Question

我已附加代码以从DB获取动态下拉列表。当用户使用onChange（）从第一个下拉列表中选择时，将根据第一个下拉列表的选择检索第二个下拉列表的数据。但是，我可以获得第一个下拉列表但没有第二个下拉列表的结果。似乎ajax没有调用。

另一个问题：

我可以为第二个下拉列表制作单选按钮

谢谢

<?php

echo"1"
if(isset($_POST['get_option']))
{
 $host = 'localhost';
 $user = 'root';
 $pass = 'root';
 mysql_connect($host, $user, $pass);
 mysql_select_db('db_photography');
 $state = $_POST['get_option'];
 $find=mysql_query("select city from places where state='$state'");
 while($row=mysql_fetch_array($find))
 {
	 echo"1";
  echo "<option>".$row['city']."</option>";
 }
 exit;
}
?>

<html>
<head>
<link rel="stylesheet" type="text/css" href="select_style.css">
<script type="text/javascript" src="jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script type="text/javascript">
function fetch_select(val)
{
 $.ajax({
 type: 'post',
 url: 'fetch_data.php',
 data: {
  get_option:val
 },
 success: function (response) {
  document.getElementById("new_select").innerHTML=response; 
 }
 });
}

</script>

</head>
<body>
<p id="heading">Dynamic Select Option Menu Using Ajax and PHP</p>
<center>
<div id="select_box">
 <select onchange="fetch_select(this.value);">
  <option>Select state</option>
  <?php
  $host = 'localhost';
  $user = 'root';
  $pass = 'root';
  mysql_connect($host, $user, $pass);
  mysql_select_db('db_photography');

  $select=mysql_query("select state from places group by state");
  while($row=mysql_fetch_array($select))
  {
   echo "<option>".$row['state']."</option>";
  }
 ?>
 </select>

 <select id="new_select">
 </select>
	  
</div>     
</center>
</body>
</html>

动态下拉列表没有结果

0 个答案: