CODEWITHSUNDEEP
javaspringspring-mvcjetty
Blankman
Blankman

Reputation: 267280

Jetty isn't picking up my spring application

I have a Spring MVC application where I use hibernate, freemarker. It is setup as a multi-maven project. I am using IntelliJ ultimate.

Jetty starts fine, but when I go to

http://localhost:8080/

It simply outputs the folders of my project, and I can view my source code in the browser!

Here is my setup currently:

    final Server server = new Server(8080);

    ProtectionDomain domain = HttpServer.class.getProtectionDomain();
    URL location = domain.getCodeSource().getLocation();

    WebAppContext webAppContext = new WebAppContext();
    webAppContext.setResourceBase(location.toExternalForm());
    webAppContext.setDescriptor(location.toExternalForm() + "/WEB-INF/web.xml");
    webAppContext.setContextPath("/");
    webAppContext.setParentLoaderPriority(true);

    server.setHandler(webAppContext);

    server.start();
    server.join();

My project layout is a multi-maven project (using intelli J), the layout is like:

/myapp/src/main/java/main.java (this contains the above code to start jetty)
/myapp/src/main/webapp
/myapp/src/main/webapp/assets
/myapp/src/main/webapp/WEB-INF
/myapp/src/main/webapp/WEB-INF/web-context.xml (spring config file)
/myapp/src/main/webapp/WEB-INF/web.xml
/myapp/src/main/webapp/WEB-INF/views/ (parent folder for my freemarker template files)
/myapp/src/main/webapp/WEB-INF/views/home/index.ftl

My web.xml is:

<context-param>
    <param-name>log4jConfigLocation</param-name>
    <param-value>classpath*:log4j.properties</param-value>
</context-param>

 <servlet>
    <servlet-name>myapp</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>WEB-INF/web-context.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>myapp</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>

When I run this in IntelliJ (11 ulimitate), I get the following output:

2012-08-15 19:17:11,611 [main] INFO  org.eclipse.jetty.server.Server - jetty-7.6.2.v20120308
2012-08-15 19:17:11,886 [main] INFO  org.eclipse.jetty.webapp.StandardDescriptorProcessor - NO JSP Support for /, did not find org.apache.jasper.servlet.JspServlet
2012-08-15 19:17:11,962 [main] INFO  org.eclipse.jetty.server.handler.ContextHandler - started o.e.j.w.WebAppContext{/,file:/Users/me/projects/myapp/myapp-web/target/classes/}
2012-08-15 19:17:12,021 [main] INFO  org.eclipse.jetty.server.AbstractConnector - Started [email protected]:8080

This obviously isn't working because when I run it using tomcat w/intelliJ I get a huge output for things like hibernate, spring, etc.

My pom.xml for the web module has:

..
 <packaging>war</packaging>
..

Upvotes: 4

Views: 6033

Answers (4)

Ted Shaw
Ted Shaw

Reputation: 2306

Have you tried to set defaultsDescriptor parameter

webAppContext.setDefaultsDescriptor(JETTY_HOME+"/etc/webdefault.xml");

JETTY_HOME is where you installed jetty, you can find JETTY_HOME/etc/webdefault.xml contains essential settings.

Upvotes: 0

Mikael Gueck
Mikael Gueck

Reputation: 5591

Instead of invoking Jetty manually from main.java, refer to the Maven Jetty plugin in your pom.xml, and start your application by running mvn jetty:run

<build>
  <plugins>
    <plugin>
      <groupId>org.mortbay.jetty</groupId>
      <artifactId>jetty-maven-plugin</artifactId>
      <version>8.1.4.v20120524</version>
    </plugin>
  </plugin>
</build>

If instead your objective is to create an executable WAR file with an embedded copy of Jetty, you'd want to approach the task as so:

    final Connector connector = new SelectChannelConnector();
    connector.setPort(8080);

    final Server server = new Server();
    server.addConnector(connector);
    server.setStopAtShutdown(true);

    final WebAppContext context = new WebAppContext();
    context.setContextPath("/");
    context.setServer(server);

    final ProtectionDomain protectionDomain = Main.class.getProtectionDomain();
    final URL location = protectionDomain.getCodeSource().getLocation();
    context.setWar(location.toExternalForm());
    server.addHandler(context);

    server.start();
    server.join();

Upvotes: 0

Marek Dec
Marek Dec

Reputation: 964

If you aim for running this in production, and you already use maven to build your application I'd suggest that you create the actual war package first:

mvn clean package

Then, once you have your webapp built create a new Jetty server that runs from a war archive (and not the source code).

As the Jetty manual states, it should be something like this:

    Server server = new Server(8080);

    WebAppContext webapp = new WebAppContext();
    webapp.setContextPath("/");
    webapp.setWar("path_to_the_war_file/your_app.war");
    server.setHandler(webapp);

    server.start();
    server.join();

Upvotes: 4

Matt
Matt

Reputation: 11805

you're probably missing your context loader listener.

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

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/applicationContext.xml</param-value>
</context-param>

Upvotes: 2

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