Question

我尝试使用下拉列表在WordPress页面中显示我的数据库中的一些信息（Adresse），但是当我从下拉列表中选择一个项目（Nom）时，它什么也没显示。我确实在这里按照了正确的答案：ajax call to populate form fields from database query when select value changes 所以我不知道我在这里犯的错误：

下拉列表代码：

<select class='Nom' onchange="showUser(this.value)" name='Nom' id='Nom'>
<option value="">--- Select ---</option>

    [insert_php]
    $servername = "localhost";
    $username = "root";
    $password = "";
    $conn = new mysqli($servername, $username, $password, "somapam_bd");

     $sql = mysqli_query($conn, "SELECT Nom FROM herboristes");

    while($ligne_liste=mysqli_fetch_array($sql)) {
        echo '<option value="'.$ligne_liste['Nom'].'">'.$ligne_liste['Nom']."</option>\n";
    }
    echo '</select>';
    [/insert_php]
    <div id="txtHint" type="text" name="txtHint"><b>Person info will be listed here...</b></div>

onchange函数我将它插入index.php：

<script>
function showUser(str)
{
if (str=="")
{
    document.getElementById("txtHint").value="";
    return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
    if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {

      document.getElementById("txtHint").value = this.responseText;

    }
}
xmlhttp.open("GET","formdata.php?q="+str,true);
xmlhttp.send();
}
</script>

formdata.php代码：

 <?php

$servername = "localhost";
$username = "root";
$password = "";
$conn = new mysqli($servername, $username, $password, "somapam_bd");



  $q = intval($_GET['q']);


 $result = mysqli_query($conn, "SELECT * FROM herboristes where Nom = '".$q."'");


while($row=mysqli_fetch_array($result))
    {
    $cID = $row['Adresse'];

}
echo $cID ;


   ?>

Onchange功能无法在wordpress页面

0 个答案: