Question

Almost as Easy as Hello World # 3- Ajax

Asked by: gazdzid

Working with ajax I am excited because my javascript can call my server code. however the response returned is not correct as the responce is the first three chrtrs of the jave file and probably for good reason. the reason may be due to the url in the java script, which is
var url = "http://localhost:8080/kdb_eldb/cfobeldb/JavaSource/AjaxResponseServlet.java";
SomehowI believe I would need to include something like:
?key=" + escape(key.value); but not sure how.

I am developing my first ajax aplication the that makes a request to the server. The Ajax project that I am trying to work on should evaluate a keystroke and convert it to a decimal value.

The app;ication should work as follows the user types a letter the letter in the field. The letter is passed to a javascript at which point a call to the server is made and a response is given, which is the decimal value of the key stroke.

Found below in the doc is appicture of my directory and screen shot

Please let me know if you need to see the web xml.

Found below is the code to my project and screenshot of my results.

Please Help!!

var req; 
  function convertToDecimal(){ 
 
    var key = document.getElementById("key"); 
    var keypressed = document.getElementById("keypressed"); 
    keypressed.value = key.value; 
 
 
var url = "http://localhost:8080/kdb_eldb/cfobeldb/JavaSource/AjaxResponseServlet.java";
 
 
   if (window.XMLHttpRequest){ 
      req = new XMLHttpRequest(); 
      //var x=1;
    } 
    else if (window.ActiveXObject){ 
      req = new ActiveXObject("Microsoft.XMLHTTP"); 
     // var x=2;
    } 
 
    alert (url)
    alert (callback)
    req.open("GET",url,true); 
 
    req.onreadystatechange = callback; 
    req.send(null);
  } 
 
  function callback(){ 
 
    if (req.readyState==4){ 
    
      if (req.status == 200){ 
     
        var decimal = document.getElementById('decimal'); 
        decimal.value = req.responseText; 
      } 
       alert("status is " + req.status);
      
    }
    clear();
  } 
  function clear(){ 
 
    var key = document.getElementById("key"); 
   
    key.value=""; 
  }
  function focusIn(){ 
 
    document.getElementById("key").focus();
  }
 
 
 
<html>
  <head>
    <link rel="stylesheet" type="text/css" href="t_ajax.css">
    
	<script language="JavaScript" src="includes/ajax.js"></script>
    <title>Ajax On Java, Chapter 2 Example</title>
  </head>
  <body onload="focusIn();">
    <h1> AJAX CHARACTER DECODER </h1>
    <h2> Press a key to find its value. </h2>
      <table>
        <tr>
          <td>
            eEnter Key Here 
            <input type="text" id="key" name="key" onkeyup="convertToDecimal();">
          </td>
      </table> 
      <br />
      <table>
        <tr>
          <td colspan="5" style="border-bottom:solid black 1px;">
            Key Pressed:
            <input type="text" readonly id="keypressed">
          </td>
        </tr>
        <tr>
          <td> Decimal </td>
        </tr>
        <tr>
          <td> <input type="text" readonly id="decimal"> </td>
        </tr>
      </table>
  </body>
</html>
 
 
Server Code
 
 
/*
 * sTakes 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.java;
 
 
 
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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ajax-screenshot.doc
- 76 KB
AJAX Sreenshot

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Asked On: 2009-10-14 at 10:26:48ID24812053
Tags: Ajax - html - javascript
Topics: Hypertext Markup Language (HTML)
,; JavaScript
,; Asynchronous Javascript and XML (AJAX)
Participating Experts: 2
Points: 500
Comments: 13

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Answers

by: hieloPosted on 2009-10-14 at 17:20:06ID: 25576373

>>SomehowI believe I would need to include something like:
?key=" + escape(key.value); but not sure how.
Correct. Instead of:

var url = "http://localhost:8080/kdb_eldb/cfobeldb/JavaSource/AjaxResponseServlet.java";
 
use:
var url = "http://localhost:8080/kdb_eldb/cfobeldb/JavaSource/AjaxResponseServlet.java?key=" + encodeURIComponent(key.value);

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by: xxli79Posted on 2009-10-15 at 00:59:21ID: 25578259

Your server is not support J2EE? Why returned the java source?
You'd better post your web.xml.
By the way, your servlet class should have a pakage name!

by: gazdzidPosted on 2009-10-15 at 08:07:59ID: 25581325

Hello hielo:
I did add the code as provded but resuls are the same. The return is the same: /* * sTakes

Hello xxli79:

Found below the web xml, which I believe has been altered in the past but seems to have been restored to the code found below. (The alteration was done manually by me but I do not currently see the alteraion)
(Code is Below)

You stated: Your server is not support J2EE? Why returned the java source?
I am not sure if you are asking a question or telling me a piece of information in either case let me tell you why I used the link found below: Because I was getting a 404 error code and the link below gave me a 200, which I believe is good and need to know how to complete:

var url = "http://localhost:8080/kdb_eldb/cfobeldb/JavaSource/AjaxResponseServlet.java";

You stated: By the way, your servlet class should have a pakage name!
Not exactly sure how to do this but maybe this will be come a new question. Can I still make the application work without a package name or is it easier to open a new question that asks how to add a new package name to a already created application?

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
	<display-name>
	cfobkdb</display-name>
	<welcome-file-list>
		<welcome-file>blank-int.jsp</welcome-file>
	</welcome-file-list>
<servlet>
<servlet-name>AjaxResponseServlet </servlet-name>
<servlet-class>JavaSource.AjaxResponseServlet </servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>AjaxResponseServlet</servlet-name>
<url-pattern>/AjaxResponseServlet </url-pattern>
</servlet-mapping>
</web-app>

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by: hieloPosted on 2009-10-15 at 08:40:46ID: 25581717

The way you worded your question made it seem as if though you didn't know how to pass the parameters via the querystring. However, after reading the response from xxli79, I see now what you are doing wrong.

Basically, if you were to have a file named hello.txt and it had only two words in it:
Greetings Stranger

Then if you used:
var url = "http://localhost:8080/kdb_eldb/cfobeldb/JavaSource/hello.txt?key=123";

the server will see that you are requesting a .txt file. Web servers typically do NOT execute txt files, so assuming that is the case with your server as well, then the server will just send the "raw" content of your .txt file. So basically your client-script will receive "Greetings Stranger". Notice, that if you changed the content of your txt file to:
<?php
$x="howdy";
echo ( $x );
?>

your client-side script will still NOT get just "howdy". It will literally get the complete "raw" content of the file - namely:

<?php
$x="howdy";
echo ( $x );
?>

Why? Again, because the server does NOT execute a .txt file. It just opens it, reads the content and sends it as is. The point I am trying to make is that NOT ALL files on the server are interpreted. In your case, it seems that your .java files are not executed. Perhaps a server misconfiguration?

I am not a JSP developer so I don't know what you would need to do to get your pages working, but from the problems I have seen here, you would typically "point" the url to a .jsp page, not to a .java file. Since you are pointing the url to a .java source code file, and the server doesn't interpret .java files, it sends the "raw" content, similarly like it would do with a .txt file. Makes sense?

by: gazdzidPosted on 2009-10-15 at 09:19:17ID: 25582121

hielo:

You wrote: The point I am trying to make is that NOT ALL files on the server are interpreted. In your case, it seems that your .java files are not executed. Perhaps a server misconfiguration?

Thank you for your kind reply and insight. I do not believe this to be a server misconfiguration issue. i believe it has to do with identifying the correct url variable and possibly the correct web.xml code. Let us see if xxli79, kadaba: or any other experts have differing opinions or suggestions.

by: hieloPosted on 2009-10-15 at 12:06:50ID: 25583660

>> I do not believe this to be a server misconfiguration issue
Understood, but the .java file is definitely not the file you need to reference. The java files are meant to be compiled and then you execute the resulting file. Most likely this is what the server needs to execute - the compiled file. Reading kadaba's other response, I believe that is what he is referring to when he references 
 
<url-pattern>/AjaxResponseServlet </url-pattern>
 
On another note, why did you NOT tag/post this question under JSP? You will get more help if you post it in the most appropriate section. I suggest you contact community support and ask them to move it to the JSP section.

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by: gazdzidPosted on 2009-10-15 at 15:35:29ID: 25585585

kadaba: comment 25480292 States: generally cfobkdb will be the project name which will also be the docbase. I hope this is the same in your case.

Honestly, I do not know the name of my project, maybe someone can tell by looking at the diagram found in the word document

by: gazdzidPosted on 2009-10-15 at 15:45:25ID: 25585634

one other observation:

found within my server code
I do not have a pakage identified. Is this a bad thing?

Thank you for your support!!!

by: xxli79Posted on 2009-10-15 at 20:05:42ID: 25586804

I'v checked you web.xml.

First, make your servlet run:
1. define a package name for your servlet class.
at the first line of your .java file, add a new line: package ajaxtest;
and move your .java file to a new directory named ajaxtest.
2. modify the web.xml:
<servlet-class>ajaxtest.AjaxResponseServlet</servlet-class>
3. redeploy your webapp and test it directly:
http://xxxx[:port]/[webapp context]/AjaxResponseServlet
And you can referrence: http://java.sun.com/j2ee/tutorial/1_3-fcs/doc/Servlets.html

Second, make ajax run:
1. replace your url as:
http://xxxx[:port]/[webapp context]/AjaxResponseServlet?key=" + encodeURI(key.value);
2. test again.

by: gazdzidPosted on 2009-10-16 at 10:46:39ID: 25591636

Hello xxli79,
1. define a package name for your servlet class. (See document attached to this post
Created package - Deleted old package created new
Path within web.xml - JavaSource/servletdg/AjaxResponseServlet.java

2. modify the web.xml: My code is as follows but still have question with observation stated below.
<servlet-class>JavaSource.serveletdg.AjaxResponseServlet</servlet-class>
The following url provides a 404 (connecting to java)

Item # 3 tried test both urls both result in 404
var url = "http://localhost:8080/kdb_eldb/cfobeldb/JavaSource/servletdg/AjaxResponseServlet?key=" + escape(key.value);

var url = "http://localhost:8080/kdb_eldb/cfobeldb/JavaSource/servletdg/AjaxResponseServlet?key=" + encodeURI(key.value);

Observation
The web.xml seems to have an error as there is a yellow spot on the globe not sure what this means or how to fix. Maybe there is a way to rebuild??

Please not prior to test, I did stop and restart tomcat.

I think I am getting close

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4"
	xmlns="http://java.sun.com/xml/ns/j2ee"
	xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
	<display-name>cfobkdb</display-name>
	<welcome-file-list>
		<welcome-file>blank-int.jsp</welcome-file>
	</welcome-file-list>
 
	<servlet>
		<servlet-name>AjaxResponseServlet</servlet-name>
		<servlet-class>JavaSource.serveletdg.AjaxResponseServlet</servlet-class>
	</servlet>
	<servlet-mapping>
		<servlet-name>AjaxResponseServlet</servlet-name>
		<url-pattern>/AjaxResponseServlet</url-pattern>
	</servlet-mapping>
 
</web-app>

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directory.doc
- 50 KB
directory showing servlet java ans error web xml

by: gazdzidPosted on 2009-10-16 at 11:31:04ID: 25592053

I think the issue is due web.xml under problems I see a warning that states
Description
CHKJ3004W: The servlet mappling refers to a servlet that is not defined

Code can be found below

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4"
	xmlns="http://java.sun.com/xml/ns/j2ee"
	xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
	<display-name>cfobkdb</display-name>
	<welcome-file-list>
		<welcome-file>blank-int.jsp</welcome-file>
	</welcome-file-list>
 
	<servlet>
		<servlet-name>AjaxResponseServlet</servlet-name>
		<servlet-class>servletdg.AjaxResponseServlet</servlet-class>
	</servlet>
	<servlet-mapping>
		<servlet-name>AjaxResponseServlet</servlet-name>
		<url-pattern>/AjaxResponseServlet</url-pattern>
	</servlet-mapping>
 
 
 
 
 
</web-app>

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by: gazdzidPosted on 2009-10-16 at 12:21:07ID: 25592461

I was able to select project/build and the error was removed now tring to work on identifying the correct url

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4"
	xmlns="http://java.sun.com/xml/ns/j2ee"
	xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
	<display-name>cfobkdb</display-name>
	<welcome-file-list>
		<welcome-file>blank-int.jsp</welcome-file>
	</welcome-file-list>
 
	<servlet>
		<servlet-name>AjaxResponseServlet</servlet-name>
		<servlet-class>servletdg.AjaxResponseServlet</servlet-class>
	</servlet>
	<servlet-mapping>
		<servlet-name>AjaxResponseServlet</servlet-name>
		<url-pattern>/JavaSource.servletdg.AjaxResponseServlet</url-pattern>
	</servlet-mapping>
 
 
 
 
 
</web-app>

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by: gazdzidPosted on 2009-11-05 at 07:07:59ID: 31641186

I did find the issue. Thanks!!

20120131-EE-VQP-002

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Almost as Easy as Hello World # 3- Ajax

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Ajax - html - javascript

Hypertext Markup Language (HTML)

JavaScript

Asynchronous Javascript and XML (AJAX)

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