问题
I've a servlet registered in web.xml as below.
<servlet>
<servlet-name>Manager</servlet-name>
<servlet-class>Manager</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Manager</servlet-name>
<url-pattern>/RequestManager</url-pattern>
</servlet-mapping>
Basically I want to call this servlet as my default home page when I open http://localhost:8080/appname. So, I tried registering it as welcome file in same web.xml as below:
<welcome-file-list>
<welcome-file>Manager</welcome-file>
</welcome-file-list>
But, when I run the project, I get an error saying "requested resource not available". However, if I write in the url with my servlet URL pattern, it works fine.
回答1:
Specify an empty string as servlet's URL pattern.
<servlet>
<servlet-name>Manager</servlet-name>
<servlet-class>Manager</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Manager</servlet-name>
<url-pattern></url-pattern>
</servlet-mapping>
See also:
- Difference between / and /* in servlet mapping url pattern
Unrelated to the concrete problem, the <welcome-file> should represent an URL path, not a servlet name. It'd have worked if you specifed <welcome-file>RequestManager</welcome-file>. But this affects all subfolders. Actually, the <welcome-file> has an entirely different meaning than "home page file" you've had in mind. It represents the default resource which should be served when a folder is been requested.
回答2:
You can use index.jsp to forward to your servlet.
<jsp:forward page="servlet_context">
and add index.jsp as welcome file in web.xml
回答3:
inside servlet class you can forward Control Using :
request.getRequestDispatcher("forward page URL").forward(req,res);
or else if you are using JSP then use
<% RequestDispatcher rd = servletContext.getRequestDispatcher("/pathToResource");
rd.forward(request, response); %>
or
<jsp:forward page="relative URL" />
来源：https://stackoverflow.com/questions/32020447/set-servlet-as-default-home-page-in-web-xml