[英]php, mysql dynamicaly dropdown using jquery
我正在尝试为我的网站创建一个动态下拉菜单,以添加地址,例如州,城市,区域和邮政编码。 当我在实时浏览器中尝试一个预览时,我正在使用代码,但是当我尝试使用Java脚本时,它却无法工作,它在下拉菜单中不显示数据。您能告诉我我在哪里犯错吗?
它是index.php
<head>
</head>
<script src="scripts/jquery.js" type="text/javascript"></script>
<script src="scripts/scripts.js" type="text/javascript"></script>
<body>
<h1>address</h1>
<hr/>
<label>please select state</label>
<select id="slctstate"></select>
<br />
<br />
<label>please select city</label>
<select id="slctcity"></select>
</body>
</head>
它是script.js
$(document).ready(function() {
$.getJSON("get_stat.php", success = function(data){
var options = "";
for(var i = 0; i < data.lenght; i++)
{
options += "<option value='" + data[i].toLowerCase() + "'>" + data[i] + "</option>";
}
$("#sltstate").append(options);
});
$("#slctstate").change(function(){
$.getJSON("get_city.php?state=" +$(this).value(), success = function(data){
var options = "";
for(var i = 0; i < data.lenght; i++)
{
options += "<option value='" + data[i].toLowerCase() + "'>" + data[i] + "</option>";
}
$("#slctcity").append(options);
});
});
});
它是get_state.php
<?php
require "Connections/dbopen.php";
$query = "SELECT state_name FROM states";
$data = mysqli_query($conn, $query);
$states = array();
while ($row = mysqli_fetch_array($data))
{
array_push($states, $row["state_name"]);
}
echo json_encode($states);
require "Connections/dbclose.php";
?>
她的是get_city.php
<?php
require "Connections/dbopen.php";
if(isset($_GET["$state"]));
{
$state = $_GET["state"];
$query = "SELECT city_name FROM city
INNER JOIN states ON
city.state_id=states.state_id
WHERE state_name LIKE '{$state}'";
$data = mysqli_query($conn, $query);
$city = array();
while ($row = mysqli_fetch_array($data))
{
array_push($city, $row["city_name"]);
}
echo json_encode($city);
require "Connections/dbclose.php";
}
?>
但在最后一步,我的下拉菜单没有任何价值,任何人都可以帮助我谢谢
您可能需要进行以下更改:
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