php count error - php

Question

As you see I have created drop down and give it name the field name, and I have also the count function.
What I want to do is, if some select drop down menu..show the how many result found in number like 10, 20,.. if the second dropdown selected it will check the two drop down selected and pass the result count..like that continuous..if you want to see example what exactly I want is go here and choose car the count will update..
http://en.comparis.ch/Carfinder/marktplatz/Search.aspx
I have the ajax and working well but I can't get the exact PHP code to count in live time.
AJAX Code..
var xmlHttp
function showCount(str)
{
xmlHttp=GetXmlHttpObject();
if (xmlHttp==null)
{
alert ("Your browser does not support AJAX!");
return;
}
var url="phpApplication.php";
url=url+"?action=count2";
url=url+"&sid="+Math.random();
xmlHttp.onreadystatechange=stateChanged;
xmlHttp.open("POST",url,true);
xmlHttp.send(null);
}
function stateChanged()
{
if (xmlHttp.readyState==4)
{
document.getElementById("countR").innerHTML=xmlHttp.responseText;
}
}
function GetXmlHttpObject()
{
var xmlHttp=null;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
}
return xmlHttp;
}
PHP Code
function dropdown($field, $table)
{
//initialize variables
$oHTML = '';
$result = '';
//check to see if the field is passed correctly
if (($field == "")||($table == ""))
{
die("No column or table specified to create drop down from!");
}
$sql = "select distinct($field) from $table";
//call the db function and run the query
$result = $this->conn($sql);
//if no results are found to create a drop down return a textbox
if ((!$result) ||(mysql_num_rows($result)==0))
{
$oHTML .= "<input type='text' name='$field' value='' size='15'>";
}elseif (($result)&&(mysql_num_rows($result)>0)){
//build the select box out of the results
$oHTML .= "<select name='$field' onchange='showCount(this.value)'>\n<option value='all'>All</option>\n";
while ($rows = mysql_fetch_array($result))
{
$oHTML .= "<option value='".$rows[$field]."' name='".$rows[$field]."'>".$rows[$field]."</option>\n";
}
$oHTML .= "</select>\n";
}
//send the value back to the calling code
return $oHTML;
}//end function
function count1(){
$sql2 = "SELECT SQL_CALC_FOUND_ROWS * from produkt_finder_table where Bauform_d ='".($_POST['Bauform_d'])."' ";
$query = $this->conn($sql2);
$result = mysql_query( $query );
$count = mysql_result( mysql_query( 'SELECT FOUND_ROWS()' ), 0, 0 );
echo $count;
//$sql2 = "SELECT COUNT(Bauform_d) from produkt_finder_table where Bauform_d = 'mobil' ";
//echo var_dump($result1);
while($row = mysql_fetch_array($query))
{
// echo $row['COUNT(Bauform_d)'];
}
//echo mysql_num_rows($query);
// if (isset($_POST['Bauform_d']))
//{
/* if (mysql_num_rows($result)==0) {
echo '0';
} else {
echo mysql_num_rows($result);
$row = mysql_fetch_array($result);
echo $result;
echo $row['COUNT(Bauform_d)'];
// }
}*/
}
$action = $_GET['action'];
$proFin = new Application();
switch($action) {
case 'show':
$proFin->show_form();
break;
case 'search':
$proFin->search();
break;
case 'searchAll':
$proFin->searchAll();
break;
case 'count2':
$proFin->count1();
break;
}

Answer 1

What parts have you debugged?
Change the count1() function to just echo back the time or something using time().
Then if it returns the correct value you know your JS is working and your PHP script is calling the correct function.
I assume that the PHP code doesnt work as the SQL query is looking for $_POST['Bauform_d'] which isnt set when you call the xmlHTTP Request. Run a simple print_r($_POST); to make sure you are passing all the data you expect in the query. If it is not then change your JS code to pass the value - when you are sure your php script is being passed all the correct variables then begin to add back in your SQL query etc.

Answer 2

debug debug
Jacob's answer is the key.