2:37 p.m.
Janet: I would do some logging on this to see what is going on in
detail. You'll find information on how to do this in the suds
documentation from the site. Also, I've run into bizarre transport
issues with certain versions of urllib2 in the past - rare, but it's
definitely happened and an upgrade fixed things. What version of suds
/ python are you running, by chance?
Also, for basic auth you should be able to simply pass the client
object username/password arguments and Suds will do the right thing
transport-wise. This may be worth another shot (see the notes on buggy
servers and the basic auth RFC).
-Matt
2010/10/8 Janet Valbuena <janet.valbuena(a)tpp.com.au>:
Dear Daniel
Thanks for assistance, but it still doesn’t make sense that I have to pass
the WSDL as parameter to connect to a URL via suds.client.Client where I
will get precisely that same WSDL.
I’m confused.
Thanks anyway. I think I’ll have to get back to SOAPpy unless I hear another
suggestion to fix my problem.
Regards
Janet
On 8/10/10 11:04, "Daniel Rodriguez" <danjrod(a)gmail.com> wrote:
Hi Janet,
I can't really tell you what the realm should be because I haven't found
myself in such situation
You pass the hacked transport (that should be reading the WSDL from the
"fileFromU2" file object) because the standard transport included in suds
assumes a plain HTTP download for WSDL files (unless someone in the list
proves me wrong and can provide a working example, which would be great)
That's why Joshua suggested that you downloaded the file to disk. I only
added the means to avoid disk storage by passing the WSDL file object
directly to a suds compliant transport.
But your problem is that you are apparently unable to download the WSDL file
(therefore the empty "read"). This is where your problem lies and I guess we
can't remotely help you there.
Best regards
Daniel
On Fri, Oct 8, 2010 at 10:37, Janet Valbuena <janet.valbuena(a)tpp.com.au>
wrote:
Hi Daniel
This still doesn’t make sense.
I understand you’re suggesting this:
WSDL = "http://localhost:18080/external/services/DomService?wsdl"
User = “user1”
Password = “pass123”
# Is this OK?
URI = "http://localhost:18080/” <http://localhost:18080/²>
# What should the realm be? Is this OK?
Realm = “external/services/DomService”
import urllib2
auth_handler = urllib2.HTTPBasicAuthHandler()
auth_handler.add_password(realm=Real, ?uri='your_url_without_the_file_name',
user=User, passwd=Password)
opener = urllib2.build_opener(auth_handler)
urllib2.install_opener(opener)
fileFromU2 = urllib2.open(WSDL)
#Note that up to this point, the code doesn’t show any errors but
fileFromU2.read () is an empty string.
# Then I’m supposed to use the transport hack class:
myClient = suds.client.Client(WSDL, transport=HackTransport(fileFromU2),
etc)
# The question is, why should I pass fileFromU2 as a transport parameter,
when what I’m trying to get through calling the Client is precisely that
WSDL?
What I’m really hoping to do (in case I haven’t been clear) is something
like this:
# Assuming the authorisation issue is solved somehow by producing a ‘t’
parameter to pass on here:
myClient = suds.client.Client(WSDL, transport=t)
# Call some of the methods available in the WSDL, for example:
Result = myClient.createAccount({“name”:”janet”, “phone”: “8989898989”,
“address”:”12 Victoria St”, “city”:”Manly”, “email”: “janet(a)blahh.xn--com}-ib7a)
Is perhaps suds.client.Client NOT the class I need?
Thanks again for your time and answers in advance.
Cheers
Janet
On 7/10/10 21:42, "Daniel Rodriguez" <danjrod(a)gmail.com> wrote:
Hi Janet,
The code provided by Joshua lets you download the WSDL to a "file" object.
That's why the last statement of his example is:
print urllib2.open(WSDL).read()
whis is basically first opening a file (open), then reading the content to a
string (read) and printing it to standard output. If you simply do:
fileFromU2 = urllib2.open(WSDL)
you get direct acess to the file object and can pass it to the hacked
transport I provided before.
Best regards
Daniel
On Thu, Oct 7, 2010 at 13:03, Janet Valbuena <janet.valbuena(a)tpp.com.au>
wrote:
Joshua, Daniel
Thanks very much for your quick response.
I tried Joshua’s suggestion but it didn’t work.
I don’t understand what you suggest Daniel. What should fileFromU2 be?
I may be wrong by using the Client method in the first place. ?? The purpose
of my script is to use an API which is set up as a SOAP service.
Why would I need to download the WSDL file and then pass it to the transport
hack when I can’t even get the WSDL and the purpose of client = Client
(WSDL) is to connect to the WSDL itself?
Cheers
Janet
On 7/10/10 09:33, "Daniel Rodriguez" <danjrod(a)gmail.com> wrote:
Hi,
I think the problem is in the connect itself.
The standard transport included in suds assumes that only the SOAP calls may
need authentication or other things. Downloading the WSDL is attempted with
a plain HTTP GET (Not even POST is allowed)
I guess that the transport covers at least 95% of standard situations, but
it didn't for example cover mine, where I wanted to embed the WSDSL as a
string in my code. Of course only with my own transport I could give suds a
"file" object with the content of the string.
It is my feeling that a better transport could be written (we have seen some
examples shared in the mailing list) although some use cases may require
re-thinking how a Client is initialized.
In the case pertaining to Janet, I guess she may download the WSDL file as
you suggest, pass it to the transport hack below. She may then use this
transport during client initialization. Obviously this is only good for 1
WSDL file. (The transport below can also load WSDL content directly from a
string)
?? ?class TransportHack(Transport):
?? ? ? ?def __init__(self, wsdlFile=None, **kwargs):
?? ? ? ? ? ?Transport.__init__(self, **kwargs)
?? ? ? ? ? ?self.wsdlcontent = wsdlFile
?? ? ? ?def open(self, request):
?? ? ? ? ? ?if self.wsdlFile is not None:
?? ? ? ? ? ? ? ?return self.wsdlFile
?? ? ? ? ? ?log.debug('opening: (%s)', request.url)
?? ? ? ? ? ?fp = None
?? ? ? ? ? ?location = request.url.lstrip()
?? ? ? ? ? ?if location.startswith('<?'):
?? ? ? ? ? ? ? ?log.debug('returning url (%s) as StringIO file')
?? ? ? ? ? ? ? ?fp = StringIO(location)
?? ? ? ? ? ?else:
?? ? ? ? ? ? ? ?parsed = urlparse(request.url)
?? ? ? ? ? ? ? ?if parsed.scheme == 'file':
?? ? ? ? ? ? ? ? ? ?log.debug('opening file (%s) with open', parsed.path)
?? ? ? ? ? ? ? ? ? ?try:
?? ? ? ? ? ? ? ? ? ? ? ?fp = open(parsed.path)
?? ? ? ? ? ? ? ? ? ?except Exception, e:
?? ? ? ? ? ? ? ? ? ? ? ?raise TransportError(str(e), 503, StringIO(''))
?? ? ? ? ? ? ? ?else:
?? ? ? ? ? ? ? ? ? ?log.debug('opening scheme (%s) over the network',
parsed.scheme)
?? ? ? ? ? ? ? ? ? ?try:
?? ? ? ? ? ? ? ? ? ? ? ?url = request.url
?? ? ? ? ? ? ? ? ? ? ? ?log.debug('opening (%s)', url)
?? ? ? ? ? ? ? ? ? ? ? ?u2request = u2.Request(url)
?? ? ? ? ? ? ? ? ? ? ? ?self.proxy = self.options.proxy
?? ? ? ? ? ? ? ? ? ? ? ?return self.u2open(u2request)
?? ? ? ? ? ? ? ? ? ?except u2.HTTPError, e:
?? ? ? ? ? ? ? ? ? ? ? ?raise TransportError(str(e), e.code, e.fp)
?? ? ? ? ? ?return fp
Then do something like
myClient = suds.client.Client(wsdl, transport=HackTransport(fileFromU2),
etc)
Best regards
Daniel
On Wed, Oct 6, 2010 at 21:22, Joshua J. Kugler <joshua(a)eeinternet.com>
wrote:
On Wednesday 06 October 2010, Janet Valbuena elucidated thus:
> Hi
>
> I'm new to suds (have just downloaded it after getting frustrated
> with SOAPpy).
>
> I'm trying to connect to a WSDL which requires basic HTTP
> authentication. I have been able to connect to it correctly using PHP
> but I need to use Python.
>
>
> This is my code:
>
> #!/usr/bin/python
>
> Username = "user"
> Password = "pass123"
> from suds.transport.http import HttpAuthenticated
> t = HttpAuthenticated(username=Username, password=Password)
>
> # WSDL is only available in internal network
> WSDL = "http://localhost:18080/external/services/DomService?wsdl"
> from suds.client import Client
> client = Client(WSDL, transport=t)
>
>
>
>
> And this is the error it returns:
>
> Traceback (most recent call last):
> ? File "./wsdltest.py", line 25, in <module>
> ? ? client = Client(wsdl, transport=t)
> ? File "build/bdist.linux-i686/egg/suds/client.py", line 112, in
> __init__ File "build/bdist.linux-i686/egg/suds/reader.py", line 152,
> in open File "build/bdist.linux-i686/egg/suds/wsdl.py", line 136, in
> __init__ File "build/bdist.linux-i686/egg/suds/reader.py", line 79,
> in open File "build/bdist.linux-i686/egg/suds/reader.py", line 101,
> in download File "build/bdist.linux-i686/egg/suds/sax/parser.py",
> line 136, in parse File "/usr/lib/python2.5/xml/sax/expatreader.py",
> line 107, in parse xmlreader.IncrementalParser.parse(self, source)
> ? File "/usr/lib/python2.5/xml/sax/xmlreader.py", line 125, in parse
> ? ? self.close()
> ? File "/usr/lib/python2.5/xml/sax/expatreader.py", line 217, in
> close self.feed("", isFinal = 1)
> ? File "/usr/lib/python2.5/xml/sax/expatreader.py", line 211, in feed
> ? ? self._err_handler.fatalError(exc)
> ? File "/usr/lib/python2.5/xml/sax/handler.py", line 38, in
> fatalError raise exception
> xml.sax._exceptions.SAXParseException: <unknown>:1:0: no element
> found
I would try to download the file with a quick urllib2 script and see
what you get. ?The error is not in the connect (that would raise its
own exception) but in trying to parse what *is* being downloaded.
Something like:
import urllib2
auth_handler = urllib2.HTTPBasicAuthHandler()
auth_handler.add_password(realm='name_of_your_realm',
?? ? ? ? ? ? ? ? ? ? ? ? ?uri='your_url_without_the_file_name',
?? ? ? ? ? ? ? ? ? ? ? ? ?user=User,
?? ? ? ? ? ? ? ? ? ? ? ? ?passwd=Password)
opener = urllib2.build_opener(auth_handler)
urllib2.install_opener(opener)
print urllib2.open(WSDL).read()
j
--
Joshua Kugler
Part-Time System Admin/Programmer
http://www.eeinternet.com - Fairbanks, AK
PGP Key: http://pgp.mit.edu/ ?ID 0x73B13B6A
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