Greetings to you all,

i have been working on ajax+php+mysql to make a data from table "NETWORK" be displayed using onkeyup(), which is working well.
diagram A
However, i also want the displayed data from the table "NETWORK" on the same form be submitted to another table "OUTWORK" in mysql using the onclick(). my problem is that it wont insert anything in the database nor display the result in the <div>result</div> as a confirmation or not.

Please kindly look at my codes in the picture and make valuable suggestion. thanks. Also please it should be strictly javascript+Ajax and not jquery because am yet to delve in but will later.

USAGE:
Dreamweaver CS5
Javascript + Ajax
Php
Mysql

codes

<script type='text/javascript' src='Finite.js'> </script>

<script language="JavaScript" type='text/javascript'>

		var http = createXmlHttpRequestObject();
		
		function createXmlHttpRequestObject() {
		var http;
		
		if (window.ActiveXobject){
				try{
				http = new ActiveXObject("Microsoft.XMLHTTP";
				}catch(e) {
					http = false;
				}
					}else{
			
				try{
				http = new XMLHttpRequest();
					}catch(e){
					http = false;
				}
				}
		if(!http)
					alert("cant create that object hoss!";
					else
					return http;
				}
				
function ajax_post(){
			
			//var http = new XMLHttpRequest();
			var url = "procform7.php";
			var Rn = document.getElementById("agid".value
			var Ln = document.getElementById("agtel".value
			var Mn = document.getElementById("agfn".value
			var Fn = document.getElementById("agsal".value
			//alert('good');
			var vars = "registrationno="+Rn+"&lastname="+Ln+"&middlename="+Mn+"&firstname="+Rn;
			//alert('good');
			
		//alert('MOVE');
			http.open("POST", url, true);
			http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
			
			http.onreadystatechange = function() {
				if(http.readyState == 4 && http.status == 200) {
					var return_data = http.responseText;
		
		document.getElementById("status".innerHTML = return_data;
			
					}
				
				}
		http.send(vars);
		document.getElementById("status".innerHTML = "processing...";
		}
		
</script>

<link rel="stylesheet" type="text/css" href="Doubld.css" />
<div class='stutDiv'>

<form name="schform">
        <table>
                <tr>
                      <td>Contact ID:</td>
                      <td><input id="agid" type="text"
                       name="contactid" onKeyUp="getagentids();"></td>
                </tr>
                <tr>
                      <td>Tel Number:</td>
                      <td><input id="agtel" type="text"
                            name="contacttel"></td>
                </tr>
                <tr>
                      <td>Name:</td>
                      <td><input id="agfn" type="text"
                            name="contactfullname"></td>
                </tr>
                <tr>
                      <td>Salutation:</td>
                      <td><input id="agsal" type="text"
                            name="contactsalutation"></td>
                </tr>
                <tr>
                      <td><input type="reset" value="Clear"></td>
                      <td> <div align="center">

               <input type="submit" name="submit" id="submit" value="Submit Data" onclick="javascript:ajax_post()"; />
                  </div></td>
                </tr>
        </table>
</form>

<br /> <br />

<div id="status"></div> //display report from the database
</div>

image

Place your codes on pastebin.com and give us the URLs
The codes you have placed here are presently not very useful or runnable because of that smiley bug.

Dual Core: Place your codes on pastebin.com and give us the URLs
The codes you have placed here are presently not very useful or runnable because of that smiley bug.

thanks dualcore, after using firebug, i discovered that i misspelt my variables. Its running perfectly thanks also for that site you posted, will check it out for future reference. Once again thank you for your willingness to assist.