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我正在創建2個下拉列表,第二個是基於第一個下拉列表的選擇。將數據從MySQL數據庫中檢索php ajax droplist
的index.php
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8" />
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Playing With Select list</title>
<link rel="stylesheet" type="text/css" href="css/demo.css" />
<link href='http://fonts.googleapis.com/css?family=Open+Sans:300,700' rel='stylesheet' type='text/css' />
<!--[if lte IE 8]><style>.main{display:none;} .support-note .note-ie{display:block;}</style><![endif]-->
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function()
{
$(".country").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "ajax_category.php",
data: dataString,
cache: false,
success: function(html)
{
$(".governorate").html(html);
}
});
});
});
</script>
</head>
<body>
<div class="container">
<header>
<h1><strong>Playing With Select List</strong></h1>
<h2>Select One List To see Output On Other</h2>
</header>
</div>
<span style="margin-left:22%">
<label>country :</label> <select name="country" class="category">
<option selected="selected">--Select Country--</option>
<?php
include('db.php');
$sql=mysql_query("select country_id,country_name from country");
while($row=mysql_fetch_array($sql))
{
$id=$row['country_id'];
$data=$row['country_name'];
echo '<option value="'.$id.'">'.$data.'</option>';
} ?>
</select>
<label>Governorate :</label> <select name="governorate" class="subcategory">
<option selected="selected">--Select governorate--</option>
</select>
</span>
<br><br><br>
<h1><center><strong>Go To-:<a href="www.tricktodesign.com">TrickToDesign</a></strong></center></h1>
</body>
</html>
ajax_category.php
<?php
include('db.php');
if($_POST['governorate_id'])
{
$id=$_POST['governorate_id'];
$sql=mysql_query("select b.governorate_id,b.governorate_name from governorate a,contry_id b where b.country_id=a.country_id and parent='$id'");
while($row=mysql_fetch_array($sql))
{
$id=$row['governorate_id'];
$data=$row['governorate_name'];
echo '<option value="'.$id.'">'.$data.'</option>';
}
}
?>
如何讓第二個下拉列表出現在它的數據是錯誤,我面對
沒有MIIB我wnatŧ他省,而不是子類別,但數據不apear如何fx它可以幫助我? – 2013-02-28 22:38:39