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用java实现springmvc配置_Java_Spring_Spring Mvc - Fatal编程技术网

用java实现springmvc配置

用java实现springmvc配置,java,spring,spring-mvc,Java,Spring,Spring Mvc,我正在使用Apache Tomcat 7.0.54和Spring MVC 4.0.1,并尝试使用Java配置DispatcherServlet: 我已经编写了配置类: public class AppInitializer extends AbstractDispatcherServletInitializer{ @Override protected WebApplicationContext createServletApplicationContext() {

我正在使用Apache Tomcat 7.0.54和Spring MVC 4.0.1,并尝试使用Java配置DispatcherServlet: 我已经编写了配置类:

public class AppInitializer extends AbstractDispatcherServletInitializer{

    @Override
    protected WebApplicationContext createServletApplicationContext() {
        XmlWebApplicationContext context = new XmlWebApplicationContext();
        context.setConfigLocation("/WEB-INF/her-servlet.xml");
        return context;
    }

    @Override
    protected String[] getServletMappings() {
        return new String[] {"*.her"};
    }

    @Override
    protected WebApplicationContext createRootApplicationContext() {
        return null;
    }
}
但它不起作用。但如果我写下以下内容:

public class AppInitializer implements WebApplicationInitializer{


    @Override
    public void onStartup(ServletContext sc) throws ServletException {
        XmlWebApplicationContext context = new XmlWebApplicationContext();
        context.setConfigLocation("/WEB-INF/her-servlet.xml");
        ServletRegistration.Dynamic registration = sc.addServlet("her", new DispatcherServlet(context));
        registration.setLoadOnStartup(1);
        registration.addMapping("*.her");
    }

}

它会很好用的。我不明白这个问题。你能帮我解决吗?

内联代码通常用于计算机识别或生成的字符串。不适用于像Apache这样的产品名称。有关详细信息,请参阅。我建议您使用此帖子中的方法:。换句话说,不要在onStartup中设置应用程序上下文,也不要使用基于XML的应用程序上下文配置。