标签：java request 获取跟路径 获取请求url tomcat
一、 获取此次请求的URL
**String requestUrl = request.getScheme() //当前链接使用的协议
+"://" + request.getServerName()//服务器地址
+ ":" + request.getServerPort() //端口号
+ request.getContextPath() //应用名称，如果应用名称为
+ request.getServletPath() //请求的相对url
+ "?" + request.getQueryString(); //请求参数**举例：
http://127.0.0.1:8080/world/index.jsp?name=lilei&sex=1
<Context path="world" docBase="/home/webapps" debug="0" reloadable="true"/>
request.getScheme() = "http";
request.getServerName() = "127.0.0.1";
request.getServerPort() = "8080";
request.getContextPath() = "world";
request.getServletPath() = "index.jsp";
request.getQueryString() = "name=lilei&sex=1";
http://127.0.0.1:8080/world/index.jsp?name=lilei&sex=1
<Context path="" docBase="/home/webapps" debug="0" reloadable="true"/>
request.getScheme() = "http";
request.getServerName() = "127.0.0.1";
request.getServerPort() = "8080";
request.getContextPath() = "";
request.getServletPath() = "world/index.jsp";
request.getQueryString() = "name=lilei&sex=1";二、获取服务器根路径
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>使用如下：
<head>
<link rel="stylesheet" type="text/css" href="<%=basePath%>static/css/framework/themes/default/easyui.css">
<link rel="stylesheet" type="text/css" href="<%=basePath%>static/css/framework/themes/icon.css">
<link rel="stylesheet" type="text/css" href="<%=basePath%>static/css/base.css">
<script src="<%=basePath%>static/javascript/framework/jquery.min.js"></script>
<script src="<%=basePath%>static/javascript/framework/jquery.easyui.min.js"></script>
<script src="<%=basePath%>static/javascript/framework/easyui-lang-zh_CN.js"></script>
<script src="<%=basePath%>static/javascript/framework/easyui-util.js"></script>
</head>版权声明：本文为博主原创文章，未经博主允许不得转载。
标签：java request 获取跟路径 获取请求url tomcat
原文地址：http://blog.csdn.net/claram/article/details/48035207
