Spring启动为嵌入式服务器配置自定义jsessionid

Question

我想配置我的servlet上下文,例如设置自定义jsessionId键(请参阅更改cookie JSESSIONID名称)

我相信我可以使用SpringBootServletInitializer运行WAR文件时,操纵servletContext在onStartup().但是,当我在嵌入式应用程序服务器上运行时 new SpringApplicationBuilder().run(),我不知道操作servlet上下文的最佳位置.

Answer 1

从Spring Boot 1.3开始,您只需设置配置属性即可;

Spring Boot 1.3,1.4,1.5

server.session.cookie.name = MYSESSIONID

Spring Boot 2.x

server.servlet.session.cookie.name = MYSESSIONID

比编写配置类简单得多.

有关更多与会话相关的属性,请参阅https://docs.spring.io/spring-boot/docs/current/reference/html/common-application-properties.html.

Answer 2

ServletContextInitializer在应用程序的配置中声明一个bean:

@Bean
public ServletContextInitializer servletContextInitializer() {
    return new ServletContextInitializer() {

        @Override
        public void onStartup(ServletContext servletContext) throws ServletException {
            servletContext.getSessionCookieConfig().setName("yourCookieName");
        }
    };

}

或者,您的应用程序类本身可以实现ServletContextInitializer:

@Configuration
@EnableAutoConfiguration
@ComponentScan
public class Application implements ServletContextInitializer {

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

    @Override
    public void onStartup(ServletContext servletContext) throws ServletException {
        servletContext.getSessionCookieConfig().setName("yourCookieName");
    }

}