如何通过PHP AJAX实现城市联动查询?

2026-04-05 06:039阅读0评论SEO资讯
  • 内容介绍
  • 文章标签
  • 相关推荐

本文共计562个文字,预计阅读时间需要3分钟。

如何通过PHP AJAX实现城市联动查询?

使用PHP和AJAX实现城市联动功能,以下是一个简化的示例:

1. 文件名:select.php

2.文件大小:约2KB

3.代码内容:

php

// 检查连接if ($conn->connect_error) { die(连接失败: . $conn->connect_error);}

// 获取父级ID$pid=$_GET['pid'];

// 查询子级城市$sql=SELECT id, title FROM web_city WHERE pid=$pid;$result=$conn->query($sql);

// 返回结果if ($result->num_rows > 0) { while($row=$result->fetch_assoc()) { echo . $row[title]. ; }} else { echo 暂无数据;}

$conn->close();?>

php+ajax城市联动

1.[文件] select.php~2KB 下载(42)

<?php /* 表结构 -- -- 表的结构 `web_city` -- CREATE TABLE IF NOT EXISTS `web_city` ( `id` mediumint(8) NOT NULL AUTO_INCREMENT, `title` varchar(60) NOT NULL, `pid` mediumint(8) NOT NULL DEFAULT '0', PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=5 ; -- -- 转存表中的数据 `web_city` -- INSERT INTO `web_city` (`id`, `title`, `pid`) VALUES (1, '北京', 0), (2, '东单', 1), (3, '西单', 1); */ $my_db = mysql_connect("localhost","root",""); mysql_select_db("city", $my_db); mysql_query("set names 'utf8'"); $sql = "select * from web_city where pid = 0"; $query = mysql_query($sql); $provice = array(); while($row = mysql_fetch_assoc($query)) { $provice[] = $row; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>php+ajax 城市联动</title> <script src="jquery-1.8.0.min.js"></script> </head> <body> <script> function select_city(){ var id = jQuery("#provice option:selected").val(); $("#city").html('<option value="">选择市</option>'); $.ajax({ type: "post", url: "ajax.php", data: "id="+id, dataType: "json", success: function(msg){ var tbody = ""; $.each(msg.optionss,function(n,value){ var trs = ""; trs += "<option value='"+ value.id +"'>"+value.title+"</option>"; tbody += trs; }) $("#city").append(tbody); } }); } </script> <div> <select name="provice" id="provice" onchange="select_city()"> <option value="">选择省/市</option> <?php foreach($provice as $val){?> <option value="<?php echo $val['id'];?>"><?php echo $val['title'];?></option> <?php }?> </select> <select name="city" id="city"> <option value="">选择市</option> </select> </div> </body> </html>

2.[文件] ajax.php~406B 下载(35)

<?php $my_db = mysql_connect("localhost","root",""); mysql_select_db("city", $my_db); mysql_query("set names 'utf8'"); if(isset($_POST['id'])){ $id = $_POST['id']; $sql_city = "select * from web_city where pid = $id"; $query_city = mysql_query($sql_city); $city = array(); while($row = mysql_fetch_assoc($query_city)) { $city[] = $row; } echo json_encode(array('optionss'=>$city)); } ?>

3.[文件] jquery-1.8.0.min.js~90KB 下载(51)

如何通过PHP AJAX实现城市联动查询?

本文共计562个文字,预计阅读时间需要3分钟。

如何通过PHP AJAX实现城市联动查询?

使用PHP和AJAX实现城市联动功能,以下是一个简化的示例:

1. 文件名:select.php

2.文件大小:约2KB

3.代码内容:

php

// 检查连接if ($conn->connect_error) { die(连接失败: . $conn->connect_error);}

// 获取父级ID$pid=$_GET['pid'];

// 查询子级城市$sql=SELECT id, title FROM web_city WHERE pid=$pid;$result=$conn->query($sql);

// 返回结果if ($result->num_rows > 0) { while($row=$result->fetch_assoc()) { echo . $row[title]. ; }} else { echo 暂无数据;}

$conn->close();?>

php+ajax城市联动

1.[文件] select.php~2KB 下载(42)

<?php /* 表结构 -- -- 表的结构 `web_city` -- CREATE TABLE IF NOT EXISTS `web_city` ( `id` mediumint(8) NOT NULL AUTO_INCREMENT, `title` varchar(60) NOT NULL, `pid` mediumint(8) NOT NULL DEFAULT '0', PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=5 ; -- -- 转存表中的数据 `web_city` -- INSERT INTO `web_city` (`id`, `title`, `pid`) VALUES (1, '北京', 0), (2, '东单', 1), (3, '西单', 1); */ $my_db = mysql_connect("localhost","root",""); mysql_select_db("city", $my_db); mysql_query("set names 'utf8'"); $sql = "select * from web_city where pid = 0"; $query = mysql_query($sql); $provice = array(); while($row = mysql_fetch_assoc($query)) { $provice[] = $row; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>php+ajax 城市联动</title> <script src="jquery-1.8.0.min.js"></script> </head> <body> <script> function select_city(){ var id = jQuery("#provice option:selected").val(); $("#city").html('<option value="">选择市</option>'); $.ajax({ type: "post", url: "ajax.php", data: "id="+id, dataType: "json", success: function(msg){ var tbody = ""; $.each(msg.optionss,function(n,value){ var trs = ""; trs += "<option value='"+ value.id +"'>"+value.title+"</option>"; tbody += trs; }) $("#city").append(tbody); } }); } </script> <div> <select name="provice" id="provice" onchange="select_city()"> <option value="">选择省/市</option> <?php foreach($provice as $val){?> <option value="<?php echo $val['id'];?>"><?php echo $val['title'];?></option> <?php }?> </select> <select name="city" id="city"> <option value="">选择市</option> </select> </div> </body> </html>

2.[文件] ajax.php~406B 下载(35)

<?php $my_db = mysql_connect("localhost","root",""); mysql_select_db("city", $my_db); mysql_query("set names 'utf8'"); if(isset($_POST['id'])){ $id = $_POST['id']; $sql_city = "select * from web_city where pid = $id"; $query_city = mysql_query($sql_city); $city = array(); while($row = mysql_fetch_assoc($query_city)) { $city[] = $row; } echo json_encode(array('optionss'=>$city)); } ?>

3.[文件] jquery-1.8.0.min.js~90KB 下载(51)

如何通过PHP AJAX实现城市联动查询?