java 服务器无法识别 HTTP 标头 SOAPAction 的值：
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/26248323/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Server did not recognize the value of HTTP Header SOAPAction:
提问by user1050619
I'm trying to call a SOAP webservice using SAAJ for this WSDL. I submitted the request for this WSDL here http://www.webservicex.net/ws/WSDetails.aspx?WSID=64and it generated the following SOAP requests..
我正在尝试为此 WSDL 使用 SAAJ 调用 SOAP web 服务。我在这里提交了对这个 WSDL 的请求http://www.webservicex.net/ws/WSDetails.aspx?WSID=64，它生成了以下 SOAP 请求。
<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns: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>
<GetGeoIP xmlns="http://www.webservicex.net/">
<IPAddress>string</IPAddress>
</GetGeoIP>
</soap:Body>
</soap:Envelope>
Using SAAJ I generated the same SOAP requests and submitted it but get this error - Server did not recognize the value of HTTP Header SOAPAction: .
使用 SAAJ，我生成了相同的 SOAP 请求并提交了它，但收到此错误 - 服务器无法识别 HTTP 标头 SOAPAction: 的值。
<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema"><SOAP-ENV:Header/>
<SOAP-ENV:Body>
<GetGeoIP xmlns="http://www.webservicex.net/">
<IPAddress>SUNW</IPAddress>
</GetGeoIP>
</SOAP-ENV:Body>
</SOAP-ENV:Envelope>
Java Code:
Java代码：
package newpackage;
import javax.xml.namespace.QName;
import javax.xml.soap.*;
public class SOAPClientSAAJ {
public static void main(String args[]) throws Exception {
// Create SOAP Connection
SOAPConnectionFactory soapConnectionFactory = SOAPConnectionFactory.newInstance();
SOAPConnection soapConnection = soapConnectionFactory.createConnection();
// Send SOAP Message to SOAP Server
String url = "http://www.webservicex.net/geoipservice.asmx";
SOAPMessage soapResponse = soapConnection.call(createSOAPRequest(), url);
// print SOAP Response
System.out.print("Response SOAP Message:");
soapResponse.writeTo(System.out);
soapConnection.close();
}
private static SOAPMessage createSOAPRequest() throws Exception {
MessageFactory messageFactory = MessageFactory.newInstance();
SOAPMessage soapMessage = messageFactory.createMessage();
SOAPPart soapPart = soapMessage.getSOAPPart();
String serverURI = "http://www.w3.org/2001/XMLSchema";
// SOAP Envelope
SOAPEnvelope envelope = soapPart.getEnvelope();
envelope.addNamespaceDeclaration("xsd", serverURI);
SOAPBody body = soapMessage.getSOAPBody();
QName bodyName = new QName("http://www.webservicex.net/", "GetGeoIP" );
SOAPBodyElement bodyElement = body.addBodyElement(bodyName);
QName name = new QName("IPAddress");
SOAPElement symbol = bodyElement.addChildElement(name);
symbol.addTextNode("SUNW");
soapMessage.saveChanges();
/* Print the request message */
System.out.print("Request SOAP Message:");
soapMessage.writeTo(System.out);
System.out.println();
return soapMessage;
}
}
回答by Renjith
Add
添加
MimeHeaders headers = soapMessage.getMimeHeaders();
headers.addHeader("SOAPAction", "http://www.webservicex.net/" + "GetGeoIP");
回答by ankit
If you are using Marshelling and Unmarsheling, Just implement WebseviceMessageCallbackinterface like:
如果您使用 Marshelling 和 Unmarsheling，只需实现如下WebseviceMessageCallback接口：
public class SOAPConnector extends WebServiceGatewaySupport
{
public Object callWebService( String url, Object request )
{
return getWebServiceTemplate().marshalSendAndReceive( url, request,
webServiceMessage -> {
(( SoapMessage )webServiceMessage).setSoapAction(
"https://www.w3schools.com/xml/CelsiusToFahrenheit" );
} );
}
}
Otherwise proceed with: https://stackoverflow.com/a/26253141/6097074
