旧版 Spring MVC 到 Spring Boot

Question

我有一个使用 web.xml 配置的旧应用程序。 web.xml 看起来像这样。

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

    <!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            classpath:spring/common-beans-context.xml
        </param-value>
    </context-param>

    <servlet>
        <servlet-name>dispatcherOne</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/dispatcherOne-context.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcherOne</servlet-name>
        <url-pattern>/test/*</url-pattern>
    </servlet-mapping>

    <servlet>
        <servlet-name>dispatcherTwo</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/dispatcherTwo-context.xml</param-value>
        </init-param>
        <load-on-startup>2</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcherTwo</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app>

要将其转换为 Spring Boot 应用程序，

我做了以下改动：

@Bean
    public XmlWebApplicationContext xmlWebApplicationContext() {
        XmlWebApplicationContext applicationContext = new XmlWebApplicationContext();
        applicationContext.setConfigLocations("classpath:spring/common-beans-context.xml");
        applicationContext.refresh();
        return applicationContext;
    }

@Bean
    public ServletRegistrationBean<DispatcherServlet> mvcTestServlet(XmlWebApplicationContext applicationContext) {
        DispatcherServlet dispatcherServlet = new DispatcherServlet();
        // create child application context
        XmlWebApplicationContext childApplicationContext = new XmlWebApplicationContext();
        childApplicationContext.setConfigLocation("classpath:spring/dispatcherOne-context.xml");
        // set parent from the previous method
        childApplicationContext.setParent(applicationContext);
        dispatcherServlet.setApplicationContext(childApplicationContext);
        childApplicationContext.refresh();
        ServletRegistrationBean<DispatcherServlet> servletRegistrationBean = new ServletRegistrationBean<DispatcherServlet>(
                dispatcherServlet, "/test/*");
        servletRegistrationBean.setName("dispatcherOne");
        servletRegistrationBean.addUrlMappings("/test/*");
        servletRegistrationBean.setLoadOnStartup(1);
        return servletRegistrationBean;
    }

    @Bean
    public ServletRegistrationBean<DispatcherServlet> mvcServlet(XmlWebApplicationContext applicationContext) {
        DispatcherServlet dispatcherServlet = new DispatcherServlet();
        // similar to the previous method..
        // 1. create local applicationContext and then set parent.
    }

我添加了 childApplicationContext.refresh() 以查看 bean 是否正确加载，但我认为发生的是两个调度程序 servlet 无法访问 parentContext 中定义的 bean。并且抛出 bean not found 异常，即使它们在 parentContext 中可用

有解决办法吗？

或者还有其他方法可以实现吗？

Answer 1

要创建 Spring Boot 应用程序并重新使用现有配置，请执行以下操作，并且不要自己创建上下文。您基本上是在尝试用您当前正在做的事情来超越或解决 Spring Boot。

创建一个带有@SpringBootApplication注解的类，并使用@ImportResource让Spring Boot创建主应用上下文。

@SpringBootApplication
@ImportResource("classpath:spring/common-beans-context.xml")
public YourApplication extends SpringBootServletInitializer {

  public static void main(String[] args) {
    SpringApplication.run(YourApplication.class, args);
  }

  protected SpringApplicationBuilder configure(SpringApplicationBuilder builder) {
    return builder.sources(YourApplication.class);
  } 
}

现在您的原始配置中有 2 个DispatcherServlets，您需要添加 2 个ServletRegistrationBean 来设置 URL 等。

@Bean
public ServletRegistrationBean<DispatcherServlet> dispatcherServletOneRegistration() {
  ServletRegistrationBean<DispatcherServlet> registration = new ServletRegistrationBean(new DispatcherServlet(), "/test/*");
  registration.setLoadOnStartup(1);
  registration.setName("dispatcherOne");
  registration.setInitParameters(Collections.singletonMap("contextConfigLocation", "classpath:spring/dispatcherOne-context.xml");
  return registration;
}

第二个 servlet 或多或少相同。

@Bean
public ServletRegistrationBean<DispatcherServlet> dispatcherServletTwoRegistration() {
  ServletRegistrationBean<DispatcherServlet> registration = new ServletRegistrationBean(new DispatcherServlet(), "/");
  registration.setLoadOnStartup(1);
  registration.setName("dispatcherTwo");
  registration.setInitParameters(Collections.singletonMap("contextConfigLocation", "classpath:spring/dispatcherTwo-context.xml");
  return registration;
}