如何轴在Java的Apache客户端调用 [英] How to call axis apache client in java

问题描述

我想连接到在Java Apache Axis的web服务，我有一些错误的参数，但我不知道是哪个：

I want to connect to the web service with apache axis in java and I have some wrong parameter but I don't know which:

import org.apache.axis.client.Call;
import org.apache.axis.client.Service;
import javax.xml.namespace.QName;

 public class Test_Web_Service
 {

 public static void main(String [] args) throws Exception {

     try {

            String endpoint =  "http://www.w3schools.com/webservices/tempconvert.asmx";

            Service  service = new Service();
            Call call= (Call) service.createCall();

            call.setProperty( Call.SOAPACTION_USE_PROPERTY, new Boolean( true ) );
            call.setProperty( Call.SOAPACTION_URI_PROPERTY, "http://tempuri.org/CelsiusToFahrenheit");

            call.setTargetEndpointAddress( new java.net.URL(endpoint) );
            call.setOperationName(new QName("http://tempuri.org/CelsiusToFahrenheit","CelsiusToFahrenheit"));

            String ret = (String) call.invoke( new Object[] {"20"} );
            System.out.println("Sent '20', got '" + ret + "'");

     } catch (Exception e) {
            System.err.println(e.toString());
    }
 }
}

Web服务的链接： http://www.w3schools.com/webservices/ tempconvert.asmx

在RET变量，我得到消息错误。这是因为我在错误的QName参数。

web service link: http://www.w3schools.com/webservices/tempconvert.asmx
In ret variable I get message Error. Is this because I have wrong parameter in QName.

推荐答案

这是由于阻抗不匹配beetween客户code和service.server不能去code您的要求，继续处理使用默认值

This is due to impedence mismatch beetween the client code and the service.server is not able to decode your request, and continue processing with default values

您可以试试这个代替

call.setOperationName(new QName("http://tempuri.org/","CelsiusToFahrenheit")); 
call.addParameter(new QName("http://tempuri.org/","Celsius"),XMLType.XSD_STRING,ParameterMode.IN);
String ret = (String) call.invoke( new Object[] {"20"} );

注意的namespaceURI变化了。

note the change in namespaceURI too.

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