如何从 HttpServletRequest 获取完整的 url？

Question

好的，我有这些代码：

HttpServletRequest httpRequest = (HttpServletRequest) request;
String requestQueryString = httpRequest.getQueryString();
System.out.println(requestQueryString);

打开网址“ http://127.0.0.1:8888/MyProject.html?gwt.codesvr=127.0.0.1:9997?_escaped_fragment_=home”时，

它打印出来：

 gwt.codesvr=127.0.0.1:9997?_escaped_fragment_=home

该网址缺少该http://127.0.0.1:8888/MyProject.html?部分。

如何解决？

Answer 1

这将是

httpRequest.getRequestURL() + "?" + httpRequest.getQueryString()

印刷

http://127.0.0.1:8888/MyProject.html?gwt.codesvr=127.0.0.1:9997?_escaped_fragment_=home