在发送SOAP请求作为HTTP POST时没有获得SOAPAction头错误

Posted 2021-04-02

由于一些项目限制，我在SOAPUI中发送SOAP请求作为HTTP POST。我的要求在这里：

POST httplink HTTP/1.1
Accept-Encoding: gzip,deflate
Content-Type: text/xml;charset=UTF-8
SOAPAction: "urn:HPD_IncidentInterface_WS/HelpDesk_Query_Service"
Content-Length: 725
Host: itsm-mt-dev
Connection: Keep-Alive
User-Agent: Apache-HttpClient/4.1.1 (java 1.5)


<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:urn="urn:HPD_IncidentInterface_WS">
   <soapenv:Header>
      <urn:AuthenticationInfo>
         <urn:userName>XXXXXXXXXX</urn:userName>
         <urn:password>XXXXXXXXX</urn:password>
         <!--Optional:-->
         <urn:authentication>?</urn:authentication>
         <!--Optional:-->
         <urn:locale>?</urn:locale>
         <!--Optional:-->
         <urn:timeZone>?</urn:timeZone>
      </urn:AuthenticationInfo>
   </soapenv:Header>
   <soapenv:Body>
      <urn:HelpDesk_Query_Service>
         <urn:Incident_Number>XXXXXXXXXX</urn:Incident_Number>
      </urn:HelpDesk_Query_Service>
   </soapenv:Body>
</soapenv:Envelope>

虽然我已经设置了SOAPAction标头，但我仍然没有收到SOAPAction标头错误。

回复如下：

<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
   <soapenv:Body>
      <soapenv:Fault>
         <faultcode xmlns:ns1="http://xml.apache.org/axis/">ns1:Client.NoSOAPAction</faultcode>
         <faultstring>no SOAPAction header!</faultstring>
         <detail>
            <ns2:hostname xmlns:ns2="http://xml.apache.org/axis/">itsm-mt-dev</ns2:hostname>
         </detail>
      </soapenv:Fault>
   </soapenv:Body>
</soapenv:Envelope>

任何人都可以建议我在这里做什么？

答案

看起来您发送的SOAPAction标头不正确。查看WSDL并找出正在测试的服务的soapAction元素的值。

在WSDL中查找类似于<soap:operation soapAction="http://example.com/GetLastTradePrice"/>的行。

另一答案

request.Headers.Add( "SOAPAction", YOUR SOAP ACTION );

另一答案

我试图在ynneh建议的答案中添加注释，但代码不可读。他的答案很有用，但太短了。

创建一个httpwebrequest然后添加标题。以下是完整示例：

    Dim bytSoap = System.Text.Encoding.UTF8.GetBytes("soap content")

        Dim wrRequest As HttpWebRequest = HttpWebRequest.Create("xxxxx")

        wrRequest.Headers.Add("your soap header", "xxx")

        wrRequest.ContentType = "text/xml; charset=utf-8"
        wrRequest.Method = "POST"
        wrRequest.ContentLength = bytSoap.Length

        Dim sDataStream As Stream = wrRequest.GetRequestStream()
        sDataStream.Write(bytSoap, 0, bytSoap.Length)
        sDataStream.Close()

        Dim wrResponse As WebResponse = wrRequest.GetResponse()
        sDataStream = wrResponse.GetResponseStream()
        Dim srReader As StreamReader = New StreamReader(sDataStream)
        Dim strResult As String = srReader.ReadToEnd()
        txtResult.Text = strResult
        sDataStream.Close()
        srReader.Close()
        wrResponse.Close()

另一答案

//德尔福

var
 xmlhttp : IXMLHTTPRequest; // unit MSXML2_TLB
begin
 xmlhttp := CoXMLHTTP.Create;
 xmlhttp.open('POST', Edit7.Text, False, EmptyParam, EmptyParam);
 xmlhttp.setRequestHeader('SOAPAction','POST')