ajax 返回json数据，(PHP获取mysql输出json)如何实现下拉框六级联动？

Question

ajax传递name值给php后台查询数据并输出demo66.php 的json数据。
如何在前台实现下拉框六级联动？

 
<input id="name" name="name" class="layui-input"> 
 
<select name="b" id="b" lay-verify="required"  lay-filter="b" class="layui-input">
                                <option value="">请选择组别</option>
                                </select>
<select name="c" id="c" lay-verify="required"  lay-filter="c" class="layui-input">
                                <option value="">请选择区域</option>
                                </select>
<select name="d" id="c" lay-verify="required"  lay-filter="d" class="layui-input">
                                <option value="">请选择负责人</option>
                                </select>
<select name="f" id="f" lay-verify="required"  lay-filter="f" class="layui-input">
                                <option value="">请选择职务</option>
                                </select>
<input id="g" name="g" class="layui-input"> 
<input id="h" name="h" class="layui-input"> 
 
 
<script type="text/javascript">
 $('#name').bind('input propertychange', function() {
        var name= $(this).val(); //传递参数
         $.ajax({
            url: 'demo66.php',
            type: 'get',
            data: 'name=' + name,            
            success:  function(data){
                //b,c,d,f如何实现下拉框联动？？？？？？？
                }
            },
            dataType: 'json'
        });
    });
</script>
 
 

demo66.php

 
<?php
//如果无返回空 
  if(empty($_GET['name'])) {
    echo 0;
    exit();
} else {
    $name= $_GET['name'];
} ;  
 try {
  $pdo = new PDO('mysql:host=127.0.0.1;dbname=demo;port=3306', 'root', 'root');
} catch (PDOException $e) {
  die('connet error :' . $e->getMessage());
};
$pdo->exec('set names utf8');
$res = $pdo->query("select id,name,team,area,mag,leader,g,h from tb where name='$name' )")->fetch(PDO::FETCH_ASSOC);
 //单一的用fetch而不是fetchAll
//如果无则返回空
if(empty($res)) {
     echo 0;
} else {
    echo json_encode($res);
}
 

Answer 1

这个适合你
https://www.jb51.net/article/166334.htm