有人可以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-database-without-refreshing-the-page
基本上我所做的与图例中的完全相同，问题是第二个下拉列表没有显示任何内容。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？
我尝试在该网站上发布问题，但已经一周没有人回答了，所以我来到了这里。
任何帮助将不胜感激。
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。
最佳答案
//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>";
?>
关于php - ajax php 下拉列表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/15415571/