2019独角兽企业重金招聘Python工程师标准>>> request.getRequestURI() /jqueryWeb/resources/request.jsp request.getRequestURL() http://localhost:8080/jqueryWeb/resources/request.jsp request.getContextPath()/jqueryWeb request.getServletPath()/resources/request.jsp 转载于:https://my.oschina.net/deepins/blog/316339