发表新帖
发表新帖

How to get correct current URL in JSP in Spring webapp

后端 未结
关注
10 577
梦谈多话
梦谈多话 2021-02-03 18:53

I\'m trying to get a correct current URL in JSP in Spring webapp. I\'m trying to use the following fragment in the JSP file:

${pageContext.request.requestURL}
相关标签:
10条回答
  • 情歌与酒
    2021-02-03 19:41

    Which Spring version are you using? I have tested this with Spring 3.1.1.RELEASE, using the following simple application:

    Folder structure
-----------------------------------------------------------------------------------

spring-web
    |
     --- src
          |
           --- main
                 |
                  --- webapp
                         |
                          --- page
                         |     |
                         |      --- home.jsp
                         |
                          --- WEB-INF
                               |
                                --- web.xml
                               |
                                --- applicationContext.xml

home.jsp
-----------------------------------------------------------------------------------

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">

<html dir="ltr" lang="en" xmlns="http://www.w3.org/1999/xhtml">
    <head>
        <title>Welcome to Spring Web!</title>
    </head>
    <body>
        Page URL: ${pageContext.request.requestURL}
    </body>
</html>

web.xml
-----------------------------------------------------------------------------------

<?xml version="1.0" encoding="UTF-8"?>

<web-app xmlns="http://java.sun.com/xml/ns/javaee" metadata-complete="true" version="2.5" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<display-name>org.example.web</display-name>

<servlet>
    <servlet-name>spring-mvc-dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/webContext.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>spring-mvc-dispatcher</servlet-name>
    <url-pattern>*.htm</url-pattern>
</servlet-mapping>
</web-app>

applicationContext.xml
-----------------------------------------------------------------------------------

<?xml version="1.0" encoding="UTF-8"?>

<beans xmlns="http://www.springframework.org/schema/beans" xmlns:context="http://www.springframework.org/schema/context" xmlns:mvc="http://www.springframework.org/schema/mvc" xmlns:util="http://www.springframework.org/schema/util" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd     http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.1.xsd http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd http://www.springframework.org/schema/util http://www.springframework.org/schema/util/spring-util-3.1.xsd">

<context:annotation-config />
<context:component-scan base-package="org.example" />

<bean id="viewResolver"
    class="org.springframework.web.servlet.view.InternalResourceViewResolver">
    <property name="prefix" value="/page/" />
    <property name="suffix" value=".jsp" />
</bean>

<mvc:annotation-driven />

    On accessing http://localhost:8080/spring-web/page/home.jsp, the URL is correctly displayed as http://localhost:8080/spring-web/page/home.jsp.

    0 讨论(0)
  • 梦如初夏
    2021-02-03 19:44

    The best way would be to use EL like this:

    ${requestScope['javax.servlet.forward.request_uri']}
    0 讨论(0)
  • 误落风尘
    2021-02-03 19:44

    Maybe you are looking for something like: 

    <%= new UrlPathHelper().getOriginatingRequestUri(request) %>

    This is not that elegant but solved my problem.

    0 讨论(0)
  • -上瘾入骨i
    2021-02-03 19:49

    Try this:

    <%@ page import="javax.servlet.http.HttpUtils.*" %>
<%= javax.servlet.http.HttpUtils.getRequestURL(request) %>
    0 讨论(0)
 看不清?
提交回复
热议问题