Ajax Programming

Hii sir,

i want to write ajax programming for retrieving information using id and displayed into dropdown.
I have write the logic for that,but i dont know how to put that into Chronoforms.
Please help me.

Hi anil_admin,

Please check the FAQs for double drop-down. I think that is what you are trying to do.

Bob

Thanx sir reply my queries,

sir but i want to fetch the database using id.
Autofill the dropdowns and radiobutton.

Actually i have done this on localhost,it works properly.
but how it could be use in chronoforms..

Hi anil_admin,

The host shouldn't make any difference - if it works on localhost then it should work similarly in a form.

The double-drop-down won't work for radio buttons (but you didn't mention those in your first post).

You can do this putting your Ajax JavaScript into a Load JavaScript action and the code to return the look-up values in a new event; then set the Ajax URL to link to that event e.g. ...&event=ajax

Bob

SIr actully i have to build 2 php file . one for ajax programming and another for fetching data code .
so, how to add these 2 file codes in chronoforms in DBRead(on found) event.In the textbox when i have to type any number it will be fired into the database and get back the database.

ajax.php:Ajax code

<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
$(document).ready(function(){
	$("#code").keyup(function(){
		console.log("keyup");
		//ajax starts
		$.ajax({
		  type: "POST", 
		  url: "ajax1.php", 
		  data: { q: $("#code").val()} 
		}).done(function( msg ){ 
					
	      console.log(msg);
		  
          obj = JSON.parse(msg);
		  console.log(obj[0]);
		  console.log(obj);
		  $("#projectname").val(obj[0]); 
	        $("#fund").val(obj[1]); 
               	});	
	});
});
</script>

fetch.php: These is my fetch code

<?php
$q = $_REQUEST['q'];
if($q!="")
{
	$con = mysql_connect("localhost","root","");
	if(!$con)
	{
		die("Error has occured");
	}
	mysql_select_db("normal");

	$query="SELECT * from normal WHERE code=$q";//selecting names of cities
	
	$result = mysql_query($query,$con);
	if(mysql_num_rows($result)>0)
		{			
		while($row=mysql_fetch_assoc($result))
			{
			 $output=array($row['projectname'],$row['fund']);
			 
			}
			echo json_encode($output);
		}
}
?>

Sir how to add these 2 files .

Hi anil_admin

You can do this putting your Ajax JavaScript into a Load JavaScript action and the code to return the look-up values in a new event; then set the Ajax URL to link to that event e.g. ...&event=ajax

You don't need to load the jquery.js file and you don't need <script> tags in the Load JavaScript box.

Bob

Hello sir,

I have to fetch value from database using ajax and json and it will be put down into dropdown .
but when i enter another code into "code" textbox the previous value is still in the dropdown and next value come down in dropdown.

var projectname = msg.projectname;
jQuery("#projectname").append($('<option>', {
 text: projectname
  }));

So how i select particular value in the dropdown that i fetch from the database.

Hi anil_admin,

I'm sorry I really don't understand what you are trying to do here.

If you want to set the new option as selected then you can do that in your jQuery code.

Bob