HttpServletRequest request获取请求url等信息
程序员文章站
2024-02-01 23:28:40
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ServletRequestAttributes requestAttributes = (ServletRequestAttributes) RequestContextHolder
.getRequestAttributes();
HttpServletRequest request = null;
if (null != requestAttributes) {
request = ((ServletRequestAttributes) RequestContextHolder.getRequestAttributes()).getRequest();
String url = request.getScheme()+"://"+ request.getServerName()+request.getRequestURI()+"?"+request.getQueryString();
System.out.println("7获取全路径(协议类型://域名/项目名/命名空间/action名称?其他参数)url="+url);
String url2=request.getScheme()+"://"+ request.getServerName();//+request.getRequestURI();
System.out.println("协议名://域名="+url2);
System.out.println("获取项目名="+request.getContextPath());
System.out.println("获取参数="+request.getQueryString());
System.out.println("获取全路径="+request.getRequestURL());
System.out.println("获取web项目的路径="+request.getSession().getServletContext().getRealPath("/"));//获取web项目的路径
System.out.println("获取类的当前目录="+this.getClass().getResource("/").getPath());//获取类的当前目录
}else{
System.out.println("--------null == requestAttributes-----------");
}
资料地址:https://blog.csdn.net/lili518/article/details/77922265