Question

我有以下控制器

@Controller
@RequestMapping("/room/{roomId:^(?!main.html$).*}")
public class Rooms {

    @RequestMapping
    public String index(@PathVariable(value = "roomId") String id) {
       // do some stuff...
       return "main.html";
    }
}

我的资源/静态文件夹中有一个main.html文件。我希望当用户进入/ room / {something}进行处理时，返回我的静态main.html页面。

我尝试的所有内容都会导致页面找不到异常或

javax.servlet.ServletException: Circular view path [main.html]: 
would dispatch back to the current handler URL 
[/room/main.html] again

我错过了什么才能让它发挥作用？

感谢您的帮助

Answer 1

如果您正在为视图解析器进行类配置

@Bean
    public UrlBasedViewResolver urlBasedViewResolver()
    {
        UrlBasedViewResolver res = new InternalResourceViewResolver();
        res.setViewClass(JstlView.class);
        res.setPrefix("/WEB-INF/");
        res.setSuffix(".jsp");
        return res;
    }

如果使用xml

<beans:bean     class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <beans:property name="prefix" value="/WEB-INF/jsp/" />
        <beans:property name="suffix" value=".jsp" />
    </beans:bean>

还可以通过以下代码更改您的控制器

@RestController
@RequestMapping("/room/")
public class Rooms {

    @RequestMapping(method = RequestMethod.GET)
    public ModelAndView index(@PathVariable(value = "roomId") String id) {
       // do some stuff...

       ModelAndView model = new ModelAndView();
        model.setViewName("main");
       return model;
    }
}

返回带有url路径参数的静态页面

1 个答案: