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我是一個新手，至於JQuery，AJAX & PHP的問題，我正在嘗試構建一個帶有鏈接選項的表單，它將爲新的mysql記錄提供初始數據。 我已經下載了JQuery鏈接選擇遠程插件，併爲我的生活，我無法弄清楚如何獲取第二個選擇下拉列表返回的數據。第一個下拉列表使用另一個函數填充，當用戶從此下拉列表中選擇一個值時，我希望將該值用作第二個下拉列表的搜索字符串。 的代碼段的主要形式是：JQuery鏈式選擇實現
<?php
<head>
<script type="text/javascript" src="../js/jquery-1.7.2.min.js"></script>
<script type="text/javascript" src="../js/jquery.chained.remote.js" charset="utf-8"></script>
</script>
<script type="text/javascript">
$('#series').remoteChained("#series", "../models/json.php");
</script>
</head>
<body>
<form action="" method="post">
<table border="1">
<tr>
<th>Season:</th>
<td><select name="season" id="mark">
<option value="">Select Season...</option>
<?php
$i = 0;
while ($i < count($showseason)){ // obtain seasons from getseason() function
?>
<option value="<?php echo $showseason[$i]; ?>">
<?php echo $showseason[$i]; ?> </option>
<?php
$i++;
} ?>
</select></td>
<th>Competition:</th><td><select name="competition" id="series">
<option value="">Competition type...</option>
</select></td>
以下是爲MySQL的代碼選擇（json.php）
<?php
include_once $_SERVER['DOCUMENT_ROOT'] . '/includes/helpers.inc.php';
include_once $_SERVER['DOCUMENT_ROOT'] . '/includes/db_connect.php';
try
{
$result = $pdo->prepare("SELECT sideid, competition FROM pennantsides");
$result->execute();
echo json_encode($result->fetchAll(PDO::FETCH_ASSOC));
}
catch (PDOException $e)
{
$output = 'Error selecting records - Please contact your Site Administrator';
include $_SERVER['DOCUMENT_ROOT'] . '/views/output.html.php';
exit();
}