Question

我正在尝试使用ZEEP来执行SOAP请求。

我收到了这个请求：

def status(request, IP):
    URL = "http://" + IP + "/Service"
    session = Session()
    session.auth = HTTPBasicAuth('username', 'password')
    client = Client(URL,
                    transport=Transport(session=session), strict=False)

    response = client.service.Method1(request={'getList'})

    return response

但是我遇到了缺少wsdl的错误：WSDL中的服务定义。

我陷入困境，无法找到更多错误搜索方式。有什么想法吗？

编辑上下文。这是一个有效的硬编码请求，我正在尝试使用zeep：

 def GetList(session, IP):        

        request = '<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/">'
        request = request + '<soapenv:Header>'
        request = request + '<session>' + session + '</session>'
        request = request + ' </soapenv:Header>'
        request = request + ' <soapenv:Body>'
        request = request + '<getList/>'
        request = request + '</soapenv:Body>'
        request = request + '</soapenv:Envelope>'

        request = u"""""" + request + """""".format()

        encoded_request = request.encode('utf-8')

        headers = {"Host": ""+ IP +"",
                   "Content-Type": "text/xml; charset=UTF-8",
                   "Content-Length": str(len(encoded_request)),
                   "SOAPAction": ""}

        response = requests.post(url="http://"+ url +"/Service",
                                 headers=headers,
                                 data=encoded_request,
                                 verify=False)

        return response.content

编辑2.添加wsdl

<?xml version="1.0" encoding="UTF-8"?>    
-<definitions xmlns="http://schemas.xmlsoap.org/wsdl/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:wsdlsoap="http://schemas.xmlsoap.org/wsdl/soap/" xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/" xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/" xmlns:intf="http://xml.apache.org/axis/wsdd/ " xmlns:impl="http://xml.apache.org/axis/wsdd/ " xmlns:apachesoap="http://xml.apache.org/xml-soap" targetNamespace="http://xml.apache.org/axis/wsdd/ ">    
-<wsdl:types>    
-<schema xmlns="http://www.w3.org/2001/XMLSchema" targetNamespace="http://xml.apache.org/axis/wsdd/ ">    
<element type="xsd:anyType" name="parseMessage"/>    
<element type="xsd:anyType" name="parseMessageReturn"/>    
</schema>    
</wsdl:types>    
-<message name="parseMessageResponse">    
<part name="parseMessageReturn" element="impl:parseMessageReturn"/>    
</message>    
-<message name="parseMessageRequest">    
<part name="part" element="impl:parseMessage"/>    
</message>    
-<portType name="MessageHandlerProxy">    
-<operation name="parseMessage">    
<input name="parseMessageRequest" message="impl:parseMessageRequest"/>    
<output name="parseMessageResponse" message="impl:parseMessageResponse"/>    
</operation>    
</portType>    
-<binding type="impl:HandlerProxy" name="ServiceBinding">    
<wsdlsoap:binding transport="http://schemas.xmlsoap.org/soap/http" style="document"/>    
-<operation name="parseMessage">    
<wsdlsoap:operation soapAction=""/>    
-<input name="parseMessageRequest">    
<wsdlsoap:body use="literal" namespace="http://xml.apache.org/axis/wsdd/ "/>    
</input>    
-<output name="parseMessageResponse">    
<wsdlsoap:body use="literal" namespace="http://xml.apache.org/axis/wsdd/ "/>    
</output>    
</operation>    
</binding>    
-<service name="ProxyService">    
-<port name="Service" binding="impl:ServiceBinding">    
<wsdlsoap:address location="http://ip.ip.ip.ip/Service"/>    
</port>    
</service>    
</definitions>

Answer 1

您必须在网址变量的末尾添加?wsdl。

此外，您拨打电话的方式也不正确。对getList的调用应如下所示：

def status(request, IP):
    URL = "http://" + IP + "/Service?wsdl"
    session = Session()
    session.auth = HTTPBasicAuth('username', 'password')
    client = Client(URL,
                    transport=Transport(session=session), strict=False)

    response = client.service.getList([add the arguments here])

    return response

缺少wsdl：WSDL中的服务定义

1 个答案: