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
测试路径是
http://localhost:8080/demo --------------------编程问答-------------------- [img=http://user.qzone.qq.com/1501796?ADUIN=1501796&ADSESSION=1340073857&ADTAG=CLIENT.QQ.4675_MyTip.0&ptlang=2052#!app=4][/img]
tomcat的启动信息 --------------------编程问答-------------------- 我今天早上也碰到过这样的问题，你先在d盘的demo下放个文件试试，比如这样
http://localhost:8080/demo/aa.jpg
如果不行的话，在Tomcat \conf\Catalina\localhost下添新建文件demo.xml
文件内容就写
<Content path="/demo" docBase="D:\demo"
debug="0" privileged="true">
</Context>
重启即可
注意：path="/demo" path的路径一定要跟xml文件去掉斜杠同名，即demo.xml --------------------编程问答-------------------- http://localhost:8080/demo/
直接进入文件夹可能会报404，直接转到文件则不会 --------------------编程问答-------------------- Content path="/demo" docBase="D:\demo"
debug="0" privileged="true">
</Context>
重启即可
--------------------编程问答--------------------
这样吗
<Content path="demo.xml" docBase="D:\demo"/> --------------------编程问答--------------------
我的意思是说，你不用在server.xml里面配置
而是在Tomcat \conf\Catalina\localhost\文件夹下，新建文件：demo.xml
然后这个demo.xml的内容是：
---------------------------------
<Content path="/demo" docBase="D:\demo"
debug="0" privileged="true">
</Context>
----------------------------------------
因为path="/demo"，所以你新建的文件要叫demo.xml
如果你的path="/aaa",那你的新建文件要叫aaa.xml
这些配好以后在D:\demo文件夹下放个jpg文件，如aa.jpg
然后你打开http://localhost:8080/demo/aa.jpg看看是否成功 --------------------编程问答-------------------- 参考一下这个
http://android.blog.51cto.com/268543/49993
补充：Java , Web 开发
