java如何在apache服务器中运行spark app

Question

I have a Spark MVC app, which is very simple.

According to spark documentation , this should be enough to run the app:

public class SparkServer {
    public static void main(String args[]) {
        Spark.staticFileLocation("src/main/webapp");
        System.out
                .println("bla bla bla");
        RService rService = new SparqlRService();
        new RController(rService);
    }
}

I put that class in a package inside my project, and I run the web app (dynamic web app) on Apache Tomcat server.

The print statement doesn't appear when run Apache Tomcat, that means this class is not being called. I know it makes sense . that is why i am asking. please how can I let Apache Tomcat run my spark app?

Update

After @mlk answer , i did the following:

public class SparkServer implements SparkApplication {

    @Override
    public void init() {
        Spark.staticFileLocation("src/main/webapp");
        System.out
                .println("bla bla lba");
        RService rService = new SparqlRService();
        new RController(rService);
    }
}

and in my web.xml i did:

    <?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">
    <display-name>SRecommender</display-name>
    <welcome-file-list>
        <welcome-file>index.html</welcome-file>
        <welcome-file>index.htm</welcome-file>
        <welcome-file>index.jsp</welcome-file>
        <welcome-file>default.html</welcome-file>
        <welcome-file>default.htm</welcome-file>
        <welcome-file>default.jsp</welcome-file>
    </welcome-file-list>
    <filter>
        <filter-name>SparkFilter</filter-name>
        <filter-class>spark.servlet.SparkFilter</filter-class>
        <init-param>
            <param-name>
    applicationClass</param-name>
            <param-value>com.srecommender.SparkServer</param-value>
        </init-param>
    </filter>

    <filter-mapping>
        <filter-name>SparkFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

</web-app>

Where my SparkServer is in a package: com.recommender that exists in a source folder: src/main/java

I run apache tomcat, but still when i call any path from spark, it is returning 404.

HITE i can run spark from the main method and call my pages, they are working. so the problem with the way i configured spark to run in apache tomcat

Update 2

This is how to set the path of the view

public RecommendationController(RecommendationService service) {
        get("/", (request, response) -> {

            Map<String, Object> model = new HashMap<>();
            model.put("data", "forza ROMA");
            model.put("data2", "It's Rome, It's home");

            return View("src/main/webapp/template/index.html", model);
        });

Answer 1

Spark comes with a container built in, so you don't need tomcat or jetty. If you want to deploy to a full blow container then you can by creating a web.xml file and implementing spark.servlet.SparkApplication .

Edit: you are missing the applicationClass property in your Web.xml.

Answer 2

WebApps don't execute static void main() methods, they bootstrap according to their web.xml from their deployed WAR file.

Do you have any experience with running webapps? You need a container like Tomcat or Jetty. Apache server is just for serving HTTP and is not an application container in itself.