Reputation: 2591
I'm pretty sure this questions is very easy. I built a app with google app engine. I use JPA and JAX-RS with Jersey. my web.xml looks like this:
<?xml version="1.0" encoding="utf-8" standalone="no"?><web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="2.5" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<servlet>
<servlet-name>Jersey Web Application</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.rest</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/resources/*</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>
30
</session-timeout>
</session-config>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>SystemServiceServlet</servlet-name>
<servlet-class>com.google.api.server.spi.SystemServiceServlet</servlet-class>
<init-param>
<param-name>services</param-name>
<param-value/>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>SystemServiceServlet</servlet-name>
<url-pattern>/_ah/spi/*</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
</web-app>
My index.html is in war/WEB-INF/index.html. What do i need to do to display it ?
Upvotes: 0
Views: 595
Reputation: 3801
keep your file in war folder itself, Don't keep inside WEB-INF.
Once you have taken it out, You just need to give localhost:8888/ it will display the welcome-file ie index.html
you can keep other Jsp's inside WEB-INF, it will just be invisible for direct URL access.
Upvotes: 0