我使用以下脚本从数据库中获取记录，并使用 jquery、ajax 和 php 将它们放入选择框中。选择框也使用 Select 2 进行样式设置和添加功能 http://ivaynberg.github.io/select2/select-2.1.html#basics
如果我从第一个选择框中选择客户，然后从第二个选择框中选择车辆，则效果很好......如果我随后改变主意并选择其他公司，则车辆框将保持打开状态最后一次注册并且不会恢复为:
<option>Select A Customers Vehicle</option>
如果我单击车辆选择框，我可以从公司中选择车辆，并且上次查询中的“幽灵车辆”消失，所以它确实有效，只是当我再次更改公司时我会喜欢它再次将车辆框重置回默认值，直到我选择车辆。
这是主页:
<script src="js/jquery/jquery.js"></script>
<script src="js/jqueryui/js/jquery-ui.js"></script>
<link href="js/select2/select2.css" rel="stylesheet"/>
<script src="js/select2/select2.js"></script>
<script>
$(document).ready(function() { $("select").select2(); });
</script>
<?php
if (session_status() !== PHP_SESSION_ACTIVE) {session_start();}
if (isset($_SESSION['key'])) {$sessionkey = $_SESSION['key'];}else {$sessionkey = '';}
if ($sessionkey == 'sbhjbKA2bsbhjbKA209bhjbKA2bsbhjbKA209KaXff19u0bsbhjbKA209KaXff19u9Ka'){
include 'connectmysqli.php';
echo '<link rel="stylesheet" href="css/template/template.css" />';
echo '<strong class="pagetitle">Add New Sale</strong>
';
$saleID = rand().rand();
$today = date("Y-m-d");
echo '<form method="post" action="addsalesubmit.php">';
echo '<input type="hidden" value="'.$saleID.'" name="saleID" id="saleID">';
echo '<input type="hidden" value="'.$today.'" name="date" id="date">';
?>
<html>
<head>
<meta http-equiv="Content-type" content="text/html; charset=utf-8">
<title>Select test</title>
<script type="text/javascript" charset="utf-8">
$(document).ready(function(){
$('#customer').on('change', function (){
$.getJSON('select.php', {customerId: $(this).val()}, function(data){
var options = '';
for (var x = 0; x < data.length; x++) {
options += '<option value="' + data[x]['id'] + '">' + data[x]['reg'] + ' - ' + data[x]['make'] + ' - ' + data[x]['model'] + '</option>';
}
$('#vehicle').html(options);
});
});
});
</script>
</head>
<body>
<br>
<select id="customer">
<option>Please Select / Search For A Customer</option>
<?php
$sql = <<<SQL
SELECT *
FROM `customers`
SQL;
if(!$result = $db->query($sql)){ die('There was an error running the query [' . $db->error . ']');}
while($row = $result->fetch_assoc()){
if ($row['bussinessname'] == ''){$name = $row['title'].' '.$name = $row['firstname'].' '.$name = $row['surname'];}else
{$name = $row['bussinessname'];}
echo '<option value="'.$row['customerID'].'">'.$name.'</option>';
}
echo '</select></p>';
?>
</select>
<br>
<br>
<select id="vehicle">
<option>Select A Customers Vehicle</option>
</select>
</body>
</html>
<?php
}
else
{echo '<h1 style="font-family:Verdana, Geneva, sans-serif; color:red;">Access Denied !</h1>';}
?>
这是执行所有获取操作的 php 脚本:
<?php include 'connectmysqli.php'; ?>
<?php
$id = $_GET['customerId'];
$sql = 'SELECT * FROM vehicles WHERE customerID = ' . (int)$id;
$result = $db->query($sql);
$json = array();
while ($row = $result->fetch_assoc()) {
$json[] = array(
'id' => $row['vehicleID'],
'reg' => $row['reg'],
'make' => $row['make'],
'model' => $row['model']
);
}
echo json_encode($json);
?>
最佳答案
每次调用onchange时，首先清空第二个下拉列表
<script type="text/javascript" charset="utf-8">
$(document).ready(function(){
$('#customer').on('change', function (){
$('#vehicle').html("<option value=''>Select</option>");// add this on each call then add the options when data receives from the request
$.getJSON('select.php', {customerId: $(this).val()}, function(data){
var options = '';
for (var x = 0; x < data.length; x++) {
options += '<option value="' + data[x]['id'] + '">' + data[x]['reg'] + ' - ' + data[x]['make'] + ' - ' + data[x]['model'] + '</option>';
}
$('#vehicle').html(options);
$("select").select2();
});
});
});
</script>
关于php - 当第一个更改时重置第二个 jquery 下拉选择框，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/17401157/