如何在类中获取request，和网站路径

Posted 2020-09-22 hoge66的专栏

@RequestMapping(value = "res/testjs")
	public ModelAndView testjs( ModelMap model ) throws Exception {
		String path = resService.getRequestPath();
		model.put("path", path);
		return new ModelAndView("res/testjs");
	}

　　上面是调用：

下面是方法：

import javax.servlet.http.HttpServletRequest;
import org.springframework.web.context.request.RequestContextHolder;
import org.springframework.web.context.request.ServletRequestAttributes; 

public String getRequestPath() {
		HttpServletRequest request  = ((ServletRequestAttributes) RequestContextHolder.getRequestAttributes()).getRequest();
		String path =request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+"/"+request.getRequestURL()+"/";  
		return path;
	}

　　还有web.xml配置监听：

<listener>
        <description>springrequestListener </description>
        <listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
     </listener>

　　效果：