just change the apache rewrite rule so that they are 
delivered straight to your zope application.


Just a thought...


Jonathan 



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Re: [Zope] Re: Sending XML straight down to a zope

2006-10-12 Thread Jonathan 


- Original Message - 
From: Jonathan [EMAIL PROTECTED]

To: Peter Bengtsson [EMAIL PROTECTED]
Cc: zope@zope.org
Sent: Thursday, October 12, 2006 1:21 PM
Subject: Re: [Zope] Re: Sending XML straight down to a zope




- Original Message - 
From: Peter Bengtsson [EMAIL PROTECTED]

To: Tres Seaver [EMAIL PROTECTED]
Cc: zope@zope.org
Sent: Thursday, October 12, 2006 1:09 PM
Subject: Re: [Zope] Re: Sending XML straight down to a zope





Tres Seaver wrote:

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Peter Bengtsson wrote:

I'm trying to send an XML straight into Zope without specifying it as
a parameter and with a Content-Length. It seems that Zope's mapply
function or whatever it's called digests the raw http body and tries
to turn it into parameters?


Here's the code on the Zope server (uploadExpenseXML()):
def uploadExpenseXML(self):
   return str(self.REQUEST.form.keys())


Here's the dummy code that sends the XML into my Zope:

xml_content = open('validxmlfile.xml').read()
http = httplib.HTTP(localhost, 8080)
http.putrequest(POST, /uploadExpenseXML)
http.putheader(User-Agent, Simple)
http.putheader(Host, localhost)
http.putheader(Content-Length, %d % len(xml_content))
http.endheaders()
http.send(xml_content)
reply, message, headers = http.getreply()
print http.getfile().read()


The result I get is:
['?xml version']

If I debug the value of that single REQUEST.form variable, it starts
like this:
'1.0 encoding=ISO-8859-1 standalone=yes ?\nXMLExpense
Version=1.0

Obviously, one solution would be to ask the XML sending company to not
post it like this but instead post it by parameter which I know will
work.
But, what if I can't change their minds?



PS. When faced with the same problem a long time ago I ended up
writing a mod_python app running one a different port that converted
the http post from mod_python into a parameter based http post. I
don't want to have to do that again.


Stock Zope can only handle POST request with XML payloads if they
conform to the XMLRPC spec:  they must have the 'Content-type' header of
'text/xml', and the document must be encoded as an xmlrpc request.

If you can get the vendor to use a PUT request instead of a POST, you
could process the result inside a custom object'c 'PUT' method.
Otherwise, you are going to need to create a custom derivative of
ZPublisher.HTTPRequest, and override its 'processInputs' to handle the
vendor's non-standard dump.  You could make that server listen on a
different port, for instance (I'm guessing you can tell them what URL to
POST to, right?)



Cool. Thanks for the suggestion. I've tried to make them change the way 
they post it to me and if that's not possible I'll consider your advice 
about either PUT or digging into ZPublisher which I was hoping not to 
have to do.


As an alternative strategy, how about:

i) set up a url just for this problem user
ii) us apache rewrite to redirect to a non-zope cgi script
iii) use the cgi script to 'fix' the xml
iv) do a redirect with the fixed xml to a 'zope' url


Sorry, 'redirect' is the wrong terminology, you would have to do an RPC call 
or use something like wget, urllib, etc to feed the new xml data to your 
zope app, and then return the response (or a response) to the originator.



Jonathan


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[Zope] Re: Sending XML straight down to a zope

2006-10-12 Thread Tres Seaver 
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Dieter Maurer wrote:
 Chris Withers wrote at 2006-10-12 08:35 +0100:
 Peter Bengtsson wrote:
 xml_content = open('validxmlfile.xml').read()
 http = httplib.HTTP(localhost, 8080)
 http.putrequest(POST, /uploadExpenseXML)
 That's not really a valid transaction...
 
 Why not?
 
 
 The real problem is that Zope wrongfully assumes that each
 POST with content type text/xml is an XML-RPC request.
 
 That's wrong at least since XML is more widely used on the browser
 side (e.g. XForms).

I think XForms uses PUT to save the model back to the server, rather
than POST.

WRT my proposed earlier solution:  Adding a custom server type does not
necessarily require deploying a hacked version of ZPublisher:  it is
possible to register new server types in Zope2 products (e.g., the
original ClockServer product).


Tres.
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[Zope] Re: Sending XML straight down to a zope

2006-10-11 Thread Tres Seaver 
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Peter Bengtsson wrote:
 I'm trying to send an XML straight into Zope without specifying it as
 a parameter and with a Content-Length. It seems that Zope's mapply
 function or whatever it's called digests the raw http body and tries
 to turn it into parameters?
 
 
 Here's the code on the Zope server (uploadExpenseXML()):
 def uploadExpenseXML(self):
return str(self.REQUEST.form.keys())
 
 
 Here's the dummy code that sends the XML into my Zope:
 
 xml_content = open('validxmlfile.xml').read()
 http = httplib.HTTP(localhost, 8080)
 http.putrequest(POST, /uploadExpenseXML)
 http.putheader(User-Agent, Simple)
 http.putheader(Host, localhost)
 http.putheader(Content-Length, %d % len(xml_content))
 http.endheaders()
 http.send(xml_content)
 reply, message, headers = http.getreply()
 print http.getfile().read()
 
 
 The result I get is:
 ['?xml version']
 
 If I debug the value of that single REQUEST.form variable, it starts
 like this:
 '1.0 encoding=ISO-8859-1 standalone=yes ?\nXMLExpense
 Version=1.0
 
 Obviously, one solution would be to ask the XML sending company to not
 post it like this but instead post it by parameter which I know will
 work.
 But, what if I can't change their minds?
 
 
 
 PS. When faced with the same problem a long time ago I ended up
 writing a mod_python app running one a different port that converted
 the http post from mod_python into a parameter based http post. I
 don't want to have to do that again.

Stock Zope can only handle POST request with XML payloads if they
conform to the XMLRPC spec:  they must have the 'Content-type' header of
'text/xml', and the document must be encoded as an xmlrpc request.

If you can get the vendor to use a PUT request instead of a POST, you
could process the result inside a custom object'c 'PUT' method.
Otherwise, you are going to need to create a custom derivative of
ZPublisher.HTTPRequest, and override its 'processInputs' to handle the
vendor's non-standard dump.  You could make that server listen on a
different port, for instance (I'm guessing you can tell them what URL to
POST to, right?)


Tres.
- --
===
Tres Seaver  +1 202-558-7113  [EMAIL PROTECTED]
Palladion Software   Excellence by Designhttp://palladion.com
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