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下面使用jquery选定状态改变是Jquery的代码城市不是动态根据我在下面的代码
$("#state").change(function(){
$.ajax
({
url: "www.myweb.com/ajax.abc.php",
type: "POST",
data: {state: function(){return $("#state").val()}}; /* i have given stateId as value*/
cache: false,
success: function(html)
{
$('#city').find('option').remove().end().append(html);
}
});
下面的代码在AJAX文件被表示(即ajax.abc.php)
if(isset($_POST['state']))
{
$getCity = mysql_query('SELECT * FROM tblCity WHERE stateId = '.$_POST['state']);
while($fetchCity = mysql_fetch_array($getCity))
{
echo '<option value="'.$fetchCity["cityName"].'">'.$fetchCity["cityName"].'</option>';
}
}
你是否进入'成功'?意味着你的ajax调用有效吗? –