Multiple Forms

[echo1951]

I have a list of search results and want now to display for each result a small form in which the user can submit data. I found the script below that can do it for one form. The problem is that I do not know how to display the feedback value ajaxDisplay for each form.

 

function ajaxFunction(){

 

var ajaxRequest;  // The variable that makes Ajax possible!

 

try{

// Opera 8.0+, Firefox, Safari

ajaxRequest = new XMLHttpRequest();

} catch (e){

// Internet Explorer Browsers

try{

ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");

} catch (e) {

try{

ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");

} catch (e){

// Something went wrong

alert("Your browser broke!");

return false;

}

}

}

// Create a function that will receive data sent from the server

ajaxRequest.onreadystatechange = function(){

if(ajaxRequest.readyState == 4){

var ajaxDisplay = document.getElementById('ajaxDiv');

ajaxDisplay.innerHTML = ajaxRequest.responseText;

}

}

var ProfileID = document.getElementById('ProfileID').value;

var invProfileID = document.getElementById('invProfileID').value;

var status = document.getElementById('status').value;

var queryString = "?ProfileID=" + ProfileID + "&invProfileID=" + invProfileID + "&status=" + status;

 

ajaxRequest.open("GET", "ajax-example.php" + queryString, true);

ajaxRequest.send(null);

}

 

<form name='myForm'>

ID: <input type='text' id='ProfileID' /> <br />

invID: <input type='text' id='invProfileID' />

<input type=hidden id='status' value='1'>

<input type='button' onclick='ajaxFunction()' value='Query MySQL' />

</form>

<div id='ajaxDiv'>Your result will display here</div>