如何根據上一個下拉選擇填充下拉列表SQL

Question

我一直在關注該解決方案1創建一個下拉列表，更改基於用戶先前選擇的另一個下拉列表中較高的頁面。不過，我不確定什麼，我需要做的選項會一直在這裏開始填充

感謝

<?php 
mysql_connect('localhost'); 
mysql_select_db("test"); 
$result = mysql_query("SELECT * FROM `contents` WHERE `parent` = 0"); 
echo "<select name='name'>"; 
while(($data = mysql_fetch_array($result)) !== false) 
    echo '<option value="', $data['id'],'">', $data['name'],'</option>' 
?> 

     <select onchange="ajaxfunction(this.value)"> 
     <!-- Options would have been initially populated here --> 
      </select> 
     <select id="sub">  
      </select> 

    <script type="text/javascript"> function ajaxfunction(parent) 
{ 
$.ajax({ 
    url: 'process.php?parent=' + parent; 
    success: function(data) { 
     $('#sub option').remove(); //// here sub is the id of second select box 
     $('#sub').append(data) 
    } 
}); 
} 
</script> 

Answer 1

**<select onchange="ajaxfunction(this.value)"> 
**<!-- Options would have been initially populated here -->** 
</select>** 


<script type="text/javascript"> 
function ajaxfunction(parent) 
{ 
    $.ajax({ 
     url: 'process.php?parent=' + parent; 
     success: function(data) { 
      $('#sub option').remove(); //// here sub is the id of second select box 
      $('#sub').append(data) 
     } 
    }); 
} 
</script> 
/// script section may be any where on your page 


<select id="sub"> ////second select box 
</select>