JSON php MySql menu not showing data in HTML page

Question

I'm trying to populate a drop-down menu from a MySql.

My problem is that the data from JASON are not showing in my HTML page.

This is what I want to achieve. ID: Select ID

JASON //This works and the output: {"article1":{ "title":"acGH2867" },"article2":{ "title":"apGS0158" }}

$jsonData = '{';
    foreach ($conn_db->query("SELECT customerID FROM customers WHERE furniture='33' ") as $result){
        $i++;
        $jsonData .= '"article'.$i.'":{ "title":"'.$result['customerID'].'" },';
    }
$jsonData = chop($jsonData, ",");
$jsonData .= '}';
echo $jsonData;

HTML

<script type="text/javascript">
    $(document).ready(function(){
    var ddlist = document.getElementById("ddlist");
    var hr = new XMLHttpRequest();
    hr.open("cData.php", true);
    hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    hr.onreadystatechange = function() {
        if(hr.readyState == 4 && hr.status == 200) {
            var d = JSON.parse(hr.responseText);
            for(var o in d){
                if(d[o].title){
                    ddlist.innerHTML += '</option><option value='+d[o].title+'</option>';
                }
            }
        }
    }
    ddlist.innerHTML = "Loading....";

    $('#dlist').on('change', function(){
        $('#val1').value() = $(this).val();
        });
    });
</script>
</head>
<div class="dlist" id="ddlist">
</div>

Don't build your JSON manually. I would also think what you are looking for as output is an array structure enclosed by [] in JSON. — Mike Brant
– Mike Brant, Commented Jun 2, 2014 at 17:44
I suspect ddlist.innerHTML += '</option> isn't going to do what you think it does. (even though it likely will work anyway in most browsers) — user400654
– user400654, Commented Jun 2, 2014 at 17:45

Harish U Warrier · Accepted Answer · 2014-06-02 18:38:38Z

2

try this

    $jsonData = array();
        foreach ($conn_db->query("SELECT customerID as title FROM customers WHERE furniture='33' ") as $result)

{
            $i++;
            $jsonData["article'.$i]=$result;
        }
   echo json_encode($jsonData);

edited Jun 2, 2014 at 18:38

Harish U Warrier

6065 silver badges13 bronze badges

answered Jun 2, 2014 at 18:36

user3700326

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