Spring启动结构

Question

/project
 -/src/main
 -- /java
 -- / resources
 --- /application.properties
 --- /public
  ---- /app
  ----- /index.html
  ----- /bower-components
  ----- /scripts
  ----- /views/**.html
  ----- /images
  ----- /styles

我的目标是，当应用程序加载时，index.html页面将从/project/src/main/resources/public/app/index.html加载，并且URL应该是“localhost:8080/#/”。目前，URL类似于localhost:8080/app/index.html。

我试图将映射放在application.properties中，如下所示：

server.context-path=/public/app

我还试图通过扩展WebMVCConfigurerAdapter覆盖应用程序配置。类是：

@Configuration
public class ApplicationConfiguration extends WebMvcConfigurerAdapter {

    private static final String[] CLASSPATH_RESOURCE_LOCATIONS = {
            "classpath:/META-INF/resources/", "classpath:/resources/", "classpath:/public/" };

    @Override
    public void addResourceHandlers(ResourceHandlerRegistry registry) {
        registry.addResourceHandler("/**").addResourceLocations(
                CLASSPATH_RESOURCE_LOCATIONS);
    }
    @Override
    public void addViewControllers(ViewControllerRegistry registry) {        
        registry.addViewController( "/" ).setViewName( "app/index" ); 
    }
}

谢谢

Answer 1

就我个人而言，我不确定通过localhost:8080/#/访问应用程序的索引页是否正确。因为“#”用于在当前页面上导航。请看输入链接描述，在这里回答关于超链接的问题。

编辑：

看起来像是CLASSPATH_Resource_Locations中的问题。您是否尝试将“classpath://resources/”更改为“classpath://resources/app/”？

@Bean 
public InternalResourceViewResolver getInternalResourceViewResolver() {
        InternalResourceViewResolver resolver = new InternalResourceViewResolver();
        resolver.setPrefix("/WEB-INF/app/");
        resolver.setSuffix(".html");
        return resolver;
}