I build a web service with ejb3 and maven (EAR File),code First, JAXWS, Without WSDL and without WAR, only ejb, with Eclipse, the service works in JBOSS but now i need put the service in Weblogic 11g.
With JBOSS i have the annotation
#WebContext(contextRoot="/wsManCa7", urlPattern="/manCA7WS")
But on Weblogic not, I found this link http://erikwramner.wordpress.com/2012/03/26/ejb3-web-service-context-path-in-weblogic-11g/ to create weblogic-webservices.xml and webservices.xml I put the files inside the META-INF of the jar (And JAr inside EAR) but web logic throws this error
Servlet: "WSEE_SERVLET" failed to preload on startup in Web application:
"/ManCA7". com.sun.xml.ws.server.ServerRtException: Port namespace
http://someserver.com/ManCA7 doesnt match Service namespace {1} at
com.sun.xml.ws.server.EndpointFactory.createEndpoint(EndpointFactory.java:160) at
com.sun.xml.ws.api.server.WSEndpoint.create(WSEndpoint.java:496) at
com.sun.xml.ws.api.server.WSEndpoint.create(WSEndpoint.java:539) at
weblogic.wsee.jaxws.JAXWSDeployedServlet.getEndpoint(JAXWSDeployedServlet.java:183) at
weblogic.wsee.jaxws.JAXWSServlet.registerEndpoint(JAXWSServlet.java:138) at
weblogic.wsee.jaxws.JAXWSServlet.init(JAXWSServlet.java:67) at ....
I understand the problem is this tag
<wsdl-port xmlns:ws="http://someserver.com/ManCA7">ws:ManCA7Port</wsdl-port>
But which is the required Service namespace???
I don't know what put inside the tag :(
my webservices.xml is
<?xml version="1.0" encoding="UTF-8"?>
<webservices xmlns="http://java.sun.com/xml/ns/javaee"xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="1.2">
<webservice-description>
<webservice-description-name>ManCA7</webservice-description-name>
<port-component>
<port-component-name>ManCA7Port</port-component-name>
<wsdl-port xmlns:ws="http://someserver.com/ManCA7">ws:ManCA7Port</wsdl-port>
<service-endpoint-interface>my.company.manCA7.sei.ManCa7SEI</service-endpoint-interface>
<service-impl-bean>
<ejb-link>ManCa7EndPoint</ejb-link>
</service-impl-bean>
</port-component>
</webservice-description>
</webservices>
And my weblogic-webservices.xml is
<?xml version = '1.0' encoding = 'UTF-8'?>
<weblogic-webservices xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.bea.com/ns/weblogic/weblogic-webservices
http://www.bea.com/ns/weblogic/weblogic-webservices/1.0/weblogic-webservices.xsd"
xmlns="http://www.bea.com/ns/weblogic/weblogic-webservices">
<webservice-description>
<webservice-description-name>ManCA7</webservice-description-name>
<webservice-type>JAXWS</webservice-type>
<port-component>
<port-component-name>ManCA7Port</port-component-name>
<service-endpoint-address>
<webservice-contextpath>/ManCA7</webservice-contextpath>
<webservice-serviceuri>/Mant</webservice-serviceuri>
</service-endpoint-address>
<wsdl>
<exposed>true</exposed>
</wsdl>
</port-component>
</webservice-description>
</weblogic-webservices>
And my EJB expose the service with this
#WebService(endpointInterface = "my.company.manCA7.sei.ManCa7SEI")
#TransactionManagement(TransactionManagementType.CONTAINER)
#Stateless
public class ManCa7EndPoint implements ManCa7SEI{
private final Logger logger = LoggerFactory.getLogger(ManCa7EndPoint.class);
.
.
.
I Found the Solution...
In the annotation webservice I put the same namespace of webservices.xml and worked !!
Inside webservices.xml
<wsdl-port xmlns:ws="http://someserver.com/ManCA7">ws:ManCA7Port</wsdl-port>
Inside EJB
#WebService(endpointInterface = "my.company.manCA7.sei.ManCa7SEI",targetNamespace = "http://someserver.com/ManCA7")
#TransactionManagement(TransactionManagementType.CONTAINER)
#Stateless
public class ManCa7EndPoint implements ManCa7SEI{
The solution was targetNamespace = "http://someserver.com/ManCA7"
Related
I'm trying implement a web service for two endpoints and getting this error
"404 Not Found: Invalid Request" when tried accessing the service after deploying onto the apache toncat 8.
Below are my web service implementation classes, sun-jaxws.xml and web.xml
WebImplementation1.java
package com.ws.soap.services;
import javax.jws.WebService;
#WebService(endpointInterface = "com.ws.soap.services.WebServiceImpl1")
public class WebServiceImpl1 {
public String printMessage() {
return "Hello from WebServiceImpl1 ";
}
}
WebServiceImplementation2.java
package com.ws.soap.services;
import javax.jws.WebService;
#WebService(endpointInterface = "com.ws.soap.services.WebServiceImpl2")
public class WebServiceImpl2 {
public String displayMessage() {
return "Hello from WebServiceImpl2 ";
}
}
sun-jaxws.xml
<?xml version="1.0" encoding="UTF-8"?>
<endpoints xmlns="http://java.sun.com/xml/ns/jax-ws/ri/runtime"
version="2.0">
<endpoint name="WebServiceImpl1" implementation="com.ws.soap.services.WebServiceImpl1"
url-pattern="/impl1" />
<endpoint name="WebServiceImpl2" implementation="com.ws.soap.services.WebServiceImpl2"
url-pattern="/impl2" />
</endpoints>
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<display-name>JAX-WS-Tomcat</display-name>
<listener>
<listener-class>
com.sun.xml.ws.transport.http.servlet.WSServletContextListener
</listener-class>
</listener>
<servlet>
<servlet-name>sayhello</servlet-name>
<servlet-class>com.sun.xml.ws.transport.http.servlet.WSServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>sayhello</servlet-name>
<url-pattern>/impl1</url-pattern>
<url-pattern>/impl2</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>30</session-timeout>
</session-config>
</web-app>
Using the exact code supplied (plus the JAX-WS RI jars downloaded from https://jax-ws.java.net/), I was able to create a webapp and successfully access the service endpoints /impl1 and /impl2. Be advised the <url-pattern> and <endpoint ... url-pattern="/impl1"> directives state the resource path to the JAX-WS endpoints within the context path of the enclosing web application.
So, if the name of the webapp is MyWebServices (MyWebServices.war with no other files/code than described in the post, deployed to Tomcat 8) and you have <url-pattern>/impl1</url-pattern> in web.xml, and with a default Tomcat instance listening on port 8080, your web service endpoint would be http://localhost:800/MyWebServices/impl1 with the WSDL available via http://localhost:800/MyWebServices/impl1?wsdl
If you want to customize your context path of your webapp (e.g. you don't want /MyWebServices/... you can use the techniques described in this SO question.
For example, my local Tomcat 8 is running on port 8081:
Can anybody tell me how to make a restful web service with Jersey 2.0 by not using maven. I have searched everywhere and found tutorial for Jersey1.x versions but not for 2.0. Please help
We provide detail answere based on the user answer user2629427. we checked this on windows 7.
Requirement: (brackets indicate version which this example is tested)
tomcat (8 zip version)
jersey (2.x)
Unzip the tomcat & create a below folder structure in tomcat's 'webapps' folder (folder names are case sensitive).
abc
|___ WEB-INF
|____ classes
|____ lib
Put 'Hello.java' and 'MyApplication.java' into 'classes' folder and 'web.xml' into 'WEB-INF' folder.
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
id="WebApp_ID"
version="3.1">
<servlet>
<servlet-name>rest</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.king.MyApplication</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>rest</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
Myapplication.java
package com.king;
import org.glassfish.jersey.server.ResourceConfig;
public class MyApplication extends ResourceConfig {
public MyApplication() {
packages("com.king");
}
}
Hello.java
package com.king;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("/hello")
public class Hello {
#GET
#Produces(MediaType.TEXT_PLAIN)
public String sayPlainTextHello() {
return "Hello Jersey";
}
// This method is called if XML is request
#GET
#Produces(MediaType.TEXT_XML)
public String sayXMLHello() {
return "<?xml version=\"1.0\"?><hello>Hello Jersey</hello>";
}
// This method is called if HTML is request
#GET
#Produces(MediaType.TEXT_HTML)
public String sayHtmlHello() {
return "<html><title>Hi Jersey</title><body><h1>Hello Jersey this is laksys</body></h1></html>";
}
}
Unzip jersey and copy all jar files from api, ext, and lib (not folders) into your apps 'lib' folder.
Now compile the two java files using following command
D:\apache-tc-8\webapps\abc\WEB-INF\classes>javac -d . -cp ..\lib\javax.ws.rs-api-2.0.1.jar;..\lib\jersey-server.jar;..\l ib\jersey-common.jar *.java
Next run the tomcat server
D:\apache-tc-8\bin>startup
In browser address bar type this: http://localhost:8080/abc/rest/hello
I found the answer
package com.hellowebservice;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("/hello")
public class Hello {
#GET
#Produces(MediaType.TEXT_PLAIN)
public String sayPlainTextHello() {
return "Hello Jersey";
}
// This method is called if XML is request
#GET
#Produces(MediaType.TEXT_XML)
public String sayXMLHello() {
return "<?xml version=\"1.0\"?>" + "<hello> Hello Jersey" + "</hello>";
}
// This method is called if HTML is request
#GET
#Produces(MediaType.TEXT_HTML)
public String sayHtmlHello() {
return "<html> " + "<title>" + "Hello Jersey" + "</title>"
+ "<body><h1>" + "Hello Jersey" + "</body></h1>" + "</html> ";
}
}
Web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.5"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<display-name>FirstRestWebService</display-name>
<servlet>
<display-name>Rest Servlet</display-name>
<servlet-name>RestServlet</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.hellowebservice.MyApplication</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>RestServlet</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
MyApplication.java
package com.hellowebservice;
import org.glassfish.jersey.server.ResourceConfig;
public class MyApplication extends ResourceConfig {
public MyApplication() {
packages("com.hellowebservice");
}
}
run with localhost:8080/FirstRestWebService/rest/hello
Just to add to the previous answer. If you aren't using Maven and just building using Eclipse with a Dynamic Web Project and deploying to web app server like Tomcat.
Just download the Jersey JAX-RS 2.0 RI bundle Jersey Downloads, unzip and add all the jars in the lib, api and ext folders to your build path. (I tried without ext jars but got classnotfound when starting the server).
Also add all the jars to the Deployment Assembly of your Dynamic Web Project so they get automatically copied to the WEB-INF/lib directory when deployed to your web app server. Along with the code & web.xml in the above answer, you should have a RESTful api using Jersey 2 up and running.
I'm trying to set disableCNCheck to true for my web service. I'm using Grails 2.2.0 with the Cxf Client plugin.
I found this question:
wsdl2java CXF command line error about disableCNCheck option
with this piece of code:
protected void disableCNCheck(Object port) {
Client client = ClientProxy.getClient(port)
TLSClientParameters params = new TLSClientParameters()
params.setDisableCNCheck(true)
HTTPConduit httpConduit = (HTTPConduit) client?.getConduit()
httpConduit?.setTlsClientParameters(params)
}
In which class would this code belong and where would the method be called? Is there a configuration parameter for the Cxf Client plugin that I could set instead?
Instead of the code you can achieve the disableCNCheck by adding the below xml config to grails-app/conf/spring/resources.xml [ Grails 2.2.3 , cxf-client plugin 1.6.1
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:http="http://cxf.apache.org/transports/http/configuration"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd
http://cxf.apache.org/transports/http/configuration
http://cxf.apache.org/schemas/configuration/http-conf.xsd
">
<http:conduit name="*.http-conduit">
<http:tlsClientParameters disableCNCheck="true" />
</http:conduit>
</beans>
I am trying to evaluate Spring Integration specially interested in exposing a simple POJO based service into a SOAP based webservice via service activator. Currently I am stuck & have issue generating dynamic wsdl. The WSDL is not loaded & browser shows 404 error.
I try to access with the following url on my local
http://localhost:8080/ws-inbound-gateway/echoService
http://localhost:8080/ws-inbound-gateway/echoService/echoService.wsdl
Below is the configuration
inbound-gateway-config.xml
<int:channel id="inbound" />
<bean id="marshaller" class="org.springframework.oxm.jaxb.Jaxb2Marshaller">
<property name="contextPath" value="com.manish.schema.generated" />
</bean>
<int-ws:inbound-gateway id="empServiceGateway"
request-channel="inbound" marshaller="marshaller"
unmarshaller="marshaller" />
<int:service-activator input-channel="inbound"
requires-reply="true" ref="employeeServiceActivator" method="getEmployeeDetails">
</int:service-activator>
<bean id="employeeServiceActivator"
class="org.springframework.integration.samples.ws.EmployeeServiceResponder" />
<bean id="employeeService" class="com.manish.service.EmployeeService" />
EmployeeService is just a pojo class while EmployeeServiceResponder is a service activator that invokes method on a service class.
For dynamic wsdl generation
spring-ws-config.xml
<import resource="classpath:/META-INF/spring/integration/inbound-gateway-config.xml" />
<sws:dynamic-wsdl id="echoService" portTypeName="empServiceGateway" locationUri="/echoService" targetNamespace="http://manish.niyati.com/echo">
<sws:xsd location="/WEB-INF/echo.xsd"/>
</sws:dynamic-wsdl>
<bean
class="org.springframework.ws.server.endpoint.mapping.UriEndpointMapping">
<property name="defaultEndpoint" ref="empServiceGateway"></property>
</bean>
web.xml
<servlet>
<servlet-name>spring-ws</servlet-name>
<servlet-class>org.springframework.ws.transport.http.MessageDispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>WEB-INF/spring-ws-config.xml</param-value>
</init-param>
<init-param>
<param-name>transformWsdlLocations</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring-ws</servlet-name>
<url-pattern>/echoService</url-pattern>
</servlet-mapping>
Kindly let me know what else is missing in order to make this service accessible as web-service via SI.
Also when I try to access the service from using WebService template I get the SOAPFAULT
02:18:59.436 INFO [main][org.springframework.ws.soap.saaj.SaajSoapMessageFactory] Creating SAAJ 1.3 MessageFactory with SOAP 1.1 Protocol
02:18:59.437 DEBUG [main][org.springframework.ws.soap.saaj.SaajSoapMessageFactory] Using MessageFactory class [com.sun.xml.internal.messaging.saaj.soap.ver1_1.SOAPMessageFactory1_1Impl]
02:18:59.484 DEBUG [main][org.springframework.ws.client.core.WebServiceTemplate] Opening [org.springframework.ws.transport.http.HttpUrlConnection#249fa95c] to [http://localhost:8080/ws-inbound-gateway/echoService]
02:18:59.519 TRACE [main][org.springframework.ws.soap.saaj.support.SaajUtils] SOAPElement [com.sun.xml.internal.messaging.saaj.soap.ver1_1.Envelope1_1Impl] implements SAAJ 1.3
02:18:59.535 TRACE [main][org.springframework.ws.soap.saaj.support.SaajUtils] SOAPElement [com.sun.xml.internal.messaging.saaj.soap.ver1_1.Body1_1Impl] implements SAAJ 1.3
02:18:59.562 TRACE [main][org.springframework.ws.client.MessageTracing.sent] Sent request [<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/"><SOAP-ENV:Header/><SOAP-ENV:Body><ed:employeeRequest xmlns:ed="http://manish.niyati.com/echo" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <ed:empid>100</ed:empid> </ed:employeeRequest></SOAP-ENV:Body></SOAP-ENV:Envelope>]
02:18:59.604 TRACE [main][org.springframework.ws.client.MessageTracing.received] Received response [<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/"><SOAP-ENV:Header/><SOAP-ENV:Body><SOAP-ENV:Fault><faultcode>SOAP-ENV:Server</faultcode><faultstring xml:lang="en">**java.lang.NullPointerException**</faultstring></SOAP-ENV:Fault></SOAP-ENV:Body></SOAP-ENV:Envelope>] for request [<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/"><SOAP-ENV:Header/><SOAP-ENV:Body><ed:employeeRequest xmlns:ed="http://manish.niyati.com/echo" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <ed:empid>100</ed:empid> </ed:employeeRequest></SOAP-ENV:Body></SOAP-ENV:Envelope>]
02:18:59.605 DEBUG [main][org.springframework.ws.client.core.WebServiceTemplate] Received Fault message for request [SaajSoapMessage {http://manish.niyati.com/echo}employeeRequest]
02:18:59.607 TRACE [main][org.springframework.ws.soap.saaj.support.SaajUtils] SOAPElement [com.sun.xml.internal.messaging.saaj.soap.ver1_1.Fault1_1Impl] implements SAAJ 1.3
Thanks in Advance
-MS
To get the WSDL, change the web.xml url-pattern to <url-pattern>/*</url-pattern>.
and the URL is http://localhost:8080/ws-inbound-gateway/echoService.wsdl.
Everything else looks good.
Regarding the WebServiceTemplateQuestion, what are you sending? It looks like you are using the ws sample app, which uses a WebServiceTemplate...
#Test
public void testWebServiceRequestAndResponse() {
StringResult result = new StringResult();
Source payload = new StringSource(
"<?xml version=\"1.0\" encoding=\"UTF-8\"?>" +
"<echoRequest xmlns=\"http://www.springframework.org/spring-ws/samples/echo\">hello</echoRequest>");
template.sendSourceAndReceiveToResult(WS_URI, payload, result);
logger.info("RESULT: " + result.toString());
assertThat(result.toString(), equalTo(
"<?xml version=\"1.0\" encoding=\"UTF-8\"?>" +
"<echoResponse xmlns=\"http://www.springframework.org/spring-ws/samples/echo\">hello</echoResponse>"));
}
And that works just fine.
Looks like your NPE is on the server - take a look at the server logs to see what happened.
You are doing it in a much more complex way. With a simple configuration and using some basic annotations, you can develop a web service in a short time.
I have myself used Apache CXF for web service development and it is quite good for spring based configuration. You can also take a look at this blog. It has shown all the steps with figures to develop a web service with the generation of wsdl and a client to consume the web service.
I need to create a WS with Spring 3.0.4.RELEASE to run in a Tomcat with Axis2. I'm following this doc: http://static.springsource.org/spring/docs/3.0.x/spring-framework-reference/html/remoting.html#remoting-web-services-jaxws-export-ri (if that paragraph can be called "doc")
Ok, here are the details:
The java class:
package foo;
#WebService(serviceName="MyService")
public class MyService{
#WebMethod
public String getString(){
return "Hello StackOverflow";
}
}
The WEB-INF/spring-ws.xml:
<?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:ws="http://jax-ws.dev.java.net/spring/core"
xmlns:wss="http://jax-ws.dev.java.net/spring/servlet"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://jax-ws.dev.java.net/spring/core https://jax-ws.dev.java.net/spring/core.xsd
http://jax-ws.dev.java.net/spring/servlet https://jax-ws.dev.java.net/spring/servlet.xsd">
<wss:binding url="/myService" service="#myService" />
<ws:service id="myService"
impl="foo.MyService" />
</beans>
The WEB-INF/web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="myService" version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>my Service</display-name>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-ws.xml</param-value>
</context-param>
<!-- this is for Spring -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- these are for JAX-WS -->
<servlet>
<servlet-name>jaxws-servlet</servlet-name>
<servlet-class>com.sun.xml.ws.transport.http.servlet.WSSpringServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>jaxws-servlet</servlet-name>
<url-pattern>/myService</url-pattern>
</servlet-mapping>
And last, but not less important, the error when I start tomcat 6.0.29:
Context initialization failed
org.springframework.beans.factory.parsing.BeanDefinitionParsingException: Configuration problem: Unable to locate Spring NamespaceHandler for XML schema namespace [http://jax-ws.dev.java.net/spring/servlet]
Offending resource: ServletContext resource [/WEB-INF/spring-ws.xml]
Someone has any clue of what is happening? Is all the configuration correct? Does anyone have a simple (working) WS to show how to deploy a WS using Spring?
Thanks in advance
I also experience this issue a while back and figured out the problem is with the "https://". Change it back to http:// and you should be good to go. But when you use http:// you get a schema validation error in eclipse because eclipse can't automatically redirect schema url from http:// to https://. And apparently netbeans is capable of it.
One more thing. You'll have to have the xbeans-spring as well. I honestly think that's a pretty stupid dependency.