Javascript 单板、中型和中型的动态相关下拉列表;标准
我正在尝试创建动态相关下拉列表。在这种情况下,用户在加载该板介质和加载选择介质标准后首先选择板。它类似于国家、州和城市的下拉列表 这里是附加代码。下拉菜单不能正常工作。当我选择线路板时,介质下拉列表未加载,显示为空白。请任何人发现错误并给我解决方法。谢谢你 index.phpJavascript 单板、中型和中型的动态相关下拉列表;标准,javascript,php,ajax,dropdown,Javascript,Php,Ajax,Dropdown,我正在尝试创建动态相关下拉列表。在这种情况下,用户在加载该板介质和加载选择介质标准后首先选择板。它类似于国家、州和城市的下拉列表 这里是附加代码。下拉菜单不能正常工作。当我选择线路板时,介质下拉列表未加载,显示为空白。请任何人发现错误并给我解决方法。谢谢你 index.php <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script> <s
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('#board').on('change',function(){
var boardID = $(this).val();
if(boardID){
$.ajax({
type:'POST',
url:'ajaxData.php',
data:'board_id='+boardID,
success:function(html){
$('#medium').html(html);
$('#standard').html('<option value="">Select medium first</option>');
}
});
}else{
$('#medium').html('<option value="">Select board first</option>');
$('#standard').html('<option value="">Select medium first</option>');
}
});
$('#medium').on('change',function(){
var mediumID = $(this).val();
if(mediumID){
$.ajax({
type:'POST',
url:'ajaxData.php',
data:'medium_id='+mediumID,
success:function(html){
$('#standard').html(html);
}
});
}else{
$('#standard').html('<option value="">Select medium first</option>');
}
});
});
</script>
<?php
//Include the database configuration file
include 'db_connect.php';
//Fetch all the country data
$query = $db->query("SELECT * FROM ag_board WHERE board_status = 1 ORDER BY board_name ASC");
//Count total number of rows
$rowCount = $query->num_rows;
?>
<select id="board">
<option value="">Select Board</option>
<?php
if($rowCount > 0){
while($row = $query->fetch_assoc()){
echo '<option value="'.$row['board_unique_id'].'">'.$row['board_name'].'</option>';
}
}else{
echo '<option value="">board not available</option>';
}
?>
</select>
<select id="medium">
<option value="">Select board first</option>
</select>
<select id="standard">
<option value="">Select medium first</option>
</select>
$(文档).ready(函数(){
$('#board')。on('change',function(){
var boardID=$(this.val();
if(boardID){
$.ajax({
类型:'POST',
url:'ajaxData.php',
数据:'board_id='+boardID,
成功:函数(html){
$('#medium').html(html);
$('#standard').html('Select medium first');
}
});
}否则{
$('#medium').html('Select board first');
$('#standard').html('Select medium first');
}
});
$('#medium')。on('change',function(){
var mediumID=$(this.val();
if(mediumID){
$.ajax({
类型:'POST',
url:'ajaxData.php',
数据:'medium_id='+mediumID,
成功:函数(html){
$('标准').html(html);
}
});
}否则{
$('#standard').html('Select medium first');
}
});
});
选择板
先选板
先选择中等
ajaxData.php
<?php
//Include the database configuration file
include 'db_connect.php';
if(!empty($_POST["board"])){
//Fetch all state data
$query = $db->query("SELECT * FROM ag_medium WHERE board_unique_id = ".$_POST['board_unique_id']." AND medium_status = 1 ORDER BY medium_name ASC");
//Count total number of rows
$rowCount = $query->num_rows;
//State option list
if($rowCount > 0){
echo '<option value="">Select medium</option>';
while($row = $query->fetch_assoc()){
echo '<option value="'.$row['medium_unique_id'].'">'.$row['medium_name'].'</option>';
}
}else{
echo '<option value="">Medium not available</option>';
}
}elseif(!empty($_POST["medium_unique_id"])){
//Fetch all city data
$query = $db->query("SELECT * FROM ag_standard WHERE medium_unique_id = ".$_POST['medium_unique_id']." AND status = 1 ORDER BY std_name ASC");
//Count total number of rows
$rowCount = $query->num_rows;
//City option list
if($rowCount > 0){
echo '<option value="">Select Standard</option>';
while($row = $query->fetch_assoc()){
echo '<option value="'.$row['std_unique_id'].'">'.$row['std_name'].'</option>';
}
}else{
echo '<option value="">Std not available</option>';
}
}
?>
你好,我正在看你的帖子。你能解释一下你所说的“下拉菜单不能正常工作”是什么意思吗?具体一点更有帮助。@rainer我的意思是,当我在第一个下拉列表中选择board时,第二个下拉列表中的媒体没有加载。信息很好。理想情况下,您可以编辑您的问题并将澄清放在那里。这只是为了帮助你开始,所以。。。
<?php
//Database credentials
$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = '';
$dbName = 'db_aneri';
//Connect and select the database
$db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName);
if ($db->connect_error) {
die("Connection failed: " . $db->connect_error);
}
?>