怎样从HttpServletRequest中得到完整的请求URL-JSP编程

Question

怎样从HttpServletRequest中得到完整的请求URL-JSP编程

Answer 1

现在给你解答
String getFullURL(HttpRequest request) {

StringBuffer url = request.getRequestURL();
if (request.getQueryString() != null) {
url.append(’?’);
url.append(request.getQueryString());
}
return url.toString();
}

Answer 2

String uri=request.getRequestURI();
String path=uri.substring(uri.lastIndexOf("/"),uri.lastIndexOf("."));
如果你的web.xml 中请求路径是" *.do ",那么在servlet中可以根据这个判断应该能获取到
比如：if（path=“/list”）{
}

Answer 3

1	String header_referer = request.getHeader("Referer");

通过上述的Java代码即可获取到请求的引用页面地址