Java Package (error)

based on the link provided
 http://forums.devshed.com/java-help-9/eclipse-sdk-the-declared-package-does-not-match-the-expected-317204.html

specifically this line
 Create a new package in Eclipse that matches the top level directory of your java package. For example, if you're importing "com.foo.bar.Bas", create a package called "com".


if my url works which reads  http://localhost:8080/a/b/c//t_ajax_add_keyword.jsp

My Package should read
com.(Need to properly complete, can it be completed based on the url above)

ur class in com.oreilly.ajax.servlet !!
but ur imported package is different !!
check that ??

Venabili: thanks for your link

dravidnsr:
I agree com.oreilly.ajax.servlet is from book example how do Identify my package?

which ide is ur using???

If it helps, found with the attachment is a view of directory

dravidnsr:

which ide is ur using??? IBM Websphere Application Server Toolkit V6.0

if u use AjaxResponseServlet means

the ist line should be

package javasources;

sorry JavaSources

if importclass have java  program means

package imported_classes.com.jguru;

I am working on learning ajax  i changed the code to as follows
package JavaSource; .  vs.  package JavaSources; (as stated in your code) I still have the same original  issue found below is a consideration idea that I had, my code, and a screen shot of what appears if i double click the error.

for consideration only
the book has it identified as a servlet.  I have it as a .java and i do see the complementing class (maybe this is part of the issue).

/*
 * Takes a character and converts it to decimal and sends back the 
 * value in the response.
 */
//package com.oreilly.ajax.servlet;
 //package imported_classes.JavaSource;
package javasource;
 
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
 
public class AjaxResponseServlet extends HttpServlet {
 
	private static final long serialVersionUID = 1L;
 
	public void doGet(HttpServletRequest req, HttpServletResponse res)
            throws ServletException, IOException {
 
        String key = req.getParameter("key");
        if (key != null) {
            // extract the first character from key
            int keyInt = key.charAt(0);
            String decimalString = Integer.toString(keyInt); 
            // setup the response
            res.setContentType("text/xml");
            res.setHeader("Cache-Control", "no-cache");
            // write out the response string
            res.getWriter().write(decimalString);
        }
        else {
            // If key comes back as a null, return a question mark.
            res.setContentType("text/xml");
            res.setHeader("Cache-Control", "no-cache");
            res.getWriter().write("?");
        }
    }
}

                                              

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<<<<<<<<< package javasource;

in ur code  chnage JavaSource

tell me ur pblm clearly ??

Thank you for your quick response and your patience in helping me.

I am trying to implement Ajax into my JSP page which is really straight html but inside of a jsp). My jsp page and my javascript run error free.  now I am working on the serverside code.  and not sure how to code the following code or where to place with respect to the rest of the application.

<<<<<<<<< package javasource; (produces same error)  corrected code can be found below

/*
 * Takes a character and converts it to decimal and sends back the 
 * value in the response.
 */
//package com.oreilly.ajax.servlet;
 //package imported_classes.JavaSource;
package JavaSource;
 
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
 
public class AjaxResponseServlet extends HttpServlet {
 
	private static final long serialVersionUID = 1L;
 
	public void doGet(HttpServletRequest req, HttpServletResponse res)
            throws ServletException, IOException {
 
        String key = req.getParameter("key");
        if (key != null) {
            // extract the first character from key
            int keyInt = key.charAt(0);
            String decimalString = Integer.toString(keyInt); 
            // setup the response
            res.setContentType("text/xml");
            res.setHeader("Cache-Control", "no-cache");
            // write out the response string
            res.getWriter().write(decimalString);
        }
        else {
            // If key comes back as a null, return a question mark.
            res.setContentType("text/xml");
            res.setHeader("Cache-Control", "no-cache");
            res.getWriter().write("?");
        }
    }
}

                                              

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if ur pblm(this question ) sloved means please close this question !!

if u have dobut in ur ajax,or servlet and other thing post new question we will try to give correct sloution ...

lot of expect is there !!

press accept button and clsoe this question !!

thanks ....

I believe the main problem is the package name vs the directory structure.

I am still having problems with the package, You have provided some greatful hints and suggestions,  however my original  issue has not been solved, which is as follows:

1.)   Can someone tell me how to fix this error and why did I get this error? (see original question)

when you asked "tell me ur pblm clearly ??" I was stating the ultimate goal, which I will be happy to post in but it will be in smaller sections and with points values equal to the complexity. first I need to solve this issue.  can you make any further suggestions?

rg20
I believe you could be right but if that is true, can justify your observation I would be most appreciative. i did post the suggestion that was provided by my IDE, which can be found in the document labeled "error.doc" (see above")

Thank you for the help!! your link did assist and allowed for more questions to be posted.  ps I will agree programmers may not be able to spell however not to be confused with typing error, just food for thought.  kidest regards  :).