Migrating Spring 5 WAR to javax.servlet | Generated by AI

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You’re almost there. Because this project is Spring 5 (javax.* era) and a WAR, the cleanest way is: use javax.servlet-api 4.x, add a minimal web.xml, a Java config, and a tiny controller.

0) Fix the servlet API (important)

Spring 5.2.x uses javax.servlet.*, but your POM pulls jakarta.servlet-api:5.0.0. Swap it to javax:

<!-- replace the jakarta dependency with this -->
<dependency>
  <groupId>javax.servlet</groupId>
  <artifactId>javax.servlet-api</artifactId>
  <version>4.0.1</version>
  <scope>provided</scope>
</dependency>

(Keep Jetty 9.4 — it’s Servlet 3.1/4.0 friendly with javax.)

1) web.xml (since you set failOnMissingWebXml=true)

Create src/main/webapp/WEB-INF/web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee
                             http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd"
         version="4.0">

  <!-- Root Spring context (services/repos if you add them later) -->
  <listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
  </listener>
  <context-param>
    <param-name>contextClass</param-name>
    <param-value>org.springframework.web.context.support.AnnotationConfigWebApplicationContext</param-value>
  </context-param>
  <context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>com.lzw.config.RootConfig</param-value>
  </context-param>

  <!-- DispatcherServlet for MVC -->
  <servlet>
    <servlet-name>dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
      <param-name>contextClass</param-name>
      <param-value>org.springframework.web.context.support.AnnotationConfigWebApplicationContext</param-value>
    </init-param>
    <init-param>
      <param-name>contextConfigLocation</param-name>
      <param-value>com.lzw.config.WebConfig</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
  </servlet>

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

</web-app>

2) Java config

Create two config classes.

src/main/java/com/lzw/config/RootConfig.java:

package com.lzw.config;

import org.springframework.context.annotation.Configuration;

@Configuration
public class RootConfig {
    // Add @Bean definitions for services/repositories if/when you have them.
}

src/main/java/com/lzw/config/WebConfig.java:

package com.lzw.config;

import org.springframework.context.annotation.ComponentScan;
import org.springframework.context.annotation.Configuration;
import org.springframework.web.servlet.config.annotation.EnableWebMvc;

@Configuration
@EnableWebMvc
@ComponentScan(basePackages = "com.lzw.web")
public class WebConfig {
    // Keep empty for REST-only. Add view resolvers/static resource handlers later if needed.
}

3) Hello controller

src/main/java/com/lzw/web/HelloController.java:

package com.lzw.web;

import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RestController;

@RestController
public class HelloController {

    @GetMapping("/hello")
    public String hello() {
        return "Hello, Spring MVC!";
    }
}

4) Directory layout sanity check

5) Run it

Since you already have Jetty in the POM:

mvn clean package
mvn jetty:run

Open: http://localhost:8080/hello → should print Hello, Spring MVC!.

Notes & pitfalls

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