soap WSDL客户端身份验证

Question

我正在尝试使用WSDL和身份验证进行简单的SOAP调用，但不知道如何将凭据传递给调用。使用WSDL是否可以做到这一点，或者我是否应该以其他方式做到这一点?

from SOAPpy import WSDL
WSDLFILE = 'link/to/wsdl/file'
server = WSDL.Proxy(WSDLFILE)
result = server.methodCall()

现在，我得到这个错误:

File "/usr/lib/pymodules/python2.7/SOAPpy/Client.py", line 263, in call
raise HTTPError(code, msg)
SOAPpy.Errors.HTTPError: <HTTPError 401 Unauthorized>

Answer 1

这是一个相当老的问题，但由于它在google搜索"SOAPpy身份验证"时首先显示-我将留下我的发现，这样也许你不必敲你的头10个小时。回馈社会。

谷歌没有提供太多帮助，但这个示例(http://code.activestate.com/recipes/444758-how-to-add-cookiesheaders-to-soappy-calls/)让我编写自己的帮助器(版本0.0 beta)

from SOAPpy import (
    WSDL,
    HTTPTransport,
    Config,
    SOAPAddress,
)
import urllib2
import base64
class AuthenticatedTransport(HTTPTransport):
    _username = None
    _password = None
    def call(self, addr, data, namespace, soapaction = None, encoding = None, http_proxy = None, config = Config, timeout = 10):
        if not isinstance(addr, SOAPAddress):
            addr = SOAPAddress(addr, config)
        content_type = 'text/xml'
        if encoding != None:
            content_type += '; charset="%s"' % encoding
        encoded_auth = None
        if ( isinstance(AuthenticatedTransport._username, str) != False ):
            if ( isinstance(AuthenticatedTransport._password, str) == False ):
                AuthenticatedTransport._password = ""
            encoded_auth = base64.b64encode('%s:%s' % (self._username, self._password))
        this_request = None
        if ( encoded_auth is not None ):
            this_request = urllib2.Request(
                url=addr.proto + "://" + addr.host + addr.path,
                data=data,
                headers={
                    "Content-Type":content_type,
                    "SOAPAction":"%s" % soapaction,
                    "Authorization":"Basic %s" % encoded_auth
                }
            )
        else:
            this_request = urllib2.Request(
                url=addr.proto + "://" + addr.host + addr.path,
                data=data,
                headers={
                    "Content-Type":content_type,
                    "SOAPAction":"%s" % soapaction,
                }
            )
        response = urllib2.urlopen(this_request)
        data = response.read()
        # get the new namespace
        if namespace is None:
            new_ns = None
        else:
            new_ns = self.getNS(namespace, data)
        # return response payload
        return data, new_ns
WSDL.Config.strictNamespaces = 0
username = "your_username"
password = "your_password"
WSDLFile = "https://%s:%s@some_host/your_wsdl.php?wsdl" % (username, password)
namespace = "http://futureware.biz/mantisconnect"
proxy = WSDL.Proxy(WSDLFile, namespace=namespace, transport = AuthenticatedTransport(username,password))
print(proxy.mc_get_version()) # This is my WSDL call, your will be different

无论出于何种原因，仅仅使用AuthenticatedTransport类是不够的，WSDL url本身也必须包含用户名和密码。也许这是因为这里由WSDL调用的SOAP包装器正在创建单独的HTTP会话(请记住在SOAP邮件列表中阅读它)。