肥皂水SOAP調用對Python

Question

我很絕望......不能在我的代碼用肥皂水......我試試這個

#!/usr/bin/python2 

import suds 
url = "http://scsadx02:8080/lrs/webconnect/vpsx?trid=vpsx" 
client = suds.client.Client(url) 

Server = client.factory.create("ns0:string") 
Serverin = client.factory.create("ns0:string") 
UserID = client.factory.create("ns0:string") 
Password = client.factory.create("ns0:string") 
NewPassword = client.factory.create("ns0:string") 
Server.value = 'SPLVSVK2' 
Serverin.value = 'SPLVPSK2' 
UserID.value = 'serviceudp' 
Password.value = 'dpuecivres' 
NewPassword.value = '' 

#method I have to call 
#Suds (https://fedorahosted.org/suds/) version: 0.4 GA build: R699-20100913 
# 
#Service (VPSXService) tns="http://www.lrs.com" 
# Prefixes (2) 
#  ns0 = "http://schemas.xmlsoap.org/soap/encoding/" 
#  ns1 = "http://www.lrs.com" 
# Ports (1): 
#  (VPSXPort) 
#   Methods (95): 
#   Logoff(xs:string SessID,) 
#   Logon(xs:string Server, xs:string UserID, xs:string Password, xs:string NewPassword,) 

client.service.Logon(Server, UserID, Password, NewPassword) 

，但我得到這個錯誤...

[davide@archy PYTHON]$ ./testina.py 
    Traceback (most recent call last): 
    File "./testina.py", line 31, in <module> 
    client.service.Logon(Server, UserID, Password, NewPassword) 
    File "/usr/lib/python2.7/site-packages/suds/client.py", line 542, in __call__ 
    return client.invoke(args, kwargs) 
    File "/usr/lib/python2.7/site-packages/suds/client.py", line 602, in invoke 
    result = self.send(soapenv) 
    File "/usr/lib/python2.7/site-packages/suds/client.py", line 637, in send 
    reply = transport.send(request) 
    File "/usr/lib/python2.7/site-packages/suds/transport/https.py", line 64, in send 
    return HttpTransport.send(self, request) 
    File "/usr/lib/python2.7/site-packages/suds/transport/http.py", line 77, in send 
    fp = self.u2open(u2request) 
    File "/usr/lib/python2.7/site-packages/suds/transport/http.py", line 118, in u2open 
    return url.open(u2request, timeout=tm) 
    File "/usr/lib/python2.7/urllib2.py", line 404, in open 
    response = self._open(req, data) 
    File "/usr/lib/python2.7/urllib2.py", line 422, in _open 
    '_open', req) 
    File "/usr/lib/python2.7/urllib2.py", line 382, in _call_chain 
    result = func(*args) 
    File "/usr/lib/python2.7/urllib2.py", line 1214, in http_open 
    return self.do_open(httplib.HTTPConnection, req) 
    File "/usr/lib/python2.7/urllib2.py", line 1184, in do_open 
    raise URLError(err) 
    urllib2.URLError: <urlopen error [Errno -2] Name or service not known> 

我該如何解決？

Answer 1

嘗試

C:\Python27>python -c "import urllib2; print urllib2.urlopen('http://scsadx02:8080/lrs/webconnect/vpsx?trid=vpsx').read()" 

andimport os; os.environ['http_proxy']=''導入其他模塊之前。