从web.xml迁移到基于Java的配置-无法启动Tomcat 8

Question

我正在尝试将我的应用程序的web.xml迁移到基于Java的配置。我们使用的是spring4.1、Java7、Servlet 3.1、Tomcat8和EclipseLuna。web服务框架是Jersey 2.14。

我主要使用以下指南：http://www.robinhowlett.com/blog/2013/02/13/spring-app-migration-from-xml-to-java-based-config/

我已经创建了遵循web.xml配置的WebApplicationInitializer，删除了web.xml，配置Maven不查找web.xml，并成功地进行了mvn清理安装。

当我尝试启动tomcat时，我得到以下错误：

'Publishing to Tomcat v8.o Server at localhost...' has encountered a problem. Resource '/sb-server/target/m2e-wtp/web-resources/WEB-INF/web.xml' does not exist.

我尝试过清理tomcat目录，但没有帮助，而且我似乎错过了一些东西，因为AFAIK Tomcat8应该是基于Java的配置友好型。

我是不是错过了迁移的一步？

以前的web.xml（如预期那样工作）：

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0"> 
    <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.sb.configuration.ServerConfiguration
        </param-value>
    </context-param>
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>
    <servlet>
        <servlet-name>Jersey REST Service</servlet-name>
        <servlet-class>
            org.glassfish.jersey.servlet.ServletContainer
        </servlet-class>
        <init-param>
            <param-name>javax.ws.rs.Application</param-name>
            <param-value>com.sb.configuration.RestJaxRsApplication</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.server.tracing.type</param-name>
            <param-value>ALL</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Jersey REST Service</servlet-name>
        <url-pattern>/api/*</url-pattern>
    </servlet-mapping>
    <!-- Spring Security -->
    <filter>
        <filter-name>springSecurityFilterChain</filter-name>
        <filter-class>org.springframework.web.filter.DelegatingFilterProxy
        </filter-class>
    </filter>
    <filter-mapping>
        <filter-name>springSecurityFilterChain</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>
</web-app>

WebApplicationInitializerImplementation:

@Order(Ordered.HIGHEST_PRECEDENCE)
public class WebAppInitializer implements WebApplicationInitializer {
    @Override
    public void onStartup(final ServletContext container) throws ServletException {
        // Set up application context
        final AnnotationConfigWebApplicationContext appContext = new AnnotationConfigWebApplicationContext();
        appContext.register(ServerConfiguration.class);
        container.addListener(new ContextLoaderListener(appContext));
        // Register listeners
        container.addListener(ContextLoaderListener.class);
        // Jersey Servlet configuration
        final ServletRegistration.Dynamic dispatcher =
                container.addServlet("Jersey REST Service", ServletContainer.class);
        dispatcher.setInitParameter("javax.ws.rs.Application", "com.sb.configuration.RestJaxRsApplication");
        dispatcher.setInitParameter("jersey.config.server.tracing.type", "ALL");
        dispatcher.setLoadOnStartup(1);
        dispatcher.addMapping("/api/*");
        // Filters
        final Dynamic filterRegistration = container.addFilter("springSecurityFilterChain", DelegatingFilterProxy.class);
        filterRegistration.addMappingForUrlPatterns(null, false, "/*");
    }
}

Answer 1

有一种更简单的方法。

在pom文件中添加对Servlet 3.1的依赖

<dependency>
    <groupId>javax.servlet</groupId>
    <artifactId>javax.servlet-api</artifactId>
    <version>3.1.0</version>
</dependency>

首先创建Servlet初始值设定项。以下示例：

public class WebAppInit implements WebApplicationInitializer {
    @Override
    public void onStartup(ServletContext servletContext) throws ServletException {
        AnnotationConfigWebApplicationContext context = new AnnotationConfigWebApplicationContext();
        context.scan(AppConfiguration.class.getPackage().getName());
        servletContext.addListener(new ContextLoaderListener(context));
        ServletRegistration.Dynamic appServlet = servletContext.addServlet("appServlet", new DispatcherServlet(context));
        appServlet.setLoadOnStartup(1);
        Set<String> mappings = appServlet.addMapping("/");
        if (!mappings.isEmpty()) {
            throw new IllegalStateException("Conflicting mappings found! Terminating. " + mappings);
        }
    }
}

您的应用程序配置也需要一些设置。以下示例：

@Configuration
@EnableWebMvc
@ComponentScan("com.your.package.root")
public class AppConfiguration extends WebMvcConfigurerAdapter {
    @Bean
    public ViewResolver viewResolver() {
        InternalResourceViewResolver viewResolver = new InternalResourceViewResolver();
        viewResolver.setOrder(0);
        return viewResolver;
    }
    /**
     * Spring and WEB settings from WebConfAdapt.
     */
    @Override
    public void configureDefaultServletHandling(DefaultServletHandlerConfigurer configurer) {
        configurer.enable();
    }
}

除此之外，不需要更多的配置。我用这个成功地启动了Tomcat。（但是，这是在Tomcat 7上）