สอบถาม การ insert ครับ โดยถ้า insert ข้อมูลที่มีอยู่ใน table แล้ว จะไม่สามารถ inset ซํ้าได้ ต้องใช้หลักการไหนครับ

<form id="form1" name="form1" method="post" action="">
  ID <input name="id" type="text" id="id" />
     <input type="button" name="Button" value="Check" onclick="doCallAjax(document.getElementById('id').value)" />
</form>

<script>
function doCallAjax(search_id) {
HttPRequest = false;
if (window.XMLHttpRequest) { // Mozilla, Safari,...
HttPRequest = new XMLHttpRequest();
	if (HttPRequest.overrideMimeType) {
	HttPRequest.overrideMimeType('text/html');
	}
} else if (window.ActiveXObject) { // IE

	try {
	HttPRequest = new ActiveXObject("Msxml2.XMLHTTP");
	} catch (e) {
		try {
		HttPRequest = new ActiveXObject("Microsoft.XMLHTTP");
		} catch (e) {}
	}
}

if (!HttPRequest) {
alert('Cannot create XMLHTTP instance');
return false;
}

var url = 'chk_same.php';
var pmeters ="id="+search_id;

HttPRequest.open('POST',url,true);
HttPRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
HttPRequest.setRequestHeader("Content-length", pmeters.length);
HttPRequest.setRequestHeader("Connection", "close");
if(search_id!=""){
HttPRequest.send(pmeters);
}else{
alert("Please insert id")
}
	HttPRequest.onreadystatechange = function(){
		if(HttPRequest.readyState == 4) // Return Request
		{
			if(HttPRequest.responseText==0){
			alert("Not same");
			}else{
			alert("Same");
			}
		}
	}
}
</script>

<?php
$host="localhost";
$username="root";
$passwd="";
$dbname="test";
$conn = mysql_pconnect($host,$username,$passwd)or die("Can't connect to Server");
mysql_select_db($dbname)or die("Can't connect to Database");
$sql = "SELECT COUNT(good_id) AS cid FROM good WHERE good_id=".$_POST["id"];
$query = mysql_query($sql)or die("Can't query");
$objquery = mysql_fetch_array($query);
echo $objquery['cid'];
?>