Issue when passing variable with AJAX to PHP to Populate a Dropdown menu

Question

I'm trying to populate two dropdown menus with AJAX that also link to a MySQL database. Dropdown 1 = Brand and Dropdown 2 = Model

The only issue I'm having is passing the variable from Dropdown 1 to another page so that a MySQL search can be run.

Here is the AJAX function:

function AjaxFunction()
{
var httpxml;
try
  {
  // Firefox, Opera 8.0+, Safari
  httpxml=new XMLHttpRequest();
  }
catch (e)
  {
  // Internet Explorer
          try
                    {
                 httpxml=new ActiveXObject("Msxml2.XMLHTTP");
                    }
            catch (e)
                    {
                try
            {
            httpxml=new ActiveXObject("Microsoft.XMLHTTP");
             }
                catch (e)
            {
            alert("Your browser does not support AJAX!");
            return false;
            }
            }
  }
function stateck()
    {
    if(httpxml.readyState==4)
      {
//alert(httpxml.responseText);
var myarray = JSON.parse(httpxml.responseText);
// Remove the options from 2nd dropdown list
for(j=document.testform.model.options.length-1;j>=0;j--)
{
document.testform.model.remove(j);
}
for (i=0;i<myarray.data.length;i++)
{
var optn = document.createElement("OPTION");
optn.text = myarray.data[i].ModelName;
optn.value = myarray.data[i].ModelName;  // You can change this to subcategory
document.testform.model.options.add(optn);

}
      }
    } // end of function stateck
    var url="dd.php";
var cat_id=document.getElementById('s1').value;
url=url+"?cat_id="+cat_id;
url=url+"&sid="+Math.random();
httpxml.onreadystatechange=stateck;
//alert(url);
httpxml.open("GET",url,true);
httpxml.send(null);
  }
</script>

here is the code for the relevant form. This works in generating the First field.

<form name="testform" method='POST' action='mainck.php'>
Name:<input type=text name=fname>
<?php
echo "<br>Select Brand <select name='brand' id='s1' onchange=AjaxFunction();>
<option value=''>Select One</option>";

$sql="SELECT DISTINCT Id, BrandName FROM tbl_brands order by BrandName"; // Query to collect data from table

foreach ($dbo->query($sql) as $row) {
echo "<option value=$row[Id]>$row[BrandName]</option>";
}
?>
</select>

Then it uses a page called dd.php which GETS the variable from first dropdown selection, runs a MySQL query and then outputs information into the second drowpdown.

<?php
@$cat_id=$_GET['Id'];
//$cat_id=2;
/// Preventing injection attack ////
if(!is_numeric($cat_id)){
echo "Data Error";
exit;
 }
/// end of checking injection attack ////
require "config.php";

$sql="SELECT ModelName FROM tbl_model WHERE tbl_model.Id IN (SELECT DISTINCT ModelId FROM tbl_products WHERE BrandId=$brand)";
$row=$dbo->prepare($sql);
$row->execute();
$result=$row->fetchAll(PDO::FETCH_ASSOC);

$main = array('data'=>$result);
echo json_encode($main);
?>

the issue is the page is not getting the variable. If I comment out the @$cat_id=$_GET['Id']; and then replace that with a set variable of $cat_id=2; the whole thing works perfectly.

The Problem: is passing the variables. The main page is not sending the variable to my page dd.php

Swaraj Giri · Accepted Answer · 2015-08-13 09:50:37Z

4

While making the xmlhttp request your url is something like url=url+"?cat_id="+cat_id; but in your dd.php you are trying to get $_GET['Id']; which wont be there.

Replace $_GET['Id']; with $_GET['cat_id'];.

Note- Never suppress errors using @. Hiding something does not make it non existent. Also use a linter in your ide/editor.

answered Aug 13, 2015 at 9:50

Swaraj Giri

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1 Comment

Jesse Over a year ago

Thank you. Cannot believe I missed this. I spent 2hrs on this!

Hasan Tareque · Accepted Answer · 2015-08-13 09:53:19Z

1

From your url on ajax function i can see you sending the variable as cat_id but you tring catch it on php side as Id

The variable name have to be same on both side.

Try $_GET['cat_id'] instead of $_GET['Id']. should work.

answered Aug 13, 2015 at 9:53

Hasan Tareque

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