这是我的代码，我编写了一个脚本，通过 php 中的查询从“list_cust_name”中的选定项目中获取“list_cust_city”中的值。我没有在“list_cust_city”中获得任何城市值。我制作了city.php。
<script>
$('#list_cust_name').change(function(){
alert("heyyy");
$.ajax({
url:'city.php',
data:{cust_name:$( this ).val()},
success: function( data ){
$('#list_cust_city').html( data );
}
});
});
</script>
<label style="color:#000">Name </label>
<?php
$query_name = "SELECT DISTINCT cust_name FROM customer_db ORDER BY cust_name"; //Write a query
$data_name = mysql_query($query_name); //Execute the query
?>
<select id="list_cust_name" name="list_cust_name">
<?php
while($fetch_options_name = mysql_fetch_assoc($data_name)) { //Loop all the options retrieved from the query
$customer=$fetch_options_name['cust_name'];
?>
<option value="<?php echo $fetch_options_name['cust_name']; ?>"><?php echo $fetch_options_name['cust_name']; ?></option>
<?php
}
?>
</select>
city.php
<body>
<?php
include('dbconnect.php');
db_connect();
$cust_name1=$_GET['cust_name']; //passed value of cust_name
$query_city = "SELECT DISTINCT cust_city FROM customer_db WHERE cust_name='$cust_name1'ORDER BY cust_city"; //Write a query
$data_city = mysql_query($query_city); //Execute the query
while($fetch_options_city = mysql_fetch_assoc($data_city)) { //Loop all the options retrieved from the query
?>
<option value="<?php echo $fetch_options_city['cust_city']; ?>"><?php echo $fetch_options_city['cust_city']; ?></option>
<?php
}
?>
</body>
最佳答案
您必须使用文档就绪，因为 DOM 未加载。
$( document ).ready(function() {
$('#list_cust_name').change(function(){
alert("heyyy");
$.ajax({
url:'city.php',
data:{cust_name:$( this ).val()},
success: function( data ){
$('#list_cust_city').html( data );
}
});
});
});
关于php - 在 php 中查询下拉列表中的选定项目，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21722502/