如何在JSP中获取完整的URL

我如何获得JSP页面的完整URL。

例如，url可能是http://www.example.com/news.do/?language=nl&country=NL

如果我做以下事情，我总是得到news.jsp而不是.do

out.print(request.getServletPath()); out.print(request.getRequestURI()); out.print(request.getRequest()); out.print(request.getContextPath());

Answer 1

鉴于URL = http：/ localhost：8080 / sample / url.jsp？id1 = something＆id2 = something＆id3 = something

 request.getQueryString(); 

它返回id1 = something＆id2 = something＆id3 = something

看到这个

Answer 2

您需要调用request.getRequestURL() ：

重构客户端用于发出请求的URL。 返回的URL包含协议，服务器名称，端口号和服务器路径，但不包括查询字符串参数。

Answer 3

 /** * Creates a server URL in the following format: * * 
scheme://serverName:serverPort/contextPath/resourcePath
 * * 
 * If the scheme is 'http' and the server port is the standard for HTTP, which is port 80, * the port will not be included in the resulting URL. If the scheme is 'https' and the server * port is the standard for HTTPS, which is 443, the port will not be included in the resulting * URL. If the port is non-standard for the scheme, the port will be included in the resulting URL. * 
 * * @param request - a javax.servlet.http.HttpServletRequest from which the scheme, server name, port, and context path are derived. * @param resourcePath - path to append to the end of server URL after scheme://serverName:serverPort/contextPath. * If the path to append does not start with a forward slash, the method will add one to make the resulting URL proper. * * @return the generated URL string * @author Cody Burleson (cody at base22 dot com), Base22, LLC. * */ protected static String createURL(HttpServletRequest request, String resourcePath) { int port = request.getServerPort(); StringBuilder result = new StringBuilder(); result.append(request.getScheme()) .append("://") .append(request.getServerName()); if ( (request.getScheme().equals("http") && port != 80) || (request.getScheme().equals("https") && port != 443) ) { result.append(':') .append(port); } result.append(request.getContextPath()); if(resourcePath != null && resourcePath.length() > 0) { if( ! resourcePath.startsWith("/")) { result.append("/"); } result.append(resourcePath); } return result.toString(); }