【发布时间】:2019-07-10 06:24:41
【问题描述】:
从下拉列表中选择数据后,我尝试使用数据库中的数据更改输入类型文本中的值
但数据没有输入文本输入值
<?php
//Include the database configuration file
include 'dbConfig.php';
//Fetch all the country data
$query = $db->query("SELECT * FROM countries WHERE status = 1 ORDER BY country_name ASC");
//Count total number of rows
$rowCount = $query->num_rows;
?>
<select id="country">
<option value="">Select Country</option>
<?php
if($rowCount > 0){
while($row = $query->fetch_assoc()){
echo '<option value="'.$row['country_id'].'">'.$row['country_name'].'</option>';
}
}else{
echo '<option value="">Country not available</option>';
}
?>
</select>
<div id="state">
</div>
<?php
//Include the database configuration file
include 'dbConfig.php';
if(!empty($_POST["country_id"])){
//Fetch all state data
$query = $db->query("SELECT * FROM states WHERE country_id = ".$_POST['country_id']." AND status = 1 ORDER BY state_name LIMIT 1");
//Count total number of rows
$rowCount = $query->num_rows;
//State option list
if($rowCount > 0){
$row = $query->row();
echo '<input type="text" id="state" value="'.$row['state_name'].'">';
}
}
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/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:'index.php',
data:'country_id='+countryID,
success:function(html){
$('#state').html(html);
}
});
}
});
});
</script>
我希望根据数据库中的数据输出
这是我的数据库:
数据库国家
数据库状态
错误
【问题讨论】:
-
您遇到的任何错误。
标签: javascript php html ajax