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有人能告诉我这个网站上的示例代码有什么问题吗http://www.x-developer.com/php-scripts/loading-drop-downs-with-ajax-php-and-fetching-values-from-没有刷新页面的数据库

基本上我做的和教程中的完全一样，问题是第二个下拉列表没有显示任何东西。我读到有人忘记在页面上添加一些 javascript 的评论之一。我该怎么做呢？

我曾尝试在该网站上发布一个问题，但一个星期没有人回答，所以我来到了这里。

任何帮助将非常感激。

这是我的 index.php 页面

<?php
include('cn.php');

$sql_country = "SELECT * FROM COUNTRY";
$result_country = mysql_query($sql_country);

echo "<select name='country' onChange='get_cities(this.value)'>"; //get_cities is defined below

while($row_country = mysql_fetch_array($result_country))
{
echo "<option value='".$row_country['id']."'>".$row_country['country']."</option>";
}

echo "</select>";

echo "<select name='city' id='city'></select>"; //We have given id to this dropdown

?>

这是我的 get_cities.js 页面

function get_cities(country_id)
{
$.ajax({
   type: "POST",
   url: "cities.php", /* The country id will be sent to this file */
   beforeSend: function () {
  $("#city").html("<option>Loading ...</option>");
    },
   data: "country_id="+country_id,
   success: function(msg){
     $("#city").html(msg);
   }
   });
 }

这是我的 city.php 页面

<?php

include('cn.php');

// Code for cities.php
$country_id = $_REQUEST['country_id'];

$sql_city = "SELECT * FROM CITY WHERE country_id = '".$country_id."'";
$result_city = mysql_query($sql_city);
echo "<select name='city'>";

while($row_city = mysql_fetch_array($result_city))
{
echo "<option value='".$row_city['id']."'>".$row_city['city']."</option>";
}

echo "</select>";

?>

包含的“cn.php”只是我与数据库的连接。

4

1 回答 1

3 
//Index.php
<?php
$conn = mysql_connect("localhost", "root", "root");
$db = mysql_select_db("country_example", $conn);

$sql_country = "SELECT * FROM country";
$result_country = mysql_query($sql_country);

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Country List</title>
</head>
<body>
<?php 

echo "<select name='country' onChange='get_cities(this.value)'>";

while($row_country = mysql_fetch_array($result_country))
{
    echo "<option value='".$row_country['id']."'>".$row_country['country']."</option>";
}
echo "</select>";
echo "<div id='cityLayer'><select name='city' id='city'></select></div>";
?>
    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
    <script type="text/javascript"> 
    function get_cities($country_id){
     $.ajax({
         url : "city.php?country_id="+$country_id,
         cache : false,
         beforeSend : function (){
              //Show a message
         },
         complete : function($response, $status){
             if ($status != "error" && $status != "timeout") {
                 $('#cityLayer').html($response.responseText);
             }
         },
         error : function ($responseObj){
             alert("Something went wrong while processing your request.\n\nError => "
                 + $responseObj.responseText);
         }
     }); 
    }
 </script>
</body>
</html>

//City.php
<?php

$conn = mysql_connect("localhost", "root", "root");
$db = mysql_select_db("country_example", $conn);

$country_id = $_REQUEST['country_id'];
$sql_city = "SELECT * FROM cities WHERE country_id = '".$country_id."'";
$result_city = mysql_query($sql_city);

echo "<select name='city'>";
while($row_city = mysql_fetch_array($result_city))
{
    echo "<option value='".$row_city['id']."'>".$row_city['city']."</option>";
}
echo "</select>";
?>
于 2013-03-14T17:12:50.883 回答