此ps_search.php页面包含笔记本电脑,行李等类别的下拉列表以及城市下拉列表。如果用户想要发布广告意味着他/她想要在网站上销售某些东西,首先选择类别如笔记本电脑,手提包或其他任何类别。然后从广告发布的位置选择城市。例如,如果用户从city1发布了笔记本电脑的广告,我们就没有其他城市的笔记本电脑广告。然后在选择笔记本电脑后,城市1应出现在下拉菜单中。我有两个文件ps_search.php ang get_city.php
// ps_search.php
<form name="search" method="post" action="<?php echo $base_url ?>search_rides.php?go">
<table width="100%" border="0" cellpadding="4" cellspacing="0" id="tbsrch-engine">
<tbody>
<tr>
<td height="61">
<strong>Choose Category:</strong>
</td>
<td>
<select name="category_id" id="category_id" style="width: 155px;" onChange="get_city(this.value,'<?php echo $base_url ?>dropdown/get_city.php')">
<option value="0">Any</option>
<?php $query = "SELECT category_id as id, category_name as name FROM tbl_ps_category ";
$result = mysql_query($query);
?>
<?php while ( $row = mysql_fetch_array($result)) {?>
<option value="<?php echo $row['id'].'-'.$row['name']; ?>"><?php echo $row['name']?></option>
<?php }?>
</select>
</td>
</tr>
<tr>
<td height="40">
<strong>City:</strong>
</td>
<td>
<div id="models">
<select name="city_id" id="city_id">
<option value="0">Any</option>
</select>
</div>
</td>
</tr>
<tr>
<td height="32"/>
<td>
<input type="submit" name="button2" id="button2" value="Search an Ad" class="fbutton"/>
</td>
</tr>
</tbody>
</table>
</form>
// get_city.php
<?php
include('../Connections/photohive.php');
$id = $_REQUEST['id'];
$explode = explode('-',$id);
$id = $explode[0];
$sql = "SELECT city_id FROM ".$ps_prefix."product WHERE category_id=".$id;
$query = mysql_query($sql);
//exit;
?>
<select name="city_id" id="city_id" style="width:155px;">
<option value="0">Any</option>
<?php
while ($row = mysql_fetch_array($query)){
$q1= sprintf("Select cityname from tbl_city where city_id='%s'" , mysql_real_escape_string($row['city_id']), " ORDER BY cityoreder ASC");
$r1= mysql_query($q1);
while($row2= mysql_fetch_assoc($r1))
{
?>
<option value="<?php echo $row['city_id']?>"><?php echo $row2['cityname']?></option>
<?php
}
}
?>
</select>
我也包含了js文件
<script src="<?php echo $base_url;?>js/jquery.ajaxq-0.0.1.js" type="text/javascript"></script>
请帮助我....
答案 0 :(得分:0)
您的javascript应如下所示:
(function ($) {
// When category select changes
$('#category_id').change(function() {
// make ajax query to get cities
$.get('get_city.php', {id: $(this).val())
.done(function(response) {
// get the cities dropdown from response and insert into page
$('#models').html(response);
});
});
})(jQuery);
我个人不喜欢在我的ajax调用中返回HTML。我通常更喜欢返回一个JSON对象,然后解析出HTML客户端(通常使用Handlebars)。我认为它更清洁。