asked on

Notice: Undefined index: func in .........include/forms/func.php on line 21

Hi Experts,

I am trying to re-create chained select boxes using Ajax from http://www.blueicestudios.com/chained-select-boxes-using-php-mysql-ajax/, but I am getting the following error. It is probably something very simple, but I can't seem to locate the problem. Any suggestions?

Notice: Undefined index: func in ......include/forms/func.php on line 21

reviewMarkup.php

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN http: //www.w3.org/TR/html4/loose.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en">
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
	$('#wait_1').hide();
	$('#drop_1').change(function(){
	  $('#wait_1').show();
	  $('#result_1').hide();
      $.get("func.php", {
		func: "drop_1",
		drop_var: $('#drop_1').val()
      }, function(response){
        $('#result_1').fadeOut();
        setTimeout("finishAjax('result_1', '"+escape(response)+"')", 400);
      });
    	return false;
	});
});

function finishAjax(id, response) {
  $('#wait_1').hide();
  $('#'+id).html(unescape(response));
  $('#'+id).fadeIn();
}
</script>

</head>
<body>
    <form action="" method="post">

    <?php
    error_reporting(E_ALL);
    include "include/userConnect.php";
    include "func.php";
    if (array_key_exists('loanAuditKey', $_POST)) {
          $loanAuditKey = $_POST['loanAuditKey'];
          } else {
          $loanAuditKey = $_GET['loanAuditKey'];
          }
        ?>
    <div align="center">
        </br>
          </br>  
   <select id="drop_1" name="drop_1" class="select">
                    <option value="" selected="selected" disabled="disabled">Select a Category</option>
      
      <?php getTierOne(); ?>
               </select>
    
    
    <span id="wait_1" style="display: none;">
    <img alt="Please Wait" src="../../images/ajax-loader.gif"/>
    </span>
    <span id="result_1" style="display: none;"></span> 
<?php if(isset($_POST['submit'])){
	$drop = $_POST['drop_1'];
	$tier_two = $_POST['tier_two'];
	echo "You selected ";
	echo $drop." & ".$tier_two;
}

?>
</form> 
</body> 
</html>

Open in new window

func.php

<?php
//**************************************
//     Page load dropdown results     //
//**************************************
function getTierOne()
{
	$result = mysql_query("SELECT DISTINCT findingCategoryKey, findingCategoryName FROM findingCategory WHERE active=1") 
	or die(mysql_error());

	  while($tier = mysql_fetch_array( $result )) 
  
		{
		   echo '<option value="'.$tier['findingCategoryKey'].'">'.$tier['findingCategoryName'].'</option>';
		}

}

//**************************************
//     First selection results     //
//**************************************
if($_GET['func'] == "drop_1" && isset($_GET['func'])) { 
   drop_1($_GET['drop_var']); 
}

function drop_1($drop_var)
{  
    include "include/userConnect.php";
	$result = mysql_query("SELECT findingKey, description, findingVerbiage FROM finding WHERE findingCategoryKey='$drop_var'AND active=1") 
	or die(mysql_error());
	
	echo '<select name="tier_two" id="tier_two">
	      <option value=" " disabled="disabled" selected="selected">Choose one</option>';

		   while($drop_2 = mysql_fetch_array( $result )) 
			{
			  echo '<option value="'.$drop_2['findingKey'].'">'.$drop_2['description'].'</option>';
			}
	
	echo '</select> ';
    echo '<input type="submit" name="submit" value="Submit" />';
}
?>

Open in new window

Ray Paseur

This is a Notice, not an Error. You may be able to suppress the message by adding the @ to the function call. Start the exploration here:
http://php.net/manual/en/function.error-reporting.php

rcowen00

ASKER

I guess I then have a slightly different question, the process works until I select the first drop-down, then the image is view-able in func.php line 54, then nothing. I am very inexperienced with Ajax, how do I go about trouble shooting it? thanks!

Jagadishwor Dulal

Can you post your table structure? I think you have problem in your select query, it seems there is no problem.... If I have created right fields.

SELECT findingKey, description, findingVerbiage FROM finding WHERE findingCategoryKey='$drop_var' AND active=1

Open in new window

Again Ray already write above your problem it's the problem by

 error_reporting(E_ALL);

Open in new window

comment this line and test.

ASKER CERTIFIED SOLUTION

Ray Paseur

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

Chris Stanyon

Change the order of line 21. Currently you're checking a value and then checking if it's set! You should check if it's set and the check the value...

if(isset($_GET['func']) && $_GET['func'] == "drop_1" ) { 
...
}

Open in new window

This way, if it's not set, it won't check the value, and therefore won't give you the NOTICE!

rcowen00

ASKER

Here is the table structure

I also commented out error_reporting(E_ALL); with no change except the notice doesn't show.

I am reviewing the links from Ray now.

Chris Stanyon

To prevent the NOTICE and fix your problem, read my previous post - I've already given you the answer.

DO NOT remove the error reporting directive while degugging - hiding error mressages is not a valid way of fixing your code ;)