[英]How to send 2 select box dynamically generated values via ajax to get 3rd select box values
我試圖根據前兩個選擇框選擇(動態值)獲得第三個選擇框值;
jQuery ajax代碼是(僅適用於機場選擇框)
$("#airport").change(function() {
var aid=$(this).val();
var dataString = 'aid='+ aid;
$.ajax
({
type: "POST",
url: "booking/findcompany.php",
data: dataString,
cache: false,
success: function(data) {
$("#company").html(data);
}
});
});
<select name="site" id="site" class="site">
<option value="" selected="selected">Select</option>
<option value="1">Site One</option>
<option value="2">Site Two</option>
<option value="3">Site Three</option>
</select>
<select name="airport" id="airport" class="airport">
<option value="1">Airport One</option>
<option value="2">Airport Two</option>
<option value="3">Airport Three</option>
</select>
<select name="company" id="company" class="company">
//Options here based on above 2 selected values
</select>
PHP代碼是findcompany.php
<?php
if($_POST['aid']) {
$aid=$_POST['aid'];
$site=$_POST['sid']; <<<<<< How Can I pass This ID
$compsql=mysql_query("select * from tbl_company Where air_id='$aid' and site_id='$sid'");
while($rows=mysql_fetch_array($compsql)){
$cid=$rows['comp_id'];
$cdata=$rows['comp_title'];
?>
<option value="<?php echo $cid; ?>"><?php echo $cdata; ?></option>
<?php } } ?>
使用$('#site').val()
獲取第一個選擇框的值,然后將其添加到dataString
。 所以現在您的js代碼看起來像-
$("#airport").change(function()
{
var aid=$(this).val();
var sid=$('#site').val();
var dataString = 'aid='+ aid +'&sid='+sid;
...
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.