Question

我目前正在为使用Axis2 WS-Security

的应用程序开发python webservice

简化的相关代码是

from SOAPpy import SOAPProxy
from SOAPpy import WSDL

file = 'path/to/my/file?wsdl'
server = WSDL.Proxy(file)

server.foo(bar)

这样做的时候我得到：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python27\lib\site-packages\SOAPpy\Client.py", line 471, in __call__
    return self.__r_call(*args, **kw)
  File "C:\Python27\lib\site-packages\SOAPpy\Client.py", line 493, in __r_call
    self.__hd, self.__ma)
  File "C:\Python27\lib\site-packages\SOAPpy\Client.py", line 407, in __call
    raise p
SOAPpy.Types.faultType: <Fault soapenv:Client: WSDoAllReceiver: Incoming message
 does not contain required Security header: >

读取axis2 WS-security的文档和提供webservices的应用程序我猜测它要求我在

之类的用户令牌认证

<wsse:Security xmlns:wsse="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd" soapenv:mustUnderstand="1">
<wsu:Timestamp xmlns:wsu="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-utility-1.0.xsd" wsu:Id="Timestamp-12468716">
<wsu:Created>
2008-06-23T13:17:13.841Z
</wsu:Created>
<wsu:Expires>
2008-06-23T13:22:13.841Z
</wsu:Expires>
</wsu:Timestamp>
<wsse:UsernameToken xmlns:wsu="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-utility-1.0.xsd" wsu:Id="UsernameToken-31571602">
<wsse:Username>
alice
</wsse:Username>
<wsse:Password Type="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-username-token-profile-1.0#PasswordText">
bobPW
</wsse:Password>
</wsse:UsernameToken>
</wsse:Security>

在执行请求时，问：我如何将其附加到SOAPpy请求？

Answer 1

查看有关向肥皂请求添加标题的文档（https://svn.origo.ethz.ch/playmobil/trunk/contrib/pywebsvcs/SOAPpy/docs/UsingHeaders.txt）。这样，您就可以将自己的自定义标头添加到任何请求中。

希望这有帮助

SOAPpy - 如何传递安全标头？

1 个答案: