ajax php下拉列表 [英] ajax php drop down list
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问题描述
基本上我做的完全一样，在turorial和问题是第二个下拉列表没有显示任何内容。我读了一个有人忘了在页面上添加一些javascript的评论之一。我如何做到这一点？
我已经尝试在该网站上发布问题，但没有人回答一个星期，所以我来到这里。
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定义在
之下（$ row_country = mysql_fetch_array（$ result_country））
{
echo< option value ='$ row_country ['id' ] '> 中$ row_country [。 '国家'。 < /选项> 中;
}
echo< / select>;
echo< select name ='city'id ='city'>< / select>; //我们给了这个下拉列表的id
？>
这是我的get_cities.js页面
函数get_cities（country_id）
{
$ .ajax（{
type：POST，
url：cities.php ，/ *国家ID将被发送到这个文件* /
beforeSend：function（）{
$（＃city）。html（< option> Loading ...< /选项>）;
}，
data：country_id =+ country_id，
success：function（msg）{
$（＃city）。 ）;
}
}）;
}
这是我的cities.php页面
<？php
include（'cn.php'）;
//代码为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 [ '城市']。 ; /选项> 中;
}
echo< / select>;
？>
包含的'cn.php'只是我与数据库的连接。
解决方案
// 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 // ENhttp://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd\">
< html xmlns =http://www.w3.org/1999/xhtml>
< head>
< title>国家/地区列表< / 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 [ '国家']。 ; /选项> 中;
}
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>
函数get_cities（$ country_id）{
$ .ajax（{
url：city.php？country_id =+ $ country_id，
cache：false，
beforeSend：function（）{
//显示消息
}，
完成：function（$ response，$ status）{
if（$ status！=error& ;& $ status！=timeout）{
$（'＃cityLayer'）。html（$ response.responseText）;
}
}，
错误：function （$ responseObj）{
alert（处理您的请求时发生错误。\\\
\\\
Error =>
+ $ 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 [ '城市'] < /选项> 中。
}
echo< / select>;
？>
Can some one tell me what's wrong with this example code on this site http://www.x-developer.com/php-scripts/loading-drop-downs-with-ajax-php-and-fetching-values-from-database-without-refreshing-the-page
Basically i did exactly the same as in the turorial and the problem is that the 2nd drop down list is no showing anything. I read one of the comments that someone forgot to add in some javascript on the page. How do i do this?
I have tried posting a question on that site but no one answered for a week now so I came here.
Any help would be much appreciated.
this is my index.php page
<?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
?>
this is my get_cities.js page
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);
}
});
}
This is my cities.php page
<?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>";
?>
The included 'cn.php' is just my connection to the database.
解决方案
//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>";
?>
这篇关于ajax php下拉列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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