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">
demo66.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);
}