小编use*_*703的帖子

我很难通过Spring-ws WebServiceTemplate调用SOAP 1.2 WebService.正在进行的请求缺少Http Header中的SOAPAction,并且服务器抛出错误"无法处理没有有效操作参数的请求.请提供有效的soap操作." 通过wireshark监控,我能够发现SOAP Action丢失了.我也不支持任何代理.

通过TCP Mon(像SOAP UI这样的工具)运行请求,我确保我尝试发送的SOAP XML是有效的,并且能够获得响应.

这是我的春季配置:

<?xml version="1.0" encoding="UTF-8"?>
 <beans xmlns="http://www.springframework.org/schema/beans" 
   xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
   xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:util="http://www.springframework.org/schema/util"
xsi:schemaLocation="http://www.springframework.org/schema/beans  
                        http://www.springframework.org/schema/beans/spring-beans-2.5.xsd
                        http://www.springframework.org/schema/util
                        http://www.springframework.org/schema/util/spring-util-3.0.xsd">

<bean id="messageFactory" class="org.springframework.ws.soap.saaj.SaajSoapMessageFactory">
    <property name="soapVersion">
    <util:constant static-field="org.springframework.ws.soap.SoapVersion.SOAP_12" />
    </property>
</bean>

<bean id="webServiceTemplate" class="org.springframework.ws.client.core.WebServiceTemplate">
 <constructor-arg ref="messageFactory" />
<property name="defaultUri" value="https://ecomapi.networksolutions.com/soapservice.asmx" />
<property name="messageSender">
    <bean class="org.springframework.ws.transport.http.CommonsHttpMessageSender" />     </property>
</bean>

这是我的java代码:

            public void simpleSendAndReceive() {
            try{
            StreamSource source = new StreamSource(new StringReader(MESSAGE));
            StreamResult result = new StreamResult(System.out);
            SoapActionCallback actionCallBack = new SoapActionCallback("https://ecomapi.networksolutions.com/soapservice.asmx") {
                public void doWithMessage(WebServiceMessage msg) {
                    SoapMessage …