获取由Servlet转发的JSP中的请求URL

Question

如何在JSP中获取由Servlet转发的请求URL？

如果我在JSP中运行以下代码,

System.out.println("servlet path= " + request.getServletPath());
System.out.println("request URL= " + request.getRequestURL());
System.out.println("request URI= " + request.getRequestURI());

然后我得到JSP的服务器端路径.但是我想在浏览器的地址栏中看到URL.我可以在转发到JSP的Servlet中获取它,但我希望在JSP中获取它.

Answer 1

如果您使用RequestDispatcher.forward()将请求从控制器路由到视图,则请求URI将作为名为的请求属性公开javax.servlet.forward.request_uri.所以,你可以使用

request.getAttribute("javax.servlet.forward.request_uri")

要么

${requestScope['javax.servlet.forward.request_uri']}

Answer 2

试试这个:

String scheme = req.getScheme();             
String serverName = req.getServerName(); 
int serverPort = req.getServerPort();    
String uri = (String) req.getAttribute("javax.servlet.forward.request_uri");
String prmstr = (String) req.getAttribute("javax.servlet.forward.query_string");
String url = scheme + "://" +serverName + ":" + serverPort + uri + "?" + prmstr;

注意:您无法从您的网址获取HREF锚点.例如,如果您有网址"toc.html #top",那么您只能获得"toc.html"

注意:req.getAttribute("javax.servlet.forward.request_uri")仅适用于JSP.如果你在JSP之前在控制器中运行它,那么结果为null

您可以为两个变体使用代码:

public static String getCurrentUrl(HttpServletRequest req) {
    String url = getCurrentUrlWithoutParams(req);
    String prmstr = getCurrentUrlParams(req);
    url += "?" + prmstr;
    return url;
}

public static String getCurrentUrlParams(HttpServletRequest request) {
    return StringUtil.safeString(request.getQueryString());
}

public static String getCurrentUrlWithoutParams(HttpServletRequest request) {
    String uri = (String) request.getAttribute("javax.servlet.forward.request_uri");
    if (uri == null) {
        return request.getRequestURL().toString();
    }
    String scheme = request.getScheme();
    String serverName = request.getServerName();
    int serverPort = request.getServerPort();
    String url = scheme + "://" + serverName + ":" + serverPort + uri;
    return url;
}

Answer 3

要从JSP文件中获取当前路径,您只需执行以下操作之一:

<%= request.getContextPath() %>
<%= request.getRequestURI() %>
<%= request.getRequestURL() %>

Answer 4

试试这个,

<c:set var="pageUrl" scope="request">
    <c:out value="${pageContext.request.scheme}://${pageContext.request.serverName}"/>
    <c:if test="${pageContext.request.serverPort != '80'}">
        <c:out value=":${pageContext.request.serverPort}"/>
    </c:if>
    <c:out value="${requestScope['javax.servlet.forward.request_uri']}"/>
</c:set>

我想将它放在我的基本模板中,并在需要时在整个应用程序中使用.