Django soaplib错误

Question

我正在尝试使用Django在一些教程之后创建一个“Hello World”Web服务，但我一遍又一遍地遇到同样的障碍。我已经定义了一个view.py和soaplib_handler.py：

view.py：

from soaplib_handler import DjangoSoapApp, soapmethod, soap_types

class HelloWorldService(DjangoSoapApp):

    __tns__ = 'http://saers.dk/soap/'

    @soapmethod(_returns=soap_types.Array(soap_types.String))
    def hello(self):
      return "Hello World"

soaplib_handler.py：

from soaplib.wsgi_soap import SimpleWSGISoapApp
from soaplib.service import soapmethod
from soaplib.serializers import primitive as soap_types

from django.http import HttpResponse


class DjangoSoapApp(SimpleWSGISoapApp):

    def __call__(self, request):
        django_response = HttpResponse()
        def start_response(status, headers):
            status, reason = status.split(' ', 1)
            django_response.status_code = int(status)
            for header, value in headers:
                django_response[header] = value
        response = super(SimpleWSGISoapApp, self).__call__(request.META, start_response)
        django_response.content = "\n".join(response)

        return django_response

似乎“响应=超级......”这一行给了我麻烦。当我加载在url.py中映射的/hello_world/services.wsdl时，我得到：

/hello_world/service.wsdl上的AttributeError 'module'对象没有属性'tostring'

有关完整的错误消息，请参阅此处： http://saers.dk:8000/hello_world/service.wsdl

您对我收到此错误的原因有什么建议吗？ ElementTree在哪里定义？

Answer 1

@zdmytriv该行

soap_app_response = super(BaseSOAPWebService, self).__call__(environ, start_response)

应该看起来像

soap_app_response = super(DjangoSoapApp, self).__call__(environ, start_response)

那么你的例子就可以了。

Answer 2

不确定这是否会解决您的问题，但函数问候中的装饰器说它假设返回一个String Array，但实际上是在返回一个String

尝试_returns = soap_types.String而不是

雷

Answer 3

从我的服务中复制/粘贴：

# SoapLib Django workaround: http://www.djangosnippets.org/snippets/979/
class DumbStringIO(StringIO):
    """ Helper class for BaseWebService """
    def read(self, n): 
        return self.getvalue()

class DjangoSoapApp(SimpleWSGISoapApp):
    def __call__(self, request):
        """ Makes Django request suitable for SOAPlib SimpleWSGISoapApp class """

        http_response = HttpResponse()

        def start_response(status, headers):
            status, reason = status.split(' ', 1)
            http_response.status_code = int(status)

            for header, value in headers:
                http_response[header] = value

        environ = request.META.copy()
        body = ''.join(['%s=%s' % v for v in request.POST.items()])
        environ['CONTENT_LENGTH'] = len(body)
        environ['wsgi.input'] = DumbStringIO(body)
        environ['wsgi.multithread'] = False

        soap_app_response = super(BaseSOAPWebService, self).__call__(environ, start_response)

        http_response.content = "\n".join(soap_app_response)

        return http_response

Django snippet有一个错误。阅读该网址的最后两条评论。