JSP新手问个tomcat小问题 请高手指教
配置TOMCAT的虚拟目录
server.xml
<Valve className="org.apache.catalina.valves.AccessLogValve" directory="logs"
prefix="localhost_access_log." suffix=".txt" pattern="common" resolveHosts="false"/>
-->
<Content path="/demo" docBase="D:\demo"/>
</Host>
</Engine>
</Service>
</Server>
WEB-INF
下的XML配置是这样
<?xml version="1.0" encoding="ISO-8859-1"?>
<!--
Licensed to the Apache Software Foundation (ASF) under one or more
contributor license agreements. See the NOTICE file distributed with
this work for additional information regarding copyright ownership.
The ASF licenses this file to You under the Apache License, Version 2.0
(the "License"); you may not use this file except in compliance with
the License. You may obtain a copy of the License at
http://www.apache.org/licenses/LICENSE-2.0
Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
-->
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
version="2.5">
<display-name>Welcome to Tomcat</display-name>
<description>
Welcome to Tomcat
</description>
</web-app>
conf下配置了
<servlet>
<servlet-name>default</servlet-name>
<servlet-class>org.apache.catalina.servlets.DefaultServlet</servlet-class>
<init-param>
<param-name>debug</param-name>
<param-value>0</param-value>
</init-param>
<init-param>
<param-name>listings</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
我觉得都没配置错误啊 为什么还是提示
HTTP Status 404 - /demo
type Status report
message /demo
description The requested resource (/demo) is not available.
Apache Tomcat/6.0.35
[解决办法]
<Content path="/" docBase="D:\demo"/> 试试
