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Posted to users@tomcat.apache.org by Thom Hehl <th...@nowhereatall.com> on 2006/01/09 01:17:10 UTC
First webapp/JSP
OK, I've looked through the docs and googled and I'm not having any luck
with this.
I have a brand new servlet for my webapp. I have it connected up and
running OK.
The problem is the forward to the JSP. I'm getting this error:
HTTP Status 404 - /visioneer/WEB-INF/SearchErrors.jsp
------------------------------------------------------------------------
*type* Status report
*message* _/visioneer/WEB-INF/SearchErrors.jsp_
*description* _The requested resource
(/visioneer/WEB-INF/SearchErrors.jsp) is not available._
------------------------------------------------------------------------
Here is the URL that calls my servlet:
http://localhost:81/visioneer/controller
And here is the code in my servlet:
public void doGet(HttpServletRequest request,
HttpServletResponse response)
throws IOException, ServletException
{
//retrieve the session
HttpSession session = request.getSession();
String act=request.getParameter(PARM_ACTION);
if(act==null)
{
String url = "/visioneer/WEB-INF/SearchErrors.jsp";
RequestDispatcher dispatcher =
getServletContext().getRequestDispatcher(url);
dispatcher.forward(request, response);
return;
}
}
And in my tomcat directory, I have
webapps/visioneer/WEB-INF/SearchErrors.jsp.
I have also tried my forward as: "/visioneer/SearchErrors.jsp"
I don't know what to try next and need advice.
Thanks.
Thom Hehl
Heavyweight Software for Heavyweight Needs
www.heavyweightsoftware.com
--
"In every revolution, there is one man with a vision."--Jerome Bixby
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Re: First webapp/JSP
Posted by Thom Hehl <th...@nowhereatall.com>.
Wow! Thanks, that's it. I started without that, but added it when I was
having problems.
Thanks again!
Wendy Smoak wrote:
>On 1/8/06, Thom Hehl <th...@nowhereatall.com> wrote:
>
>
>
>> String url = "/visioneer/HomeSearch.jsp";
>> RequestDispatcher dispatcher =
>> getServletContext().getRequestDispatcher(url);
>> dispatcher.forward(request, response);
>>
>>And I now get this error:
>>
>>*type* Status report
>>
>>*message* _/visioneer/HomeSearch.jsp_
>>
>>
>
>In general, you should not be hard-coding the name of the context in your app.
>
>http://tomcat.apache.org/tomcat-5.5-doc/servletapi/javax/servlet/ServletContext.html#getRequestDispatcher(java.lang.String)
>
>"The pathname must begin with a "/" and is interpreted as relative to
>the current context root."
>
>So leave off the 'visioneer' part, (which is the context name,) start
>with / and then give the path to the resource, depending on where you
>moved it to.
>
>HTH,
>--
>Wendy
>
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>
>
>
>
Thom Hehl
Heavyweight Software for Heavyweight Needs
www.heavyweightsoftware.com
--
"In every revolution, there is one man with a vision."--Jerome Bixby
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Re: First webapp/JSP
Posted by Wendy Smoak <ws...@gmail.com>.
On 1/8/06, Thom Hehl <th...@nowhereatall.com> wrote:
> String url = "/visioneer/HomeSearch.jsp";
> RequestDispatcher dispatcher =
> getServletContext().getRequestDispatcher(url);
> dispatcher.forward(request, response);
>
> And I now get this error:
>
> *type* Status report
>
> *message* _/visioneer/HomeSearch.jsp_
In general, you should not be hard-coding the name of the context in your app.
http://tomcat.apache.org/tomcat-5.5-doc/servletapi/javax/servlet/ServletContext.html#getRequestDispatcher(java.lang.String)
"The pathname must begin with a "/" and is interpreted as relative to
the current context root."
So leave off the 'visioneer' part, (which is the context name,) start
with / and then give the path to the resource, depending on where you
moved it to.
HTH,
--
Wendy
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Re: First webapp/JSP
Posted by Thom Hehl <th...@nowhereatall.com>.
Thanks for the response. I moved my JSPs to /visioneer and modified my
forward to:
String url = "/visioneer/HomeSearch.jsp";
RequestDispatcher dispatcher =
getServletContext().getRequestDispatcher(url);
dispatcher.forward(request, response);
And I now get this error:
*type* Status report
*message* _/visioneer/HomeSearch.jsp_
*description* _The requested resource (/visioneer/HomeSearch.jsp) is not
available.
_
Mark Thomas wrote:
>Thom Hehl wrote:
>
>
>>OK, I've looked through the docs and googled and I'm not having any luck
>>with this.
>>
>>I have a brand new servlet for my webapp. I have it connected up and
>>running OK.
>>
>>The problem is the forward to the JSP. I'm getting this error:
>>
>>
>> HTTP Status 404 - /visioneer/WEB-INF/SearchErrors.jsp
>>
>>
>
>The servlet spec requires that no resources are served from the
>WEB-INF directory for security reasons. Move the jsp to
>/visioneer/SearchErrors.jsp and all should be fine.
>
>mark
>
>
>---------------------------------------------------------------------
>To unsubscribe, e-mail: users-unsubscribe@tomcat.apache.org
>For additional commands, e-mail: users-help@tomcat.apache.org
>
>
>
>
Thom Hehl
Heavyweight Software for Heavyweight Needs
www.heavyweightsoftware.com
--
"In every revolution, there is one man with a vision."--Jerome Bixby
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Re: First webapp/JSP
Posted by Wendy Smoak <ws...@gmail.com>.
On 1/8/06, Mark Thomas <ma...@apache.org> wrote:
> The servlet spec requires that no resources are served from the
> WEB-INF directory for security reasons. Move the jsp to
> /visioneer/SearchErrors.jsp and all should be fine.
But he's forwarding to it, which is allowed.
I think it should be:
String url = "/WEB-INF/SearchErrors.jsp";
(I would move it to WEB-INF/jsp/SearchErrors.jsp though.)
--
Wendy
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Re: First webapp/JSP
Posted by Mark Thomas <ma...@apache.org>.
Thom Hehl wrote:
> OK, I've looked through the docs and googled and I'm not having any luck
> with this.
>
> I have a brand new servlet for my webapp. I have it connected up and
> running OK.
>
> The problem is the forward to the JSP. I'm getting this error:
>
>
> HTTP Status 404 - /visioneer/WEB-INF/SearchErrors.jsp
The servlet spec requires that no resources are served from the
WEB-INF directory for security reasons. Move the jsp to
/visioneer/SearchErrors.jsp and all should be fine.
mark
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