HttpInvoker without web container?

**karldmoore** · Jan 21st, 2007, 04:57 AM

OK, heres the working example I promised........ I would zip it up an attach it, but my computer isn't playing ball. The project in this example is called timeService.

Sooooo

Code:

tomcat
    webapps
        timeService
            applicationContext.xml
            invoker-servlet.xml
            web.xml

            classes
                com/test/TimeService.class
                com/test/TimeServiceImpl.class
            lib
                spring.jar
                commons-logging-1.0.4.jar

The client will also require commons-codec.jar and commons-httpclient.jar on the classpath as well as the server prereqs.

Client

TimeServiceClient

Code:

package com.test;

import org.springframework.context.ApplicationContext;
import org.springframework.context.support.ClassPathXmlApplicationContext;

public class TimeServiceClient
{
	public static void main ( String [] args )
	{
		ApplicationContext context = new ClassPathXmlApplicationContext ( "client-applicationContext.xml" );
		TimeService timeService = ( TimeService ) context.getBean ( "timeService" );
		System.out.println ( timeService.getDate () );
		sleepPlease ();
		System.out.println ( timeService.getDate () );
	}
	
	private static void sleepPlease ()
	{
		try
		{
			Thread.sleep ( 5000 );
		}
		catch ( InterruptedException e )
		{
			// ignore
		}
	}
}

client-applicationContext.xml

Code:

<?xml version="1.0" encoding="UTF-8"?>

<beans xmlns="http://www.springframework.org/schema/beans"
	   xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	   xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.0.xsd">
     
    <bean id="httpInvokerRequestExecutor" class="org.springframework.remoting.httpinvoker.CommonsHttpInvokerRequestExecutor"/>

    <bean id="timeService" class="org.springframework.remoting.httpinvoker.HttpInvokerProxyFactoryBean">
		<property name="serviceUrl" value="http://localhost:8080/timeService/invoker/timeService"/>
		<property name="serviceInterface" value="com.test.TimeService"/>
        <property name="httpInvokerRequestExecutor">
            <ref bean="httpInvokerRequestExecutor"/>
        </property>
    </bean>
    
</beans>

Server

TimeService

Code:

package com.test;

import java.util.Date;

public interface TimeService
{
	Date getDate ();
}

TimeServiceImpl

Code:

package com.test;

import java.util.Date;

public class TimeServiceImpl implements TimeService
{
	public Date getDate ()
	{
		return new Date ();
	}
}

applicationContext.xml

Code:

<?xml version="1.0" encoding="UTF-8"?>

<beans xmlns="http://www.springframework.org/schema/beans"
	   xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	   xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.0.xsd">
       
       <bean id="timeService" class="com.test.TimeServiceImpl"/>
       
</beans>

invoker-servlet.xml

Code:

<?xml version="1.0" encoding="UTF-8"?>

<beans xmlns="http://www.springframework.org/schema/beans"
	   xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	   xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.0.xsd">
       
	<bean name="/timeService" class="org.springframework.remoting.httpinvoker.HttpInvokerServiceExporter">
        <property name="service">
            <ref bean="timeService"/>
        </property>
        <property name="serviceInterface">
            <value>com.test.TimeService</value>
        </property>
    </bean>
    
</beans>

web.xml

Code:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE web-app PUBLIC
	"-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
	"http://java.sun.com/dtd/web-app_2_3.dtd">

<web-app>
	<display-name>Time Service</display-name>
	<description>Time Service</description>

	<listener>
		<listener-class>
			org.springframework.web.context.ContextLoaderListener
		</listener-class>
	</listener>

	<servlet>
		<servlet-name>invoker</servlet-name>
		<servlet-class>
			org.springframework.web.servlet.DispatcherServlet
		</servlet-class>
		<load-on-startup>1</load-on-startup>
	</servlet>

	<servlet-mapping>
		<servlet-name>invoker</servlet-name>
		<url-pattern>/invoker/*</url-pattern>
	</servlet-mapping>

</web-app>

**mark_in_gr** · Jan 22nd, 2007, 09:06 AM

Originally Posted by karldmoore

I'm guessing your problem might be to do with the serviceURL.

Code:

<property name="serviceUrl" value="http://localhost:8080/httpGroupDao"/>

If your exposed service is called groupDao, the servlet name is remoting and the webapp is called test......

Code:

<property name="serviceUrl" value="http://localhost:8080/test/remoting/groupDao"/>

I think part of my problem is I am using the Netbeans IDE and I have a multi-project app. We added a new sub-project which includes all of the dao/service layer code and basically modified this, adding a web/WEB-INF directory, which contains the web.xml file(no classes or lib YET). However, seeing that this sub-project was NOT set up as a web app, a jar is being created instead of a war. So, I think the problem is two fold; project structure and directory structure.

At this point, I am going to replace this sub-project w/a true web app, move in the needed interfaces, classes(build to class dir) and dependencies(build to lib) and then try this again.

Have you ever tried to use HttpInvoker w/Jetty, however? Do I need to set the ServletContext, like I would have to do w/Tomcat?

**karldmoore** · Jan 22nd, 2007, 10:14 AM

I've never tried this with Jetty. The example I posted was run with Tomcat. There were no other changes required though then creating the project in webapps and putting the files in the location I stated. Hope this helps.

**mark_in_gr** · Jan 22nd, 2007, 10:16 AM

It should help. I didn't have much like getting a web-app structured sub-project to play nice w/an application parent project, so I am mimicking a web app directory and placing the files in the proper spot for the build.

I will let you know shortly if that works.

**mark_in_gr** · Jan 22nd, 2007, 10:59 AM

so, you did NOT create a war file like you normally would, but manually moved the classes and libs into their respective directories?

I am still getting 404s when using Jetty. Not sure what is going on.

**karldmoore** · Jan 22nd, 2007, 11:03 AM

That right, I didn't create a WAR file. I don't think thats anything to do with it however.

**mark_in_gr** · Jan 22nd, 2007, 11:04 AM

I don't either. I believe it has something to do w/the relationship between jetty.xml, web.xml and the applicationcontext.xml.

**jwray** · Jan 22nd, 2007, 12:25 PM

I'm developing and deploying applications using HttpInvoker using Jetty as the servlet container. I've looked through the above files and I can't see a difference in how I configure things.

Basically, we have a bean '/Service' exposed via a servlet with mapping url pattern '/remote/*' under a context 'dataaccess' on server odin. The serviceUrl we use is

http://odin:80/dataaccess/remote/Service

That seems to agree with the other configuration and my web.xml is basically the same. My jetty (v5) configuration is:

Code:

<Call name="addWebApplication">
	<Arg>/dataaccess/*</Arg>
	<Arg>path to war file</Arg>
	<Set name="extractWAR">true</Set>
  </Call>

Jonny

**mark_in_gr** · Jan 22nd, 2007, 12:49 PM

Originally Posted by jwray

I'm developing and deploying applications using HttpInvoker using Jetty as the servlet container. I've looked through the above files and I can't see a difference in how I configure things.

Basically, we have a bean '/Service' exposed via a servlet with mapping url pattern '/remote/*' under a context 'dataaccess' on server odin. The serviceUrl we use is

http://odin:80/dataaccess/remote/Service

That seems to agree with the other configuration and my web.xml is basically the same. My jetty (v5) configuration is:

Code:

<Call name="addWebApplication">
	<Arg>/dataaccess/*</Arg>
	<Arg>path to war file</Arg>
	<Set name="extractWAR">true</Set>
  </Call>

Jonny

A couple of questions; I don't have the addWebApplication in my jetty.xml. Do I need this? Second, can I do this if the app is not packaged as a WAR file, like a normal web app?

**mark_in_gr** · Jan 22nd, 2007, 01:13 PM

Originally Posted by jwray

I'm developing and deploying applications using HttpInvoker using Jetty as the servlet container. I've looked through the above files and I can't see a difference in how I configure things.

Basically, we have a bean '/Service' exposed via a servlet with mapping url pattern '/remote/*' under a context 'dataaccess' on server odin. The serviceUrl we use is

http://odin:80/dataaccess/remote/Service

That seems to agree with the other configuration and my web.xml is basically the same. My jetty (v5) configuration is:

Code:

<Call name="addWebApplication">
	<Arg>/dataaccess/*</Arg>
	<Arg>path to war file</Arg>
	<Set name="extractWAR">true</Set>
  </Call>

Jonny

I believe that the following in a Jetty 6 jetty.xml should accomplish the same thing as the addWebApplication above, correct?

Code:

<property name="handler">
            <bean id="handlers"
                  class="org.mortbay.jetty.handler.HandlerCollection">
                <property name="handlers">
                    <list>
                        <bean id="contexts"
                              class="org.mortbay.jetty.handler.ContextHandlerCollection">
                            <property name="handlers">
                                <list>
                                    <bean class="org.mortbay.jetty.webapp.WebAppContext">
                                        <property name="contextPath" value="/"/>
                                        <!--<property name="war" value="../web">-->
                                            <property name="resourceBase" value="../backcheck">
                                        </property>
                                    </bean>
                                </list>
                            </property>

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