AJAX，PHP和jQuery下拉

Question

我有這個文件名爲ajax.php '

<?php 

    class AJAX { 

     private $database = NULL; 
     private $_query = NULL; 
     private $_fields = array(); 
     public $_index = NULL; 
     const DB_HOST = "localhost"; 
     const DB_USER = "root"; 
     const DB_PASSWORD = ""; 
     const DB_NAME = "dd"; 


     public function __construct(){ 
      $this->db_connect();     // Initiate Database connection 
      $this->process_data(); 
     } 

     /* 
     * Connect to database 
     */ 
     private function db_connect(){ 
      $this->database = mysql_connect(self::DB_HOST,self::DB_USER,self::DB_PASSWORD); 
      if($this->database){ 
       $db = mysql_select_db(self::DB_NAME,$this->database); 
      } else { 
       echo mysql_error();die; 
      } 
     } 

     private function process_data(){ 
      $this->_index = ($_REQUEST['index'])?$_REQUEST['index']:NULL; 
      $id = ($_REQUEST['id'])?$_REQUEST['id']:NULL; 
      switch($this->_index){ 
       case 'country': 
        $this->_query = "SELECT * FROM country GROUP BY Name"; 
        $this->_fields = array('Code','Name'); 
        break; 
       case 'state': 
        $this->_query = "SELECT * FROM city WHERE CountryCode=$id"; 
        $this->_fields = array('CountryCode','District'); 
        break; 
       case 'city': 
        $this->_query = "SELECT * FROM city WHERE CountryCode=$id"; 
        $this->_fields = array('CountryCode','Name'); 
        break; 
       default: 
        break; 
      } 
      $this->show_result(); 
     } 

     public function show_result(){ 
      echo '<option value="">Select '.$this->_index.'</option>'; 
      $query = @mysql_query($this->_query); 
      while($result = @mysql_fetch_array($query)){ 
       $entity_id = $result[$this->_fields[0]]; 
       $enity_name = $result[$this->_fields[1]]; 
       echo "<option value='$entity_id'>$enity_name</option>"; 
      } 
     } 
    } 

    $obj = new AJAX; 

    ?> 

和index.html這個JS腳本

和形式

<form> 
    <label>Select Country</label> 
    <select id="country" onchange="load_options(this.value,'state');"> 
     <option value="">Select country</option> 
    </select> 
    &nbsp;&nbsp;&nbsp; 
    <label>Select State</label> 
    <select id="state" onchange="load_options(this.value,'city');"> 
     <option value="">Select state</option> 
    </select> 
    &nbsp;&nbsp;&nbsp; 
    <label>Select city</label> 
    <select id="city"> 
     <option value="">Select City</option> 
    </select> 
    <img src="loader.gif" id="loading" align="absmiddle" style="display:none;"/> 
</form> 

國家選項可見但是當我想選擇它不是的狀態時

另外我有兩個表cou ntry和城市，並在表城市包括區名

Answer 1

你應該使用jQuery更容易實現ajax回調。如果頻繁更改，可以將數據緩存在客戶端計算機上。

你可以參考這個話題 jQuery set selected value in option box once the box has been loaded