我正在使用SpringWS-Security为一个Web服务创建一个java客户端消费者。
我的请求SOAP(我在SOAP UI中使用)
<soapenv:Envelope
xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:sch="http://myws.mycompany.com/myws/schema">
<soapenv:Header>
<wsse:Security soapenv:mustUnderstand="1"
xmlns:wsse="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd">
<wsse:UsernameToken xmlns:wsu="http://schemas.xmlsoap.org/ws/2003/06/utility">
<wsse:Username>myUsernameString</wsse:Username>
<wsse:Password Type="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-username-token-profile-1.0#PasswordText">123</wsse:Password>
</wsse:UsernameToken>
</wsse:Security>
</soapenv:Header>
<soapenv:Body>
<sch:GetUserDetails idSender="5"/>
</soapenv:Body>
</soapenv:Envelope>
我在WS中的servlet.xml。
<bean name="endpointMapping"
class="org.springframework.ws.server.endpoint.mapping.PayloadRootQNameEndpointMapping">
<property name="interceptors">
<list>
<ref local="wsSecurityInterceptor" />
</list>
</property>
<bean id="wsSecurityInterceptor"
class="org.springframework.ws.soap.security.wss4j.Wss4jSecurityInterceptor">
<property name="validationActions" value="UsernameToken" />
<property name="validationCallbackHandler" ref="springSecurityCallbackHandler" />
</bean>
<bean id="springSecurityCallbackHandler"
class="org.springframework.ws.soap.security.wss4j.callback.SpringPlainTextPasswordValidationCallbackHandler">
<property name="authenticationManager" ref="authenticationManager"/>
</bean>
<bean id="authenticationProvider" class="ws.security.CustomAuthenticationProviderImpl">
<property name="userCommonService" ref="userCommonService" />
<security:custom-authentication-provider/>
</bean>
<security:authentication-manager alias="authenticationManager" />.
在我的Java客户端中-应用程序上下文.xml
<bean name="webserviceTemplate" class="org.springframework.ws.client.core.WebServiceTemplate">
<property name="defaultUri" value="http:/localhost:8080/myws-ws/" />
<property name="marshaller" ref="marshaller" />
<property name="unmarshaller" ref="unmarshaller" />
<property name="interceptors">
<list>
<ref local="wsSecurityInterceptor" />
</list>
</property>
</bean>
<oxm:jaxb2-marshaller id="marshaller"
contextPath="org.example.bean.schema" />
<oxm:jaxb2-marshaller id="unmarshaller"
contextPath="org.example.org.bean.schema" />
<bean id="client" class="example.client.impl.EfactClientImpl">
<property name="webServiceTemplate" ref="webserviceTemplate" />
</bean>
<bean id="wsSecurityInterceptor" class="org.springframework.ws.soap.security.wss4j.Wss4jSecurityInterceptor">
<property name="securementActions" value="UsernameToken"/>
</bean>
当我使用SOAP UI来使用服务时,一切都很顺利,但我想我需要Java客户端及其上下文的帮助,因为当我运行它时,出现了以下错误:
The security token could not be authenticated or authorized; nested exception is:
javax.security.auth.callback.UnsupportedCallbackException; nested exception is org.apache.ws.security.WSSecurityException: The security token could not be authenticated or authorized; nested exception is:
javax.security.auth.callback.UnsupportedCallbackException
当我调试我的应用程序时,我可以注意到这个元素崩溃了:
GetUserRequest request = new GetUserRequest();
request.setIdentifier(user.getIdentifier());
request.setPassword(user.getPassword());
GetUserResponse response = new GetUserResponse();
/* Crashing here. */
response = (GetUserResponse) getWebServiceTemplate().marshalSendAndReceive(request);
FYI:我总是在SpringWS中看到这个安全的用户列表,但是如果有很多用户试图访问该怎么办?
WS - [服务小程序名称]-服务小程序.xml
<bean id="callbackHandler" class="org.springframework.ws.soap.security.wss4j.callback.SimplePasswordValidationCallbackHandler">
<property name="users">
<props>
<prop key="Bert">Ernie</prop>
<prop key="Mickey">Mouse</prop>
</props>
</property>
</bean>
如何解决此UnsupportedCallbackException异常?
2条答案
按热度按时间acruukt91#
必须在调用时指定SecurementUsername和SecurementPassword。这是一个编程示例:
vpfxa7rd2#
当我从cfx 2.2.X升级到2.7.X时,我遇到了这个错误。由于功能升级,服务器端代码无法读取密码,因此服务器收到的密码为空。请确保服务器收到了正确的用户名和密码,这将解决此问题。