0
我有三個下拉列表，分別是國家，州和城市。首先，國家下拉將與所有國家一起展示。當選擇一個國家時，相應的狀態將從MySQL數據庫中獲取並顯示在州下拉菜單中。與選擇州時相似，各個城市將從MySQL數據庫中提取並顯示在城市下拉菜單中。如何在提交後保留所選的onchange下拉列表值
下面是我選擇國家，州，市和點擊提交按鈕之前的默認顯示。
後，我選擇的國家，州，市，點擊提交按鈕，如下圖所示。它將刷新並返回到默認顯示。
所以，我怎麼能保持選定值（英國，英格蘭，倫敦）在下拉列表中，而不是它跳回後點擊提交按鈕默認顯示的顯示？
的index.php
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<style type="text/css">
.select-boxes{width: 280px;text-align: center;}
</style>
<script src="jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('#country').on('change',function(){
var countryID = $(this).val();
if(countryID){
$.ajax({
type:'POST',
url:'ajaxData.php',
data:'country_id='+countryID,
success:function(html){
$('#state').html(html);
$('#city').html('<option value="">Select state first</option>');
}
});
}else{
$('#state').html('<option value="">Select country first</option>');
$('#city').html('<option value="">Select state first</option>');
}
});
$('#state').on('change',function(){
var stateID = $(this).val();
if(stateID){
$.ajax({
type:'POST',
url:'ajaxData.php',
data:'state_id='+stateID,
success:function(html){
$('#city').html(html);
}
});
}else{
$('#city').html('<option value="">Select state first</option>');
}
});
});
</script>
</head>
<body>
<form id="form1" name="form1" method="get" action="<?php echo $_SERVER['PHP_SELF'];?>">
<?php
//Include database configuration file
include('dbConfig.php');
//Get all country data
$query = $db->query("SELECT * FROM countries WHERE status = 1 ORDER BY country_name ASC");
//Count total number of rows
$rowCount = $query->num_rows;
?>
<select name="country" id="country">
<option value="">Select Country</option>
<?php
if($rowCount > 0){
while($row = $query->fetch_assoc()){
echo '<option value="'.$row['country_id'].'">'.$row['country_name'].'</option>';
}
}else{
echo '<option value="">Country not available</option>';
}
?>
</select>
<select name="state" id="state">
<option value="">Select country first</option>
</select>
<select name="city" id="city">
<option value="">Select state first</option>
</select>
<input type="submit" name="Submit" id="Submit" value="Submit" />
</form>
</body>
</html>
ajaxData.php
<?php
//Include database configuration file
include('dbConfig.php');
if(isset($_POST["country_id"]) && !empty($_POST["country_id"])){
//Get all state data
$query = $db->query("SELECT * FROM states WHERE country_id IN (".$_POST['country_id'].")");
//Count total number of rows
$rowCount = $query->num_rows;
//Display states list
if($rowCount > 0){
echo '<option value="">Select state</option>';
while($row = $query->fetch_assoc()){
echo '<option value="'.$row['state_id'].'">'.$row['state_name'].'</option>';
}
}else{
echo '<option value="">State not available</option>';
}
}
if(isset($_POST["state_id"]) && !empty($_POST["state_id"])){
//Get all city data
$query = $db->query("SELECT * FROM cities WHERE state_id IN(".$_POST["state_id"].")");
//Count total number of rows
$rowCount = $query->num_rows;
//Display cities list
if($rowCount > 0){
echo '<option value="">Select city</option>';
while($row = $query->fetch_assoc()){
echo '<option value="'.$row['city_id'].'">'.$row['city_name'].'</option>';
}
}else{
echo '<option value="">City not available</option>';
}
}
?>
dbConfig.php
<?php
//db details
$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = '';
$dbName = 'location_db';
//Connect and select the database
$db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName);
if ($db->connect_error) {
die("Connection failed: " . $db->connect_error);
}
?>
使用在變化事件Ajax調用，而不是一個完整的回發 – dave
的是什麼意思。使用在變化事件，而不是一個完整的回發一個Ajax調用的？對不起，我是php的新手 – eric