Question

我试图从MySql填充下拉菜单。

我的问题是来自JASON的数据没有显示在我的HTML页面中。

这就是我想要实现的目标。 ID：选择ID

JASON //这是有效的和输出：{＆＃34; article1＆＃34;：{＆＃34; title＆＃34;：＆＃34; acGH2867＆＃34; }，＆＃34; article2＆＃34;：{＆＃34; title＆＃34;：＆＃34; apGS0158＆＃34; }}

$jsonData = '{';
    foreach ($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>

Answer 1

试试这个

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

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

JSON php MySql菜单没有在HTML页面中显示数据

1 个答案: