jQuery:选择Box没有响应

时间:2015-06-12 16:35:43

标签: javascript php jquery

我想在选择选择框时从数据库中输入<input type="text"></input>个文本。这是我的jQuery代码

<script type="text/javascript" src="jquery.js"></script>
<script>
    $(document).ready(function() {
    $('#noID').change(function() {
        var noID = $(this).val(); 
            $.ajax({
                type: 'POST',
                url: 'ambilusername.php', 
                data: 'noID=' + noID,
                success: function(response) { 
                $('#username').val(response); 
                }
            });
        });
    });
</script>

这是我的&#34; ambilusername.php&#34;

<?php
include "../include/koneksi_db.php";

$noID = $_POST["noID"];
$username = mysql_query("SELECT * FROM booking_tamu WHERE noID='$noID'");
while($k = mysql_fetch_array($username)){
    echo "$k[email]";
}
?>

这是我的完整index.php代码:

<?php 

include "../include/koneksi_db.php";

?>
<html>
<head>
    <script type="text/javascript" src="jquery.js"></script>
    <script>
        $(document).ready(function() {
        $('#noID').change(function() {
            var noID = $(this).val();
                $.ajax({
                    type: 'POST',
                    url: 'ambilusername.php',
                    data: 'noID=' + noID, 
                    success: function(response) { 
                    $('#username').val(response); 
                    }
                });
            });
        });
    </script>
</head>
<body>
<form method="post" action="?page=prosesTambahAkun">
<table class="table-data" width=100% border=1>
    <tr><td colspan="2" class="head-data">Tambah Data Akun</td></tr>
    <tr>
    <td class="pinggir-data">ID Pemesan</td>
    <td>
        <select name="noID" id="noID" required>
            <option value="">------Pilih ID Pemesan------</option>
                <?php
                    $query = "SELECT no_booking, COUNT(no_booking) AS 'jumlah' FROM booking_tamu GROUP BY no_booking ASC";
                    $hasil = mysql_query($query);
                    while ($data = mysql_fetch_array($hasil)){
                    echo "<option value='".$data['no_booking']."'>".$data['no_booking']."</option>";
                    }
                ?>
        </select>
    </td>
    </tr>
    <tr>
    <td class="pinggir-data">Username</td>
    <td>
        <input type="text" id="username" name="username" size="30px" readonly></input>
    </td>
    </tr>

    <tr><td colspan="2" align="center" class="head-data">
        <input type="submit" value="Tambah" name="tambah">
    </td></tr>
</table>
</form>
</body>
</html>

问题是当我在选择框中选择值时,文本框不会在其上放置一些内容。你能帮帮我吗?

1 个答案:

答案 0 :(得分:0)

问题是表格列的命名:

改变这个:

$username = mysql_query("SELECT * FROM booking_tamu WHERE noID='$noID'");

到此:

$username = mysql_query("SELECT * FROM booking_tamu WHERE no_booking='$noID'");