Debugging MySQL query in PHP when called from other page

Question

Page1 has an input form. I validate the input field with a JavaScript:

<input type="text" name="frmBrand" size="50" onkeyup="BrandCheck();" maxlength="100" id="frmBrand"   />
<span id="frmBrand_Status">Enter existing or new brand</span>

In the JavaScript I then call a PHP script:

function BrandCheck()
{
    var jsBrandName = document.forms["AddPolish"]["frmBrand"].value;

    if (jsBrandName !==null || jsBrandName !== "")
        {
        document.getElementById("frmBrand_Status").textContent = jsBrandName
        // alert(jsBrandName);
        var xmlhttp = new XMLHttpRequest();
        var url = "CheckBrand.php";
        var vars = "jsBrandName="+jsBrandName;
        xmlhttp.open("POST",url,true);
        xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        xmlhttp.onreadystatechange = function()
            {
                if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
                {
                    var return_data = xmlhttp.responseText;
                    document.getElementById("frmBrand_Status").innerHTML = return_data;
                }
            }
        xmlhttp.send(vars);
        document.getElementById("frmBrand_Status").innerHTML = "processing.....";
        }
} 

So far so good. I do get results from the CheckBrand.php because it changes the frmBrand_Status. But I can't get any database results from the PHP page.

<?php
    if(mysqli_connect_errno()) { //if connection database fails
        echo("Connection not established ");
    }  
    //by now we have connection to the database
    else 
    {
    if(isset($_POST['jsBrandName']))
        { //if we get the name succesfully
            $jsBrandName = $_POST['jsBrandName'];
            $dbBrandName = mysql_real_escape_string($jsBrandName);
            if (!empty($dbBrandName)) 
            {
                $dbBrandName = $dbBrandName . "%";
                $sqlQuery = "SELECT `BrandName` FROM `Brand` WHERE `BrandName` like '$dbBrandName'  ORDER BY `BrandName`";

                $result = mysqli_query($con, $sqlQuery);    
                $NumRows = mysqli_num_rows($result);
                // $BrandName_result = mysql_fetch_row($BrandName_query);
                echo "Result " . $dbBrandName . " ----- ". $jsBrandName . "Number rows " .$NumRows. " BrandName = " .$result. " SQL " .$sqlQuery;

                if( $BrandName_result = mysql_fetch_row($BrandName_query))
                {
                    While ($BrandName_result = mysql_fetch_row($BrandName_query))
                    {
                        echo "Brand = " .$BrandName_result[0];
                    }
                }
            }
            else
            {
                echo "dbBrandName = empty" . $dbBrandName;
            }
        }
    }
?>

When doing this, the html page shows the constant change of the normal variables. For example when the input field holds "Clu" I get the following output the span ID frmBrand_Status:

Result Clu% ----- CluNumber rows BrandName =  SQL SELECT `BrandName` FROM `Brand` WHERE `BrandName` like 'Clu%' ORDER BY `BrandName`

Which looks good as the brandname gets the % appended, but the Number of rows is not shown (empty field?), the SQL Query is shown and looks good, but I don't get any results.

And the if( $BrandName_result = mysql_fetch_row($BrandName_query)) section will not be reached, so there definitely is something going wrong in calling the query.

When I run that same query through PHPMyAdmin, i do get the result I expect, which is 1 row with a brandname.

I'm using firebug to try and troubleshoot the SQL Query, but I can't find where I can check this and I probably can't since PHP is serverside. correct? But how should I then trouble shoot this?

Answer 1

Found what was wrong. The $con string I was using to open the database was no longer available. On other pages in the site, the $con is available, I load the database using an include script on my index page. But it seems that the variable gets lost when it is called through the XMLHttpRequest(). Which is logical now I think of it, since this can also be a call to a remote server. So my CheckBrand.php page was just missing the $con var to connect to the database.