我开始学习 Spring MVC。我正在尝试摆脱所有 Spring XML 配置。这是我的 web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee" 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_3_0.xsd"
version="3.0">
<!-- Configure ContextLoaderListener to use AnnotationConfigWebApplicationContext
instead of the default XmlWebApplicationContext -->
<context-param>
<param-name>contextClass</param-name>
<param-value>
org.springframework.web.context.support.AnnotationConfigWebApplicationContext
</param-value>
</context-param>
<!-- Configuration locations must consist of one or more comma- or space-delimited
fully-qualified @Configuration classes. Fully-qualified packages may also be
specified for component-scanning -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>pl.mbrnwsk.sklep.config.AppConfiguration</param-value>
</context-param>
<!-- Bootstrap the root application context as usual using ContextLoaderListener -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Declare a Spring MVC DispatcherServlet as usual -->
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<!-- Configure DispatcherServlet to use AnnotationConfigWebApplicationContext
instead of the default XmlWebApplicationContext -->
<init-param>
<param-name>contextClass</param-name>
<param-value>
org.springframework.web.context.support.AnnotationConfigWebApplicationContext
</param-value>
</init-param>
<!-- Again, config locations must consist of one or more comma- or space-delimited
and fully-qualified @Configuration classes -->
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>pl.mbrnwsk.sklep.config.AppConfiguration</param-value>
</init-param>
</servlet>
<!-- map all requests for / to the dispatcher servlet -->
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
应用配置.java:
@Configuration
@EnableTransactionManagement
@ComponentScan("pl.mbrnwsk.sklep")
public class AppConfiguration {
public String hbm2ddl_auto = "update";
public AppConfiguration(){
System.out.println("AppConfiguration");
}
@Bean
public ViewResolver viewResolver(){
InternalResourceViewResolver viewResolver = new InternalResourceViewResolver();
viewResolver.setPrefix("/");
viewResolver.setSuffix(".jsp");
return viewResolver;
}
@Bean
public DataSource dataSource() {
DriverManagerDataSource ds = new DriverManagerDataSource();
ds.setDriverClassName("org.hsqldb.jdbcDriver");
ds.setUrl("jdbc:hsqldb:file:/SklepDB/");
ds.setUsername("SA");
ds.setPassword("");
return ds;
}
@Bean
public SessionFactory sessionFactory() {
LocalSessionFactoryBuilder ss = new LocalSessionFactoryBuilder(dataSource());
ss.scanPackages("pl.mbrnwsk.sklep.model");
ss.setProperty("hibernate.show_sql", "true");
ss.setProperty("hibernate.hbm2ddl.auto", hbm2ddl_auto);
ss.setProperty("hibernate.dialect",
"org.hibernate.dialect.HSQLDialect");
return ss.buildSessionFactory();
}
@Bean
public PlatformTransactionManager txManager(){
return new HibernateTransactionManager(sessionFactory());
}
}
AppConfiguration 的实例被创建了两次:一次是我启动 Tomcat 时,两次是我输入一些应该由调度程序处理的 url。这不是期望的行为。我想在 Tomcat 启动时创建 AppConfiguration。如何做到这一点?第二个问题,听众做什么?