<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Document</title>
<script type="text/javascript" src="js/jquery-1.12.4.min.js"></script>
<style type="text/css">
body>div{height: 40px;}
label{
display: inline-block;
width: 50px; height: 22px; background:#dc3545; border-radius: 30px ; transition: background-color .6s linear;
}
/*在标签的前面插入一小圆形*/
label::after{
content: "关";width: 17px; height: 17px; background-color:#FFFFFF;color:#dc3545;display: inline-block;text-align: center;line-height: 17px;position: relative;left:3px;
transition: transform.6s linear; border-radius: 50%;
}
/*获得复选框后面的第一个兄弟元素label*/
input[type=checkbox]:checked+label{
background-color:#28a745;
}
input[type=checkbox]:checked+label::after{
content: "开";color:#28a745;transform: translate(27px);/*向右移动27px*/
}
</style>
</head>
<body>
<form action="1.php" method="post" >
<input name="yfkg" type="submit" value="yfkg" onclick="submit"($yfkg=000)>
<input name="tfkg" type="submit" value="tfkg" onclick="submit"($tfkg=000)>
</form>
</html>
```php
<?php
header("Content-type: text/html; charset=utf-8");
//建立数据库连接*/
$servername = 'localhost';
$username ='root';
$password = '123';
$dbname = "nongyedapeng";
$conn = new mysqli($servername, $username, $password, $dbname);
function execute_sql($link, $database, $sql){
mysqli_select_db($link, $database)
or die("打开数据库失败: " . mysqli_error($link));
$result = mysqli_query($link, $sql);
return $result;
}
switch ($_POST["submit"]){
case "yfkg":
if ($conn){
$sql = "UPDATE control SET yfkg=111";
$conn1= execute_sql($conn, "nongyedapeng", $sql);
echo "
设计阈值成功
";
}
else{
echo "请重新输入
";
}
break;
case "tfkg":
if ($conn){
$sql = "UPDATE control SET yfkg=000";
$conn1= execute_sql($conn, "nongyedapeng", $sql);
echo "设计阈值成功
";
}
else{
echo "请重新输入
";
}
break;
default:
}
?> 
![[Web front-end] Node.js quick start](https://img.php.cn/upload/course/000/000/067/662b5d34ba7c0227.png)
Your PHP code cannot be executed at all. There are errors in your link data. Do you have classes? Why do you need new mysqli? Besides, the function you use to connect to the database is wrong. It is mysqli_connect()
You probably want to fill in the data in the form and then submit the data in the form to the database through a button, right?
According to your code, you need to create a new 1.php in the html directory and put your php code into it