Question

我有一个关于SOAP请求的问题。我想解释一下我在做什么。这是我的SOAP请求。

<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:web="http://www.webserviceX.NET">
   <soapenv:Header/>
   <soapenv:Body>
       <web:GetWeather>
         <!--Optional:-->
         <web:CityName>Istanbul</web:CityName>
         <!--Optional:-->
         <web:CountryName>Turkey</web:CountryName>
       </web:GetWeather>
   </soapenv:Body>
</soapenv:Envelope>

端点：http://www.webservicex.net/globalweather.asmx

WSDL链接：http：//www.webservicex.net/globalweather.asmx？WSDL

这也是我的代码。

public static void main(String args[]) {
    try {
        // Create SOAP Connection
        SOAPConnectionFactory soapConnectionFactory = SOAPConnectionFactory
                .newInstance();
        SOAPConnection soapConnection = soapConnectionFactory
                .createConnection();

        // Send SOAP Message to SOAP Server
        String url = "http://www.webservicex.net/globalweather.asmx?WSDL";
        // SOAPMessage soapResponse =
        SOAPMessage soapResponse = soapConnection.call(createSOAPRequest(),
                url);

        // Process the SOAP Response
        printSOAPResponse(soapResponse);
        soapConnection.close();
    } catch (Exception e) {
        System.err.println("Error occurred while sending SOAP Request to Server");
        e.printStackTrace();
    }
}

private static SOAPMessage createSOAPRequest() throws Exception {
    MessageFactory messageFactory = MessageFactory.newInstance();
    SOAPMessage soapMessage = messageFactory.createMessage();
    SOAPPart soapPart = soapMessage.getSOAPPart();

    String serverURL = "http://www.webserviceX.NET/";

    SOAPEnvelope envelope = soapPart.getEnvelope();
    envelope.addNamespaceDeclaration("web", serverURL);

    // SOAP Body
    SOAPBody soapBody = envelope.getBody();
    SOAPElement soapElement = soapBody.addChildElement("GetWeather", "web");
    SOAPElement soapElement1 = soapElement.addChildElement("CityName",
            "web");
    soapElement1.addTextNode("Istanbul");
    SOAPElement soapElement2 = soapElement.addChildElement("CountryName",
            "web");
    soapElement2.addTextNode("Turkey");

    MimeHeaders headers = soapMessage.getMimeHeaders();
    headers.addHeader("SOAPAction", serverURL + "GetWeather");
    soapMessage.saveChanges();
    return soapMessage;
}

/**
 * Method used to print the SOAP Response
 */
private static void printSOAPResponse(SOAPMessage soapResponse)
        throws Exception {
    TransformerFactory transformerFactory = TransformerFactory
            .newInstance();
    Transformer transformer = transformerFactory.newTransformer();
    Source sourceContent = soapResponse.getSOAPPart().getContent();
    System.out.print("\nResponse SOAP Message = ");
    StreamResult result = new StreamResult(System.out);
    transformer.transform(sourceContent, result);
}}

结果，我得到“服务器无法处理。程序或函数'getWeather'需要参数'@CountryName'，这是未提供的。”

这是什么意思？我为什么要接受这个例外？

有关解决方案的任何建议吗？

Answer 1

您使用变量serverUrl作为HTTP服务器URL和XML命名空间名称。它们很接近但不完全相同。命名空间名称为http://www.webserviceX.NET，但您的服务器URL为http://www.webserviceX.NET/（请注意尾随斜杠）。 XML名称空间的字符串必须与模式中的名称空间名称完全匹配。

建议您为命名空间创建一个单独的变量（或者只是内联它）：

   String serverURL = "http://www.webserviceX.NET/";

   SOAPEnvelope envelope = soapPart.getEnvelope();
   envelope.addNamespaceDeclaration("web", "http://www.webserviceX.NET");
   ...

通过此更改，您的代码适合我。

SOAP Server无法处理请求

1 个答案: