Question

String url = "http://e2e-soaservices:44000/3.1/StandardDocumentService?wsdl";
//createSOAPRequest(); 
SOAPMessage soapResponse = soapConnection.call(createSOAPRequest(), url);

createSOAPRequest方法中的

： -

MessageFactory messageFactory = MessageFactory.newInstance();
SOAPMessage soapMessage = messageFactory.createMessage();
SOAPPart soapPart = soapMessage.getSOAPPart();
File fXmlFile = new File("src/XML/gen_VDD7S0PLYPAS058_1409900400000_2.xml");
String xmlStr=finalXmlString(fXmlFile);
DocumentBuilderFactory docBuilderFactory=DocumentBuilderFactory.newInstance();
docBuilderFactory.setNamespaceAware(true);   
DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();   
Document doc=docBuilder.parse(xmlStr);  
System.out.println("dasdasdasd"+doc.toString());
String serverURI = "http://www.aaancnuie.com/DCS/2012/01/DocumentCreation/IStandardDocumentService/CreateDocuments";

// SOAP Envelope
SOAPEnvelope envelope = soapPart.getEnvelope();
envelope.addNamespaceDeclaration("example", serverURI);
SOAPBody soapBody = envelope.getBody();
soapBody.setTextContent(xmlStr);
soapMessage.saveChanges();
return soapMessage;

控制台说，

java.net.MalformedURLException：无协议

Answer 1

您需要encode您的网址。特殊角色必须逃脱。

逃避示例：

String url = "http://e2e-soaservices:44000/3.1/StandardDocumentService?wsdl";
String yourURLStr = "java.net.URLEncoder.encode(url, "UTF-8");
java.net.URL url = new java.net.URL(yourURLStr);

java.net.MalformedURLException：没有协议

1 个答案: