C # implements SOAP call WebService

Recently wrote a SOA service, began to feel that others get my service address, and then directly add the reference can be used, the result "Daniel" told No.

Let me write a SOAP call service sample, I am a bit stunned, because did not do this, so I got a demo, and then learn the next.

Learn as follows:

There is an object in. NET: WebRequest it can request the service directly in the background method

The first step

varWebRequest = (HttpWebRequest) webrequest.create ( This. Uri); WebRequest.Headers.Add ("SOAPAction", String.Format ("\ "{0}\"", This. SOAPAction)); Webrequest.contenttype="text/xml;charset=\ "Utf-8\""; webrequest.accept="Text/xml"; Webrequest.method="POST"; Webrequest.credentials= This. Credentials;

A: in the above code, there is a soapaction, this is when you deploy a good service in IIS, access to services, such as:

The picture tells the user: SOAPAction: "Http://tempuri.org/ProcessFlowRequest"

B:webrequest.credentials = this. Credentials;

Is the credential that invokes the service

Step Two

With this understanding, you need to splice the SOAP message that you see in the XML for the SOAP request.

<?XML version= ' 1.0 ' encoding= ' utf-8 '?><Soap:envelopeXmlns:xsi= ' http://www.w3.org/2001/XMLSchema-instance 'xmlns:xsd= ' Http://www.w3.org/2001/XMLSchema 'Xmlns:soap= ' http://schemas.xmlsoap.org/soap/envelope/'>  <Soap:body>    <{0} xmlns= ' {1} '>{2}</{0}>  </Soap:body></Soap:envelope>

Replace the corresponding information in the picture with the corresponding location of {X}, the information stitching is complete!

Step Three

            varWebRequest = (HttpWebRequest) webrequest.create ( This.            Uri); WEBREQUEST.HEADERS.ADD ("SOAPAction", String.Format ("\ "{0}\"", This.            SOAPAction)); Webrequest.contenttype="text/xml;charset=\ "Utf-8\""; Webrequest.accept="Text/xml"; Webrequest.method="POST"; Webrequest.credentials= This.            Credentials; //Write request soap Information            using(varRequeststream =Webrequest.getrequeststream ()) {                using(varTextWriter =NewStreamWriter (Requeststream)) {                    varEnvelope = Soaphelper.makeenvelope ( This. SOAPAction, This.                Arguments.toarray ()); }            }            //gets the SOAP request returned            returnWebrequest.getresponse ();

This will get the xml! to return the request.

Actually used to know, the original is very simple!

In the description of a usage, when called, the 404 error will be reported, the exception message is thrown: The server failed to recognize the HTTP header SOAPAction

Workaround:

Add an attribute to the WebService class of. NET (that is, the class under the. asmx file) [SoapDocumentService (Routingstyle=soapserviceroutingstyle.requestelement)]

Examples:

Baidu Network disk: Http://pan.baidu.com/s/1hquuXHa

csdn:http://download.csdn.net/detail/hater22/7490147

C # implements SOAP call WebService