您的位置:首页 > 运维架构 > Tomcat

Tomcat6.0 配置输入IP访问

2012-03-07 15:38 281 查看
server.xml:
<Connector port="80" redirectPort="8443" connectionTimeout="20000" protocol="HTTP/1.1"/>

<Host name="localhost" xmlNamespaceAware="false" xmlValidation="false" autoDeploy="true" unpackWARs="true" appBase="webapps">
<Context reloadable="true" path="" docBase="C:\Program Files\Apache Software Foundation\Tomcat 6.0\webapps\SldCenter"/>
</Host>

web.xml:
<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>


这样即可直接输入IP eg:   192.168.0.20
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签:  tomcat servlet apache path c