PHP&的动态下拉列表MySQL数据库 [英] Dynamic dropdown list with PHP & MYSQL
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问题描述
我一直在尝试创建一个具有动态下拉列表的表单，以从MYSQL提取数据.我的数据库很好，没有错误.
I have been trying to create a form with dynamic dropdown list fetching data from MYSQL. My database is fine without errors.
第一类下拉列表工作正常，但我想知道为什么第二个下拉列表无法正常工作.我只是无法跟踪代码中的任何错误，但这正在发生.这是我的代码:
The first category of dropdown is working fine but I am wondering why my 2nd dropdown is not working. I just cant trace any error in the code and yet this is happening. here's my code:
动态下拉列表的代码:
<?php
include_once "connection.php";
?>
<!DOCTYPE html>
<html>
<head>
<title>Dropdown Ajax</title>
</head>
<body>
<div class="country">
<label>Country</label>
<select name="country" onchange="getId(this.value);">
<option value="">Select Country</option>
//populate value using php
<?php
$query = "SELECT * FROM country";
$results=mysqli_query($con, $query);
//loop
foreach ($results as $country){
?>
<option value="<?php echo $country["cid"];?>"><?php echo $country["country"];?></option>
<?php
}
?>
</select>
</div>
<div class="city">
<label>City</label>
<select name="city" id="cityList">
<option value=""></option>
</select>
</div>
<script src="https://code.jquery.com/jquery-3.1.1.js" integrity="sha256-
16cdPddA6VdVInumRGo6IbivbERE8p7CQR3HzTBuELA=" crossorigin="anonymous">
</script>
<script>
function getId(val){
//We create ajax function
$.ajax({
type: "POST",
url: "getdata.php",
data: "cid="+val,
success: function(data){
$(#cityList).html(data);
}
});
}
</script>
</body>
</html>
数据库连接代码:
<?php
$con = mysqli_connect("localhost", "root", "kensift", "tuts");
//Check connection
if(mysqli_connect_errno()){
echo "Failed to connect:".mysqli_connect_errno();
}
?>
第二个动态下拉菜单的代码:
<?php
include_once "connection.php";
if (!empty($_POST["cid"])) {
$cid = $_POST["cid"];
$query="SELECT * FROM city WHERE cid=$cid";
$results = mysqli_query($con, $query);
foreach ($results as $city){
?>
<option value="<?php echo $city["cityId"];?>"><?php echo $city["city"];?>
</option>
<?php
}
}
?>
这三个代码部分位于不同的文件中.
These three code parts are in different files.
推荐答案
我认为您的代码正确，只是忘记了引号id"#cityList".
I think your code is correct except forgot the quotations id "#cityList" .
应该是
$("#cityList").html(data);
这篇关于PHP&的动态下拉列表MySQL数据库的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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